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author: pgww <pgww@lists.pglaf.org> 2026-05-22 04:26:47 -0700
committer: pgww <pgww@lists.pglaf.org> 2026-05-22 04:26:47 -0700
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+      Curiosa Mathematica: Part I | Project Gutenberg
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+
+<body>
+<div style='text-align:center'>*** START OF THE PROJECT GUTENBERG EBOOK 78586 ***</div>
+
+
+<figure class="figcenter width500" id="cover" style="width: 1640px;">
+<img src="images/cover.jpg" width="1640" height="2560" alt="Dodgson's
+1888 witty defense of Euclidean geometry, replacing the Parallel
+Postulate with a simpler axiom while resisting the growing influence of
+non-Euclidean alternatives.">
+</figure>
+
+
+<hr class="chap x-ebookmaker-drop">
+
+<div class="chapter">
+<p class="right space-above20">[TURN OVER.</p>
+</div>
+
+
+<hr class="chap x-ebookmaker-drop">
+
+<div class="chapter">
+<h2 class="nobreak" id="A_NEW_THEORY">A NEW THEORY<br>
+<br>
+<span class="allsmcap">OF</span><br>
+<br>
+PARALLELS</h2>
+</div>
+
+<hr class="chap x-ebookmaker-drop">
+
+<div class="chapter">
+<figure class="figcenter width500" id="i_f004" style="width: 150px;">
+<img src="images/i_f004.jpg" width="150" height="157" alt="decorative">
+</figure>
+</div>
+
+
+<hr class="chap x-ebookmaker-drop">
+
+<div class="chapter">
+<figure class="figcenter width500" id="i_f006" style="width: 2285px;">
+<img src="images/i_f006.jpg" width="2285" height="2876" alt="Charles Dodgson's postulate.">
+<figcaption class="caption">
+<p><span class="antiqua">In every Circle, the inscribed equilateral Tetragon<br>
+is greater than any one of the Segments which lie outside it.</span></p>
+</figcaption>
+</figure>
+</div>
+
+<hr class="chap x-ebookmaker-drop">
+
+<div class="chapter">
+<figure class="figcenter width500" id="i_f007" style="width: 2350px;">
+<img src="images/i_f007.jpg" width="2350" height="3781" alt="Title page
+of the book Curiosa Mathematica Part I written by Charles Dodgson.">
+</figure>
+</div>
+
+<hr class="chap x-ebookmaker-drop">
+
+<div class="chapter">
+<h1><i><span class="u"><span class="antiqua">Curiosa Mathematica</span></span></i></h1>
+
+
+<p class="nindc space-below2 space-above2"><i>PART I</i></p>
+
+
+<p class="nindc space-below2 space-above2"><span class="large">A NEW THEORY</span><br>
+<span class="allsmcap">OF</span><br>
+<span class="large">PARALLELS</span></p>
+
+
+<p class="nindc">BY<br>
+<span class="large">CHARLES L. DODGSON, M.A.</span></p>
+
+
+<p class="nindc space-below2 space-above2"><i>Student and late Mathematical Lecturer
+of Christ Church Oxford</i></p>
+
+
+<p class="nindc space-below2">FOURTH EDITION</p>
+
+
+<p class="nindc space-below2 space-above2"><i>PRICE TWO SHILLINGS</i></p>
+
+
+<p class="nindc space-below2 space-above2"><span class="antiqua">London</span><br>
+MACMILLAN AND CO.<br>
+1895</p>
+
+
+<p class="nindc space-below2 space-above2">[<i>All rights reserved</i>]
+</p>
+</div>
+
+
+<hr class="chap x-ebookmaker-drop">
+
+<div class="chapter">
+<p class="nindc space-below2 space-above2"><span class="antiqua">Oxford</span><br>
+HORACE HART, PRINTER TO THE UNIVERSITY</p>
+</div>
+
+<hr class="chap x-ebookmaker-drop">
+
+<div class="chapter">
+<p><span class="pagenum" id="Page_ix">[Pg ix]</span></p>
+
+<h2 class="nobreak" id="PREFACE_TO_THIRD_EDITION">PREFACE TO THIRD EDITION.</h2>
+</div>
+
+<hr class="r5">
+
+<p>The chief novelty, in the First Edition of this treatise, was the
+Axiom, by means of which I proved Euc. I. 32 without making use of his
+12th Axiom. And the chief novelty, in this Third Edition, is the change
+I have made in that Axiom, by substituting 'Tetragon' for 'Hexagon'.
+The new Figure is more simple, and more easily constructed, than its
+predecessor: while the Axiom is, I hope, as obviously true as ever.</p>
+
+<p>The proof of my "New Theory of Parallels" is, I think, greatly
+simplified and improved in this new Edition—the Propositions, which do
+<i>not</i> require any disputable Axiom, being placed by themselves in
+'Book I,' while those, which require the new Axiom for their proof, are
+placed in 'Book II.' At the end of Book II will be found a proof (so
+far as <i>finite</i> magnitudes are concerned) for Euclid's celebrated
+12th Axiom, preceded by, and dependent on, the Axiom tacitly assumed
+by him in his Book X, Prop. 1, and also assumed, I believe, by every
+subsequent writer who has attempted to <i>prove</i> his 12th Axiom. My
+proof is borrowed, with some slight alterations, from Cuthbertson's
+'Euclidean Geometry.'</p>
+
+<p>One advantage, in thus separating the Propositions into two classes,
+is that it calls attention to the very remarkable<span class="pagenum" id="Page_x">[Pg x]</span> and interesting
+fact that the Theorem "There is a Triangle whose angles are together
+not-greater than two right angles" is actually provable without any
+disputable Axiom whatever. If only it could be proved, with equal ease,
+that "there is a Triangle whose angles are together not-<i>less</i>
+than two right angles"! But alas, <i>that</i> is an <i>ignis fatuus</i>
+that has never yet been caught! The man, who first proves <i>that</i>
+Theorem, without using Euclid's 12th Axiom or any substitute for it,
+will certainly deserve a place among the world's great discoverers.</p>
+
+<p>I take this opportunity of replying to one or two criticisms, which
+have been published, on the Second Edition—earnestly assuring the
+writers of those criticisms that, in treating the questions at issue
+between us from a not-wholly-solemn point of view, I have been actuated
+by no feeling of disrespect towards them, but simply from the wish to
+lighten a subject, naturally somewhat too heavy and sombre, and thus to
+make it a little more palatable to the general Reader.</p>
+
+<p class="space-above2">
+At p. 12 of the 2nd Edition, the enunciation of Prop. VI (which
+re-appears, in a modified form, at p. 34 of the 3rd Edition) stood
+thus:—</p>
+
+<p>"<i>If the vertical angle of a Sector of a Circle be divided by radii
+into \(2^{n}\) equal angles, thus forming</i> \(2^{n}\) <i>equal Sectors;
+and if the chord of each such Sector be not less than the radius of the
+Circle: the original Sector is not less than \(2^{n}\) times the Triangle
+cut off from it by its chord.</i>" My controversy with <i>Nature</i>,
+on this enunciation, will be best given in the form of a dialogue. (Of
+course I quote <i>verbatim</i>.)</p>
+
+<p><i>Nature.</i> (Dec. 6, 1888.) "How are the figures to be constructed,
+if <i>n</i> be greater than 2?"</p>
+
+<p><span class="pagenum" id="Page_xi">[Pg xi]</span></p>
+
+<p><i>Author.</i> (In the Preface to the 2nd Edition, at p. x.) "Well,
+suppose <i>n</i> were equal to 4: i.e. we have to divide the vertical
+angle into 2<sup>4</sup> equal parts. Bisect it: that gives halves. Bisect the
+halves: that gives quarters. Bisect again: that gives eighths. Bisect
+once more: that gives sixteenths. <i>Voila tout!</i>"</p>
+
+<p><i>Nature.</i> (June 13, 1889.) "Shade of Euclid! Who knows not such
+things? We admitted the same, but stated that our difficulty in the
+construction was the condition imposed in the enunciation: viz., 'the
+chord of each such sector not less than the radius of the circle.' Take
+Mr. Dodgson's illustration of a sixteenth: this would necessitate that
+the original angle should be at least 960° ... we ... have further
+noted that no one of the chords in Mr. Dodgson's figures is even equal
+to the radius."</p>
+
+<p><i>Author.</i> "What you call 'the condition imposed' is introduced
+with an 'if': it is merely an <i>hypothesis</i>: all I undertake to
+prove is that, <i>if</i> certain things <i>were</i> true, certain other
+things <i>would be</i> true. Surely I need not remind you that the
+<i>validity</i> of a Syllogism is quite independent of the <i>truth</i>
+of its Premisses! 'I have sent for you, my dear Ducks', said the worthy
+Mrs. Bond, 'to enquire with what sauce you would like to be eaten?'
+But we don't want to be <i>killed</i>!' cried the Ducks. '<i>You are
+wandering from the point</i>' was Mrs. Bond's perfectly logical reply.
+So here. 'I beg you to observe, my dear <i>Nature</i>, that, <i>if</i>
+the chord of each Sector were not less than the radius, the logical
+result would be so-and-so.' 'But the chord <i>is</i> less than the
+radius!' you cry. All I need say, in reply, is '<i>You are wandering
+from the point.</i>'"</p>
+
+<p>"But I will be generous, and will say more. I take exception to
+<i>two</i> assertions of yours. Remember our<span class="pagenum" id="Page_xii">[Pg xii]</span> logical stand-point. We
+may use Euclid's Axioms, all but the last; and his Propositions as
+far as I. 28. Now be good enough to prove to me, with this machinery,
+first, that my hypothesis 'necessitates that the original angle should
+be at least 960°'; secondly, that 'no one of the chords' in my Figure
+'is even equal to the radius.' Your logical position is, I fear,
+this. You dispute the <i>validity</i> of a certain argument, on the
+ground that its <i>premisses</i> are false. My reply is, first, that
+you cannot <i>prove</i> them false; and secondly, that, even if you
+<i>could</i>, it wouldn't affect the question!"</p>
+
+<p class="space-above2">
+At p. 19 of the 2nd Edition, the new Axiom, on which my Theory rests,
+(which re-appears, in a modified form, at p. 14 of the 3rd Edition),
+stood thus:—"In every Circle, the inscribed equilateral Hexagon
+is greater than any one of the Segments which lie outside it." My
+controversy with the <i>Athenæum</i>, on this Axiom, shall also be
+given in the form of a dialogue.</p>
+
+<p><i>Athenæum.</i> (Oct. 27, 1888.) "... a stronger objection, in our
+opinion, is the implied assumption of the <i>possibility</i> of the
+inscribed equilateral Hexagon, a possibility which is not demonstrated
+till we reach the fifteenth Proposition of Euclid's fourth book."</p>
+
+<p><i>Author.</i> (In the Preface to the 2nd Edition, at p. xi). "But does
+it <i>need</i> demonstrating? May we not <i>assume</i> (1) that the
+Magnitude 'four right angles' contains 6-6ths of itself; (2) that it
+is <i>theoretically</i> possible to draw radii dividing it into these
+6-6ths? Once grant me this, and I ask no more. I have then the logical
+<i>right</i> to join the ends of these radii, and to prove (by Euc. I.
+4) that the chords are equal."</p>
+
+<p><span class="pagenum" id="Page_xiii">[Pg xiii]</span></p>
+
+<p><i>Athenæum.</i> (Oct. 5, 1889.) "We objected that it was not
+consistent with the spirit or practice of Euclid's reasoning to assume
+the 'theoretical possibility' of a regular Hexagon inscribed in a
+Circle, without first proving that such a figure could be actually
+constructed from his three postulates. Euclid's restrictions may be
+arbitrary, unnecessary, cramping, vexatious, absurd—indeed, we think
+they deserve these and many other epithets—but there they are, and, if
+Mr. Dodgson accepts them, he is bound to keep his assumptions within
+the boundaries which they prescribe."</p>
+
+<p><i>Author.</i> "You're particular to a shade (as Scrooge said to
+Marley's ghost): however, I'll do what I can to oblige you. I presume
+you will be satisfied if I can, without using more of Euclid than his
+first 28 Propositions, construct an angle which shall be 1-6th of 4
+right angles? Very good. First, then, with the help of his arbitrary
+Prop. I, I construct an equilateral Triangle. Next, by his unnecessary
+Prop. IX, I draw the bisectors of 2 of its angles. Next, by his
+cramping Post. 1, I join their point of intersection to the third
+vertex. Next, by his vexatious Prop. IV, I prove the 3 angles, whose
+common vertex is this point, to be equal. From which I draw the absurd
+conclusion that each of them is 1-3rd (and that therefore its half is
+1-6th) of 4 right angles. How does that strike <i>you</i>?"</p>
+
+<p class="space-above2">
+Another objection, to this same Axiom, appeared in the <i>Academy</i>
+for Feb. 9, 1889, viz. "What the Axiom practically assumes is the
+existence of similar figures." Permit me to reply, to this Reviewer,
+as follows:—"In what sense do you use the word 'similar'? In
+<i>Euclid's</i>, no doubt. That is to say, you charge me with assuming
+that, if the Circle<span class="pagenum" id="Page_xiv">[Pg xiv]</span> and its inscribed Hexagon were supposed to
+expand, the magnitude of each angle, and the ratio subsisting between
+the sides which contain it, would remain constant? The 'ratio' part
+of the question we may set aside at once: there is no doubt that,
+since the figure continues to be equilateral, the ratio continues
+to be a ratio of <i>unity</i>: hence, if I needed this assumption
+(which I don't), I should have a perfect right to make it. All, that
+remains for discussion, is the assumption, which you say I have made,
+that each <i>angle</i> of the expanding Hexagon remains constant in
+magnitude. Will you, then, be kind enough to point out, first, where
+the <i>need</i> for any such assumption arises; secondly, where I have
+<i>made</i> any such assumption? For myself I cannot in the least see
+why, in estimating the <i>area</i> of the Hexagon, I should trouble
+myself about the size of its <i>angles</i>."</p>
+
+<p>Let me take this opportunity of pointing out, once more, that <i>not
+one Proposition in this Treatise depends, in the slightest degree,
+on the speculations about Infinities, &amp;c., which occur in the
+Appendices</i>.</p>
+
+<p>The one merit, the one novelty, of my Theory (if it <i>has</i> any
+merit, or any novelty) is that, while <i>every</i> other Theory (that I
+have seen), which attempts to supersede Euclid's 12th Axiom, introduces
+the ideas of Infinities and Infinitesimals, <i>mine</i> dispenses
+<i>wholly</i> with their aid, and deals with nothing but what is, by
+universal consent, absolutely <i>within</i> the field of Human Reason.</p>
+
+
+<p class="right">C. L. D.</p>
+
+<p><span class="allsmcap">Ch. Ch., Oxford.</span><br>
+<span style="margin-left: 2em;"><i>August, 1890.</i></span></p>
+
+
+<hr class="chap x-ebookmaker-drop">
+
+<div class="chapter">
+<p><span class="pagenum" id="Page_xv">[Pg xv]</span></p>
+
+<h2 class="nobreak" id="INTRODUCTION">INTRODUCTION.</h2>
+</div>
+
+<hr class="r5">
+
+<p>It may well be doubted whether, in all the range of Science, there
+is any field so fascinating to the explorer—so rich in hidden
+treasures—so fruitful in delightful surprises—as that of Pure
+Mathematics. The charm lies chiefly, I think, in the absolute
+<i>certainty</i> of its results: for that is what, beyond almost all
+mental treasures, the human intellect craves for. Let us only be sure
+of <i>something</i>! More light, more light! Ἐν δὲ φάει, καὶ ὀλέσσον.
+'And, if our fate be death, give light and let us die!' This is the cry
+that, through all the ages, is going up from perplexed Humanity, and
+Science has little else to offer, that will really meet the demands
+of its votaries, than the conclusions of Pure Mathematics. Most other
+Sciences are in a state of constant flux—the precious truths of one
+generation being smiled at as paradoxes by the second generation, and
+contemptuously swept away as childish nonsense by the third. If you
+would see a specimen of the rapidity of this process of decomposition,
+take Biology for a sample: quote, to any distinguished Biologist you
+happen to meet, some book published thirty years ago, and observe his
+pitying smile!</p>
+
+<p><span class="pagenum" id="Page_xvi">[Pg xvi]</span></p>
+
+<p>But neither thirty years, nor thirty centuries, affect the clearness,
+or the charm, of Geometrical truths. Such a theorem as 'the square of
+the hypotenuse of a right-angled triangle is equal to the sum of the
+squares of the sides' is as dazzlingly beautiful now as it was in the
+day when Pythagoras first discovered it, and celebrated its advent, it
+is said, by sacrificing a hecatomb of oxen—a method of doing honour
+to Science that has always seemed to me <i>slightly</i> exaggerated
+and uncalled-for. One can imagine oneself, even in these degenerate
+days, marking the epoch of some brilliant scientific discovery by
+inviting a convivial friend or two, to join one in a beefsteak and a
+bottle of wine. But a <i>hecatomb</i> of oxen! It would produce a quite
+inconvenient supply of beef.</p>
+
+<p>Now this field of Mathematical research, with all its wealth of hidden
+treasure, is all too apt to yield nothing to our research: for it is
+haunted by certain <i>ignes fatui</i>—delusive phantoms, that float
+before us, and seem so fair, and are <i>all but</i> in our grasp, so
+nearly that it never seems to need more than <i>one</i> step further,
+and the prize shall be ours! Alas for him who has been turned aside
+from real research by one of these spectres—who has found a music
+in its mocking laughter—and who wastes his life and energy in the
+desperate chase!</p>
+
+<p>One of these—the chief one, I think—is the old old problem of
+'Squaring the Circle,' which has certainly wasted many a human
+life. Whether it has actually driven any one mad, I know not—most
+of its victims were, I fancy, partly crazed before they entered on
+the quest—but it clearly has the power of demolishing such slender
+reasoning powers as they may ever have chanced to possess.</p>
+
+<p><span class="pagenum" id="Page_xvii">[Pg xvii]</span></p>
+
+<p>With two of these misguided visionaries I have myself corresponded.</p>
+
+<p>The first who addressed me filled me with a great ambition—to do
+a feat I had never yet heard of as accomplished by man, namely, to
+convince a 'Circle-Squarer' of his error! The value my friend had
+selected for '\(\pi\)' was <i>not</i> an original one—being 3·2: but
+the enormous error, beginning as early as the <i>first</i> decimal
+place, tempted one with the idea that it could be easily demonstrated
+to <i>be</i> an error. I should think more than a score of letters were
+interchanged before I became sadly convinced that I had no chance.
+What man could still hope on, after receiving such a rebuff as the
+following? "You persuade yourself," so my friend wrote, "that you have
+made your circumscribed polygon equal to the circle, <i>which you know
+cannot be</i>, and have thereby pushed the quadrant beyond 90°, valuing
+the circumference at 360°." I meekly begged to be referred to the
+actual words in which I had advanced this startling assertion: but I
+never succeeded in getting the quotation verified.</p>
+
+<p>My second 'Circle-Squarer' went to work in quite another fashion.
+His object was not so much to obtain an <i>arithmetical</i> value
+for '\(\pi\),' as to construct a geometrical straight Line which,
+given the radius, should exhibit to the eye the actual length of
+the circumference. His diagram was a most imposing one—Triangles
+and Parallels were interlaced in bewildering profusion—and it
+used up no less than 23 letters of the alphabet. Some of the Lines
+had arithmetical values assigned to them: and there was one value,
+'1·8879020478639098461 &c.', which for a long time baffled all my
+endeavours to guess how in the world he had invented it. Of course one
+<i>might</i> have taken exception<span class="pagenum" id="Page_xviii">[Pg xviii]</span> to such a construction at the very
+outset, and have said "I will admit the possibility of constructing
+a Line, which shall bear to the unit-line any arithmetical ratio
+you like, so long as you express it as an <i>exact</i> decimal: but
+what <i>can</i> I do with your '&c.'?" But his was not the kind of
+mind to which the geometrical construction of an '&amp;c.' presents any
+difficulty. At length, after many failures, I chanced on the discovery
+that this portentous number was \(\dfrac{40}{3}\) of the decimal part
+of '\(\pi\).' After this it was no wonder, considering that, in the
+course of construction, he had taken \(\dfrac{3}{4}\) of this Line,
+and afterwards divided by 10, that the resulting Line, added to 3
+times the unit-Line, was triumphantly proved to represent '\(\pi\)'!
+I ventured to ask if this was the way he had obtained the long decimal
+quoted above, namely, by multiplying the decimal part of '\(\pi\)' by
+\(\dfrac{40}{3}\), and received the courteous reply "your suggestion is
+perfectly correct"!</p>
+
+<p>Another <i>ignis fatuus</i>—though not numbering so many victims as
+the 'Quadrature of the Circle'—is 'the Trisection of an Angle' (that
+is, its trisection by Euclid's machinery).</p>
+
+<p>And yet another <i>ignis fatuus</i>—the one with which the following
+treatise is concerned—is the attempt to dispense with Euclid's
+celebrated 12th Axiom.</p>
+
+<p>I may as well state briefly what the feat actually is, which
+Mathematicians have been vainly trying, since Euclid's day, to perform.</p>
+
+<p>In I. 27, 28, he proves (so far, without invoking the aid of any
+doubtful Axiom) that "two Lines, which are equally inclined to a
+certain transversal (whether by making a pair of alternate angles
+equal, or an exterior equal to its interior opposite angle, or two
+interior, on the same side of<span class="pagenum" id="Page_xix">[Pg xix]</span> the transversal, together equal to two
+right angles), will never meet."</p>
+
+<p>Next, in logical order, comes his 12th Axiom, viz. that "two Lines,
+which are <i>un</i>equally inclined to a certain transversal (he only
+<i>names</i> the case where they make two interior angles together less
+than two right angles, but he might fairly have included the others),
+<i>will</i> meet." This Axiom, as I hope to prove in Appendix III, is
+only <i>partially</i>, and not <i>universally</i>, true.</p>
+
+<p>Next, in I. 29, he proves (with the aid of this Axiom, of which it
+is what De Morgan calls the 'contranominal') the partially-true
+Theorem that "two Lines, which never meet, are equally inclined to any
+transversal."</p>
+
+<p>And from this, in I. 32, he proves that "the three angles of a Triangle
+are together equal to two right angles."</p>
+
+<p>These are only specimens of a set of Theorems which can be proved when
+once Axiom 12 is granted (e.g. there are several about 'equidistantial
+Lines,' which Euclid has altogether ignored): but they are all so
+connected as to follow easily from these.</p>
+
+<p>Now the great difficulty, which besets this subject, is that Euclid's
+Axiom (this, I think, is universally admitted) is <i>not</i>
+axiomatic—the intellect has not yet occurred, among that species of
+Vertebrates which may be defined as 'bimanous bipeds,' which accepts
+it as a genuine Axiom—and the great question to be answered is "can a
+better Axiom be found?"</p>
+
+<p>In Appendix IV, I will mention some of the substitutes that have been
+suggested, and will give some account of the 'outlook' in the direction
+of the new Axiom I have chanced on. In this place it will suffice,
+first, to explain what the task is that the long-desiderated Axiom
+has to<span class="pagenum" id="Page_xx">[Pg xx]</span> perform, and secondly, to state the grounds on which I claim
+acceptance for my Axiom.</p>
+
+<p>First, then, what is 'the coming Axiom' expected to do for us?</p>
+
+<p>It will be convenient to divide the whole class, of Theorems
+needing proof, into two sub-classes—one including those which
+are <i>universally</i> true: the other those which are only
+<i>partially</i> true—the error, if any, being <i>infinitesimal</i>
+when compared with the Magnitudes with which the Theorem is concerned.</p>
+
+<p>Euc. I. 32 is a specimen of the one kind, and Euc. I. 29 of the other.</p>
+
+<p>In proving the <i>latter</i> class, no substitute for Euclid's Axiom
+has yet been suggested, that I know of, which does not suffer from the
+same defect as Euclid's Axiom—the being only <i>partially</i>, and not
+<i>universally</i>, true—and which does not, if we attempt to modify
+the language so as to remedy this defect, in <i>some</i> way lead us
+into the bewildering region of Infinities and Infinitesimals.</p>
+
+<p>But the <i>former</i> class can, as I believe, be more easily proved.
+This is what I attempt in the following treatise—which owes its
+inspiration to a sudden thought (it occurred to me some two months ago)
+that it <i>might</i> be possible to prove Euc. I. 32 without getting
+mixed up with those spectral Infinities.</p>
+
+<p>Moreover, it is quite possible to bring into this class all that is
+valuable in Euc. I. 29. Regarding the 'separateness' of the Lines
+merely as a link between Props. 27, 28, and 29, we may combine the
+three into one grand Theorem, thus:—"Two Lines, which are equally
+inclined to a certain transversal, are so to every transversal." This
+Theorem, as well as Euc. I. 32, I prove in the following<span class="pagenum" id="Page_xxi">[Pg xxi]</span> treatise. But
+the feat of proving them, without assuming any new Axiom <i>at all</i>,
+is at present beyond my grasp. Like the goblin 'Puck,' it has led me
+"up and down, up and down," through many a wakeful night: but always,
+just as I thought I had it, some unforeseen fallacy was sure to trip me
+up, and the tricksy sprite would "leap out, laughing ho, ho, ho!"</p>
+
+<p>And now, to come to the real gist of this over-long Preface—however,
+nobody ever reads a Preface, so really it does not matter—am I not
+right in thinking that, on mere inspection of this diagram, any sane
+intellect will be ready to grant that "in any Circle, the inscribed
+Tetragon is greater than any one of the Segments that lie outside it"?</p>
+
+<figure class="figright width500" id="i_f021" style="width: 300px;">
+<img src="images/i_f021.jpg" width="300" height="309" alt="A circle
+containing an inscribed square, divided by two perpendicular diameters
+into eight triangles, illustrating Dodgson's key axiom about the
+inscribed equilateral tetragon.">
+</figure>
+
+<p>I shall be told, no doubt, that this is too <i>bizarre</i> and
+unprecedented an Axiom—that it is an appeal to the <i>eye</i>, and
+not to the reason. That it is somewhat <i>bizarre</i> I am willing to
+admit—and am by no means sure that this is not rather a <i>merit</i>
+than a defect. But, as to its being an appeal to the <i>eye</i>,
+what is "two straight Lines cannot enclose a space" but an appeal to
+the eye? What is "all right angles are equal" but an appeal to the
+<i>eye</i>?</p>
+
+<p>In all Axioms, where an appeal is made to the eye on a question of
+<i>magnitude</i>, we shall find, I think, that the whole region
+of certainty and probability may be roughly mapped out into three
+districts—an out-lying district of certainty in <i>one</i> direction,
+a similar one of certainty in the <i>opposite</i> direction, and a
+middle district of probability—the boundaries being shadowy and
+liable to be shifted hither<span class="pagenum" id="Page_xxii">[Pg xxii]</span> and thither according to the fancies or
+prejudices of each individual mind.</p>
+
+<p>Permit me to illustrate this by an example taken from ordinary life.</p>
+
+<p>You enter a room, where there is a book-case containing (say) five
+shelves, and your eye wanders carelessly along a shelf, making a rough
+estimate of the number of books in it. Now shut your eyes, and try to
+guess how many books there are altogether. Your hasty reckoning of
+one shelf gave a total (say) of 19 or 20, you are not sure which: so
+you feel safe in saying "I think there are <i>about</i> a hundred."
+"Are you <i>certain</i>," I ask, "that there are more than fifty?"
+"<i>Quite</i> certain," you reply. "And also certain that there are
+less than a hundred and fifty?" "<i>Quite</i> certain," you repeat.
+Here, then, are the three districts. The numbers up to 50 are
+<i>certainly</i> too small; the numbers over 150 are <i>certainly</i>
+too great; the intermediate numbers contain some doubtful ones—the
+most doubtful being very near 100—and this doubt shades off into
+certainty as we approach either of the out-lying districts. You
+would not risk five shillings on the chance of the true number being
+<i>under</i> 100, or on the chance of its being <i>over</i> 100, but
+you would feel quite at your ease, if told that you would forfeit a
+thousand pounds, in case the number turned out to be under 50, or over
+150.</p>
+
+<p>Another objection, that has already been raised to my Axiom, and so
+will probably be raised again, and which I may as well meet here by
+anticipation, is that, on the supposition of Euclid I. 32 <i>not</i>
+being true, it may be proved that this relationship of magnitude,
+between the Tetragon and the Segment, changes as the Circle increases,
+until, with an infinitely great Circle, the Tetragon may<span class="pagenum" id="Page_xxiii">[Pg xxiii]</span> actually be
+proved to be <i>less</i> than the Segment! This phenomenon, however,
+does not appal me so much as might be expected: for I have often
+observed it to occur that, when Theorem \(\alpha\) logically leads
+to Theorem \(\beta\), then, on the supposition of Theorem \(\beta\)
+<i>not</i> being true, it may be proved that Theorem \(\alpha\) also is
+not true. (The second sequence is, in fact, what De Morgan calls the
+'contranominal' of the first.) Hence this objection, if worth anything,
+<i>proves too much</i>: to dispute the validity of an argument, on the
+ground that, if it were valid, its contranominal would also be valid,
+is to upset the whole edifice of Logic itself: and, if you tell me,
+on such grounds as these, that <i>I</i> cannot prove what I assert,
+I may fairly retort upon you, that <i>you</i> cannot prove anything
+<i>at all</i>! You have destroyed the only machinery available for the
+purpose, and must henceforth dispense with all Logical methods, and
+console yourself with the cynical American adage "There's nothing true:
+and there's nothing new: and it don't signify!"</p>
+
+<p>To return to our Tetragon. It really contains the area of the Segment
+a little over 7 times. Hence anybody, I should suppose, would be ready
+to say "I am <i>certain</i> it contains the Segment more than twice:
+and I am equally certain it does <i>not</i> contain it twelve times."
+In guessing the <i>actual</i> number, observers would greatly differ:
+some might guess 4, others 10: but <i>all</i> would agree in putting
+it above 2. And now see how modest is the demand of my Axiom! Merely
+that you will find room in the Tetragon for one single Segment! If
+<i>that</i> is not a matter of certainty, is <i>anything</i> certain in
+this world of ours?</p>
+
+<p>I have yet one more arrow in my quiver: let me shoot it, and have done.
+If the gentle reader feels any the<span class="pagenum" id="Page_xxiv">[Pg xxiv]</span> smallest demur to granting me that
+<i>once</i> this Tetragon is greater than the Segment lying below it,
+will he grant me that <i>twice</i> it will suffice? Or four times it?
+Or eight times it? He may go on doubling as long as he likes, and, so
+long as he keeps among finite numbers, he will have granted me all I
+need for a logical proof (which will be found in Appendix I) of Euc. I.
+32. Surely he will not need to go into the Infinities? And may I add,
+in conclusion, that, if any gentle Reader be found, who thinks it just
+possible to squeeze 512 of these Tetragons into the Segment, but is
+willing to allow that no amount of skilful packing will dispose of 1024
+of them—it will give me <i>real</i> satisfaction to be supplied with
+that gentle Reader's name and address?</p>
+
+
+<p class="right">C. L. D.</p>
+
+<p><span class="allsmcap">Ch. Ch., Oxford.</span><br>
+<span style="margin-left: 2em;"><i>July, 1888.</i></span></p>
+
+
+<hr class="chap x-ebookmaker-drop">
+
+<div class="chapter">
+<p><span class="pagenum" id="Page_xxv">[Pg xxv]</span></p>
+
+<h2 class="nobreak" id="CONTENTS">CONTENTS.</h2>
+</div>
+
+<hr class="r5">
+
+<table class="autotable">
+<tbody><tr>
+<td class="tdc" colspan="2">Book I.</td>
+</tr><tr>
+<td class="tdc" colspan="2"><i>Certain universally-true Propositions,<br>
+provable from genuine Axioms.</i></td>
+</tr><tr>
+<td class="tdl"></td>
+<td class="tdr"><span class="allsmcap">PAGE</span></td>
+</tr><tr>
+<td class="tdl">Definitions, 1 to 3</td>
+<td class="tdr"><a href="#Page_1">1</a></td>
+</tr><tr>
+<td class="tdl">Axioms, 1 to 4</td>
+<td class="tdr"><a href="#Page_3">3</a></td>
+</tr><tr>
+<td class="tdl">Propositions:—</td>
+<td class="tdr"></td>
+</tr><tr>
+<td class="tdlh">I. Theorem. <i>If a Pair of Lines make, with a certain</i><br>
+<span class="tdlh3"><i>transversal, either (1) a pair of alternate angles</i></span><br>
+<span class="tdlh3"><i>equal, or (2) an exterior angle equal to its interior</i></span><br>
+<span class="tdlh3"><i>opposite angle on the same side of the transversal, or</i></span><br>
+<span class="tdlh3"><i>(3) a pair of interior angles on the same side of the</i></span><br>
+<span class="tdlh3"><i>transversal supplementary: they will make, with that</i></span><br>
+<span class="tdlh3"><i>transversal, (4), each pair of alternate angles equal,</i></span><br>
+<span class="tdlh3"><i>and (5) each of the four exterior angles equal to its</i></span><br>
+<span class="tdlh3"><i>interior opposite angle on the same side of the</i></span><br>
+<span class="tdlh3"><i>transversal, and (6) each pair of interior angles on</i></span><br>
+<span class="tdlh3"><i>the same side of the transversal supplementary.</i></span></td>
+<td class="tdr_bot"><a href="#Page_4">4</a></td>
+</tr><tr>
+<td class="tdl">Definition 5</td>
+<td class="tdr"><a href="#Page_4">"</a></td>
+</tr><tr>
+<td class="tdlh">II. Theorem. <i>If two isosceles Triangles have equal</i><br>
+<span class="tdlh3"><i>bases but unequal sides: that Triangle, which has</i></span><br>
+<span class="tdlh3"><i>the greater sides, has the greater area.</i></span></td>
+<td class="tdr_bot"><a href="#Page_5">5</a><span class="pagenum" id="Page_xxvi">[Pg xxvi]</span></td>
+</tr><tr>
+<td class="tdlh">III. Problem. <i>Given a certain angle; and given that</i><br>
+<span class="tdlh3"><i>any isosceles Triangle, whose vertical angle is not-greater</i></span><br>
+<span class="tdlh3"><i>than the given angle, has its base not-greater</i></span><br>
+<span class="tdlh3"><i>than either of its sides: to describe, on a given base,</i></span><br>
+<span class="tdlh3"><i>an isosceles Triangle having each base-angle equal to</i></span><br>
+<span class="tdlh3"><i>the given angle.</i></span></td>
+<td class="tdr_bot"><a href="#Page_6">6</a></td>
+</tr><tr>
+<td class="tdlh2">Corollary. <i>The isosceles Triangle, so described, has</i><br>
+<span class="tdlh2"><i>its vertical angle not-less than either of its base-angles.</i></span></td>
+<td class="tdr_bot"><a href="#Page_7">7</a></td>
+</tr><tr>
+<td class="tdlh">IV. Theorem. <i>Either all Triangles have the same</i><br>
+<span class="tdlh3"><i>'amount'; or else, if \(\alpha\), \(\beta\), be two 'possible amounts,'</i></span><br>
+<span class="tdlh3"><i>that is, 'amounts' belonging to existing Triangles:</i></span><br>
+<span class="tdlh3"><i>then any 'amount,' intermediate to \(\alpha\) and \(\beta\), is also</i></span><br>
+<span class="tdlh3"><i>'possible'.</i></span></td>
+<td class="tdr_bot"><a href="#Page_8">8</a></td>
+</tr><tr>
+<td class="tdlh2">Corollary 1. <i>Among angular magnitudes there is</i><br>
+<span class="tdlh2"><i>one, and only one, 'possible region'.</i></span></td>
+<td class="tdr_bot"><a href="#Page_9">9</a></td>
+</tr><tr>
+<td class="tdlh2">Corollary 2. <i>This 'possible region' either consists</i><br>
+<span class="tdlh2"><i>of one single angular magnitude, such that it,</i></span><br>
+<span class="tdlh2"><i>and it alone, is a 'possible amount'; or it consists</i></span><br>
+<span class="tdlh2"><i>of a continuous series of angular magnitudes, lying</i></span><br>
+<span class="tdlh2"><i>between 2 'limits,' which 2 limits are such that any</i></span><br>
+<span class="tdlh2"><i>magnitude, lying between them, is a 'possible amount,'</i></span><br>
+<span class="tdlh2"><i>and any magnitude, lying outside them, is an 'impossible</i></span><br>
+<span class="tdlh2"><i>amount'.</i></span></td>
+<td class="tdr_bot"><a href="#Page_9">"</a></td>
+</tr><tr>
+<td class="tdlh">V. Theorem. <i>The angles of any Triangle are together</i><br>
+<span class="tdlh3"><i>less than three right angles.</i></span></td>
+<td class="tdr_bot"><a href="#Page_10">10</a></td>
+</tr><tr>
+<td class="tdlh">VI. Theorem. <i>There is a Triangle whose angles are</i><br>
+<span class="tdlh3"><i>together not-greater than two right angles.</i></span></td>
+<td class="tdr_bot"><a href="#Page_10">"</a></td>
+</tr><tr>
+<td class="tdlh2">Corollary. <i>The 'possible region' does not lie</i><br>
+<span class="tdlh2"><i>wholly above two right angles.</i></span></td>
+<td class="tdr_bot"><a href="#Page_13">13</a><span class="pagenum" id="Page_xxvii">[Pg xxvii]</span></td>
+</tr><tr>
+<td class="tdc" colspan="2">Book II.</td>
+</tr><tr>
+<td class="tdc" colspan="2"><i>Certain universally true Propositions,
+not provable from genuine Axioms, but
+provable if the following Axiom be accepted.</i></td>
+</tr><tr>
+<td class="tdl">Axiom 1. <span class="antiqua">In any Circle, the inscribed equilateral</span><br>
+<span class="tdlh"><span class="antiqua">Tetragon is greater than any one</span></span><br>
+<span class="tdlh"><span class="antiqua">of the Segments which lie outside it.</span></span></td>
+<td class="tdr_bot"><a href="#Page_14">14</a></td>
+</tr><tr>
+<td class="tdl">Propositions:—</td>
+<td class="tdr_bot"><a href="#Page_14">"</a></td>
+</tr><tr>
+<td class="tdlh">I. Theorem. <i>An isosceles Triangle, whose vertical</i><br>
+<span class="tdlh3"><i>angle is one-eighth of a right angle, has its base</i></span><br>
+<span class="tdlh3"><i>less than either of its sides.</i></span></td>
+<td class="tdr_bot"><a href="#Page_15">15</a></td>
+</tr><tr>
+<td class="tdlh2">Corollary. <i>Hence, by Book I, Prop. III, it is</i><br>
+<span class="tdlh2"><i>possible to describe, on a given base, an isosceles</i></span><br>
+<span class="tdlh2"><i>Triangle having each base-angle equal to one-eighth</i></span><br>
+<span class="tdlh2"><i>of a right angle.</i></span></td>
+<td class="tdr_bot"><a href="#Page_17">17</a></td>
+</tr><tr>
+<td class="tdlh">II. Theorem. <i>The angles of any Triangle are together</i><br>
+<span class="tdlh3"><i>not-less than one-eighth of a right angle.</i></span></td>
+<td class="tdr_bot"><a href="#Page_18">18</a></td>
+</tr><tr>
+<td class="tdlh2">Corollary. <i>The 'possible region' does not extend</i><br>
+<span class="tdlh2"><i>below one-eighth of a right angle.</i></span></td>
+<td class="tdr_bot"><a href="#Page_18">"</a></td>
+</tr><tr>
+<td class="tdlh">III. Theorem. <i>There is a Triangle whose angles are</i><br>
+<span class="tdlh3"><i>together not-less than two right angles.</i></span></td>
+<td class="tdr_bot"><a href="#Page_19">19</a></td>
+</tr><tr>
+<td class="tdlh2">Corollary. <i>The 'possible region' does not lie</i><br>
+<span class="tdlh2"><i>wholly below two right angles.</i></span></td>
+<td class="tdr_bot"><a href="#Page_21">21</a></td>
+</tr><tr>
+<td class="tdlh">IV. Theorem. <i>There is a Triangle whose angles are</i><br>
+<span class="tdlh3"><i>together equal to two right angles.</i></span></td>
+<td class="tdr_bot"><a href="#Page_22">22</a></td>
+</tr><tr>
+<td class="tdlh">V. Theorem. <i>There is a quadrilateral Figure which</i><br>
+<span class="tdlh3"><i>is 'rectangular,' that is, which has all its angles</i></span><br>
+<span class="tdlh3"><i>right angles.</i></span></td>
+<td class="tdr_bot"><a href="#Page_22">"</a><span class="pagenum" id="Page_xxviii">[Pg xxviii]</span></td>
+</tr><tr>
+<td class="tdl">Definition.</td>
+<td class="tdr"><a href="#Page_23">23</a></td>
+</tr><tr>
+<td class="tdl">Propositions (continued):—</td>
+<td class="tdr"></td>
+</tr><tr>
+<td class="tdlh">VI. Theorem. <i>The opposite sides of a Rectangle are
+equal.</i></td>
+<td class="tdr"><a href="#Page_24">24</a></td>
+</tr><tr>
+<td class="tdlh">VII. Theorem. <i>There is a Pair of Lines, each of</i><br>
+<span class="tdlh3"><i>which is 'equidistant' from the other, that is, is such</i></span><br>
+<span class="tdlh3"><i>that all Points on it are equally distant from the</i></span><br>
+<span class="tdlh3"><i>other Line.</i></span></td>
+<td class="tdr_bot"><a href="#Page_24">"</a></td>
+</tr><tr>
+<td class="tdlh2">Corollary 1. <i>If a Pair of Lines have a common</i><br>
+<span class="tdlh2"><i>perpendicular: each of them is equidistant from the</i></span><br>
+<span class="tdlh2"><i>other.</i></span></td>
+<td class="tdr_bot"><a href="#Page_25">25</a></td>
+</tr><tr>
+<td class="tdlh2">Corollary 2. <i>It is possible to form a Rectangle</i><br>
+<span class="tdlh2"><i>of any given width and height.</i></span></td>
+<td class="tdr_bot"><a href="#Page_26">26</a></td>
+</tr><tr>
+<td class="tdlh">VIII. Theorem. <i>The angles of any Triangle are</i><br>
+<span class="tdlh3"><i>together equal to two right angles.</i></span></td>
+<td class="tdr_bot"><a href="#Page_26">"</a></td>
+</tr><tr>
+<td class="tdlh">IX. Theorem. <i>A Pair of Lines, which are equally</i><br>
+<span class="tdlh3"><i>inclined to a certain transversal, are so to any transversal.</i></span></td>
+<td class="tdr_bot"><a href="#Page_27">27</a></td>
+</tr><tr>
+<td class="tdl">Axiom 2. If two homogeneous magnitudes be both<br>
+<span class="tdlh">of them finite: the lesser may be so multiplied,</span><br>
+<span class="tdlh">by a finite number, as to exceed the greater.</span></td>
+<td class="tdr_bot"><a href="#Page_27">"</a></td>
+</tr><tr>
+<td class="tdl">Propositions (continued):—</td>
+<td class="tdr"></td>
+</tr><tr>
+<td class="tdlh">X. Theorem. <i>If a Pair of Lines make, with a certain</i><br>
+<span class="tdlh3"><i>transversal, two interior angles, on the same side of</i></span><br>
+<span class="tdlh3"><i>it, which are together less than two right angles, the</i></span><br>
+<span class="tdlh3"><i>defect being a finite angle: these Lines are intersectional</i></span><br>
+<span class="tdlh3"><i>on that side of the transversal.</i></span></td>
+<td class="tdr_bot"><a href="#Page_28">28</a><span class="pagenum" id="Page_xxix">[Pg xxix]</span></td>
+</tr><tr>
+<td class="tdc" colspan="2">Appendix I.</td>
+</tr><tr>
+<td class="tdc" colspan="2"><i>Containing an alternative Axiom, which may be
+substituted for Axiom 1 at p. 14.</i></td>
+</tr><tr>
+<td class="tdl">Definition.</td>
+<td class="tdr"><a href="#Page_32">32</a></td>
+</tr><tr>
+<td class="tdl">Propositions:—</td>
+<td class="tdr"></td>
+</tr><tr>
+<td class="tdl">(A) Theorem. <i>If, in any Sector of a Circle, its Chord</i><br>
+<span class="tdlh"><i>be not-less than its Radius: then, in a Sector whose</i></span><br>
+<span class="tdlh"><i>vertical angle is twice as great, its outer Segment is</i></span><br>
+<span class="tdlh"><i>greater than its central Triangle.</i></span></td>
+<td class="tdr_bot"><a href="#Page_33">33</a></td>
+</tr><tr>
+<td class="tdl">(B) Theorem. <i>If, in any Sector of a Circle, each of the</i><br>
+<span class="tdlh"><i>equal Sides of its inscribed isosceles Triangle be not-less</i></span><br>
+<span class="tdlh"><i>than its Radius; and if its outer Segment be</i></span><br>
+<span class="tdlh"><i>greater than a certain multiple of its central</i></span><br>
+<span class="tdlh"><i>Triangle: then, in a Sector, whose vertical angle is</i></span><br>
+<span class="tdlh"><i>twice as great, its outer Segment is greater than</i></span><br>
+<span class="tdlh"><i>twice that multiple of its central Triangle.</i></span></td>
+<td class="tdr_bot"><a href="#Page_34">34</a></td>
+</tr><tr>
+<td class="tdlh">Axiom. <span class="antiqua">In every Circle, the inscribed equilateral</span><br>
+<span class="tdlh3"><span class="antiqua">Tetragon, multiplied by \(2^{a}\) ('\(a\)' being</span></span><br>
+<span class="tdlh3"><span class="antiqua">a certain selected finite number), is greater</span></span><br>
+<span class="tdlh3"><span class="antiqua">than any one of the Segments which lie</span></span><br>
+<span class="tdlh3"><span class="antiqua">outside it.</span></span></td>
+<td class="tdr_bot"><a href="#Page_35">35</a></td>
+</tr><tr>
+<td class="tdl">Propositions (continued):—</td>
+<td class="tdr"></td>
+</tr><tr>
+<td class="tdl">(C) Theorem. <i>An isosceles Triangle, whose vertical</i><br>
+<span class="tdlh"><i>angle is \(\dfrac{1}{2^{a + 3}}\) of a right angle, has its base less than</i></span><br>
+<span class="tdlh"><i>either of its sides.</i></span></td>
+<td class="tdr_bot"><a href="#Page_36">36</a></td>
+</tr><tr>
+<td class="tdlh2">Corollary. <i>Hence, by Book I, Prop. III, it is</i><br>
+<span class="tdlh2"><i>possible to describe, on a given base, an isosceles</i></span><br>
+<span class="tdlh2"><i>Triangle having each base-angle equal to \(\dfrac{1}{2^{a + 3}}\) of a</i></span><br>
+<span class="tdlh2"><i>right angle.</i></span></td>
+<td class="tdr_bot"><a href="#Page_38">38</a><span class="pagenum" id="Page_xxx">[Pg xxx]</span></td>
+</tr><tr>
+<td class="tdl">(D) Theorem. <i>The angles of any Triangle are together</i><br>
+<span class="tdlh"><i>not-less than \(\dfrac{1}{2^{a + 3}}\) of a right angle.</i></span></td>
+<td class="tdr_bot"><a href="#Page_38">38</a></td>
+</tr><tr>
+<td class="tdl">(E) Theorem. <i>There is a Triangle whose angles are</i><br>
+<span class="tdlh"><i>together not-less than two right angles.</i></span></td>
+<td class="tdr_bot"><a href="#Page_39">39</a></td>
+</tr><tr>
+<td class="tdc" colspan="2"><span class="allsmcap">Appendix II.</span></td>
+</tr><tr>
+<td class="tdc" colspan="2"><i>Is Euclid's Axiom True?</i></td>
+</tr><tr>
+<td class="tdl">§ 1. Infinite and Finite Magnitudes.</td>
+<td class="tdr"><a href="#Page_40">40</a></td>
+</tr><tr>
+<td class="tdl">§ 2. Infinitesimal Lines and Strips.</td>
+<td class="tdr"><a href="#Page_43">43</a></td>
+</tr><tr>
+<td class="tdl">§ 3. Infinitesimal Angles and Sectors.</td>
+<td class="tdr"><a href="#Page_48">48</a></td>
+</tr><tr>
+<td class="tdl">§ 4. Pairs of Lines.</td>
+<td class="tdr"><a href="#Page_49">49</a></td>
+</tr><tr>
+<td class="tdc" colspan="2"><span class="allsmcap">Appendix III.</span></td>
+</tr><tr>
+<td class="tdc"><i>How should Parallels be defined?</i></td>
+<td class="tdr"><a href="#Page_59">59</a></td>
+</tr><tr>
+<td class="tdc" colspan="2"><span class="allsmcap">Appendix IV.</span></td>
+</tr><tr>
+<td class="tdc" colspan="2"><i>How the Question stands to-day.</i></td>
+</tr><tr>
+<td class="tdl">§ 1. Certain universally-true Theorems, provable<br>
+<span class="tdlh">from genuine Axioms (i.e. from Axioms</span><br>
+<span class="tdlh">whose self-evident character is indisputable).</span></td>
+<td class="tdr_bot"><a href="#Page_62">62</a></td>
+</tr><tr>
+<td class="tdl">§ 2. Certain universally-true Theorems, not provable<br>
+<span class="tdlh">from genuine Axioms, but provable if</span><br>
+<span class="tdlh">any one of them be accepted as an Axiom.</span></td>
+<td class="tdr_bot"><a href="#Page_63">63</a><span class="pagenum" id="Page_xxxi">[Pg xxxi]</span></td>
+</tr><tr>
+<td class="tdl">§ 3. Certain universally-true Theorems, not provable<br>
+<span class="tdlh">from genuine Axioms, but provable if</span><br>
+<span class="tdlh">any one of the 'Nine Quasi-Axioms' be</span><br>
+<span class="tdlh">accepted.</span></td>
+<td class="tdr_bot"><a href="#Page_65">65</a></td>
+</tr><tr>
+<td class="tdl">§ 4. Certain partially-true Theorems, not provable<br>
+<span class="tdlh">from any universally-true Axioms, whether</span><br>
+<span class="tdlh">genuine or 'quasi,' but provable if any one</span><br>
+<span class="tdlh">of themselves be accepted as an Axiom.</span></td>
+<td class="tdr_bot"><a href="#Page_65">"</a></td>
+</tr><tr>
+<td class="tdl">§ 5. Other methods of treatment.</td>
+<td class="tdr"><a href="#Page_67">67</a></td>
+</tr><tr>
+<td class="tdlh">Playfair's theory of 'direction'.</td>
+<td class="tdr"><a href="#Page_67">"</a></td>
+</tr><tr>
+<td class="tdlh">Fallacious proof for Euc. I. 32.</td>
+<td class="tdr"><a href="#Page_70">70</a></td>
+</tr><tr>
+<td class="tdlh">Bertrand's Theorem.</td>
+<td class="tdr"><a href="#Page_71">71</a></td>
+</tr><tr>
+<td class="tdlh">W. Hanna's fallacy.</td>
+<td class="tdr"><a href="#Page_72">72</a></td>
+</tr><tr>
+<td class="tdlh">J. Walmsley's fallacy.</td>
+<td class="tdr"><a href="#Page_72">"</a></td>
+</tr><tr>
+<td class="tdl">§ 6. The Outlook.</td>
+<td class="tdr"><a href="#Page_73">73</a></td>
+</tr><tr>
+<td class="tdlh">Advice to future explorers.</td>
+<td class="tdr"><a href="#Page_73">"</a></td>
+</tr><tr>
+<td class="tdlh">Unproved Theorems which would suffice for our<br>
+<span class="tdlh">purpose.</span></td>
+<td class="tdr_bot"><a href="#Page_75">75</a></td>
+</tr><tr>
+<td class="tdlh">New Definition needed for Right Line.</td>
+<td class="tdr_bot"><a href="#Page_75">"</a></td>
+</tr>
+ </tbody>
+</table>
+
+
+<hr class="chap x-ebookmaker-drop">
+
+<div class="chapter">
+<p><span class="pagenum" id="Page_1">[Pg 1]</span></p>
+
+<h2 class="nobreak" id="A_NEW_THEORY_OF_PARALLELS">A NEW THEORY OF PARALLELS.<br>
+<br>
+<span class="allsmcap">Book I.</span></h2>
+</div>
+
+<p class="nindc"><i>Certain universally-true Propositions, provable from genuine
+Axioms.</i></p>
+
+
+<h3><span class="allsmcap">Definitions.</span></h3>
+
+
+<p class="nindc">1.</p>
+
+<p>The sum of the angles of a Triangle is called its '<b>amount</b>.'</p>
+
+
+<p class="nindc">2.</p>
+
+<p>Any angular magnitude is called a '<b>possible amount</b>,' if there
+be a Triangle whose 'amount' is equal to it: but, if there be no such
+Triangle, it is called an '<b>impossible amount</b>.'</p>
+
+
+<p class="nindc">3.</p>
+
+<p>If any such angular magnitude vary continuously: whenever it changes
+from a 'possible amount' to an 'impossible amount,' it is said to pass
+from a '<b>possible region</b>,' to an '<b>impossible region</b>': and
+<i>vice versâ</i>.</p>
+
+<p><span class="pagenum" id="Page_2">[Pg 2]</span></p>
+
+<p class="nindc">4.</p>
+
+<p>If there be two fixed angular magnitudes such that the varying
+magnitude, while it continues between them, is always a 'possible
+amount,' but becomes an 'impossible amount' when it passes beyond them:
+they are called the '<b>superior limit</b>,' and the '<b>inferior
+limit</b>,' of the 'possible region' which lies between them.</p>
+
+
+<h3><span class="allsmcap">Axioms.</span></h3>
+
+<p><span class="pagenum" id="Page_3">[Pg 3]</span></p>
+
+
+<p class="nindc">1.</p>
+
+<figure class="figright width500" id="i_p003a" style="width: 150px;">
+<img src="images/i_p003a.jpg" width="150" height="158" alt="Two
+straight lines intersecting at point E, horizontal line AB crossed by
+diagonal line CD, forming four angles. A classic Euclidean diagram
+illustrating intersecting lines and the angles they produce.">
+</figure>
+
+<p>If a Magnitude change from one value to another; and if, in doing so,
+it vary continuously; and if a certain value, intermediate to its first
+and last values, be selected: the Magnitude must, at some moment during
+the process of change, have that selected value.</p>
+
+
+<p class="nindc">2.</p>
+
+<p>If two or more Magnitudes be such that, whenever any one of them
+varies, it varies continuously: then, whenever their sum varies, it
+varies continuously.</p>
+
+<figure class="figright width500" id="i_p003b" style="width: 150px;">
+<img src="images/i_p003b.jpg" width="150" height="148" alt="Two
+straight lines, horizontal line AB crossed by diagonal line CD, forming
+four angles.">
+</figure>
+
+<p class="nindc">3.</p>
+
+<p>If \(AB\), \(CD\), be two Lines intersecting at \(E\); and if \(CD\) be
+turned about \(E\), \(AB\) remaining stationary: each of the angles at
+\(E\) varies continuously.</p>
+
+
+<p class="nindc space-below2">4.</p>
+
+<p>If \(AB\), \(CD\), be two intersecting Lines; and if \(CD\) be turned
+about \(C\), \(AB\) remaining stationary: then, so long as the
+Lines continue to intersect, each of the 4 angles at the Point of
+intersection varies continuously.</p>
+
+
+<h3><span class="allsmcap">Propositions.</span></h3>
+<p><span class="pagenum" id="Page_4">[Pg 4]</span></p>
+
+
+<p class="nindc space-above2">PROP. I. <span class="smcap">Theorem.</span></p>
+
+<p><i>If a Pair of Lines make, with a certain transversal, either</i> (1)
+<i>a pair of alternate angles equal, or (2) an exterior angle equal to
+its interior opposite angle on the same side of the transversal, or</i>
+(3) <i>a pair of interior angles on the same side of the transversal
+supplementary: they will make, with that transversal,</i> (4), <i>each
+pair of alternate angles equal, and</i> (5) <i>each of the four exterior
+angles equal to its interior opposite angle on the same side of the
+transversal, and</i> (6) <i>each pair of interior angles on the same
+side of the transversal supplementary.</i></p>
+
+<p>This Proposition is easily deduced from Euc. I. 13, 15.</p>
+
+
+<h3><span class="allsmcap">Definitions</span> (<i>continued</i>).</h3>
+
+
+<p class="nindc">5.</p>
+
+<p>Such a Pair of Lines may be said to be '<b>equally inclined</b>' to
+that transversal.</p>
+
+
+<p class="nindc space-above2">PROP. II. <span class="smcap">Theorem.</span></p>
+<p><span class="pagenum" id="Page_5">[Pg 5]</span></p>
+
+<p><i>If two isosceles Triangles have equal bases but unequal sides: that
+Triangle, which has the greater sides, has the greater area.</i></p>
+
+<figure class="figcenter width500" id="i_p005" style="width: 300px;">
+<img src="images/i_p005.jpg" width="300" height="416" alt="Two
+nested triangles sharing base AB — outer triangle ADB with apex D,
+inner triangle ACB with apex C sitting just below D. Illustrates a
+proposition comparing triangles with a common base but differing apex
+heights.">
+</figure>
+
+<p>Let the Triangles be set on the same base, and call them \(ABC\),
+\(ABD\). And let \(AD\), \(DB\), be respectively greater than \(AC\),
+\(CB\).</p>
+
+<p>Now \(D\) cannot fall within the Triangle \(ACB\), or upon either of
+its sides; for then \(AD\), \(DB\) would be less than \(AC\), \(CB\);</p>
+
+<p class="right">[Euc. I. 21.</p>
+
+<p>neither can it fall on \(C\); for then \(AD\), \(DB\) would be equal to
+\(AC\), \(CB\);</p>
+
+<p>neither can \(AD\) intersect \(CB\), nor \(AC\) intersect \(DB\); for,
+in either case, if \(CD\) were joined, there would be two Triangles, on
+the same base \(CD\), having their coterminous sides equal; which is
+impossible;</p>
+
+
+<p class="right">[Euc. I. 7.</p>
+
+
+<p>\(\therefore\) \(AD\), \(DB\) fall outside the Triangle \(ACB\);</p>
+
+<p>\(\therefore\) Triangle \(ADB\) is greater than Triangle \(ACB\).</p>
+
+<p>Therefore, if two isosceles Triangles &amp;c.</p>
+
+
+<p class="right">Q.E.D.</p>
+<p><span class="pagenum" id="Page_6">[Pg 6]</span></p>
+
+
+<p class="nindc space-above2">PROP. III. <span class="allsmcap">Problem.</span></p>
+
+<p><i>Given a certain angle; and given that any isosceles Triangle, whose
+vertical angle is not-greater than the given angle, has its base
+not-greater than either of its sides: to describe, on a given base, an
+isosceles Triangle having each base-angle equal to the given angle.</i></p>
+
+<figure class="figcenter width500" id="i_p006" style="width: 500px;">
+<img src="images/i_p006.jpg" width="500" height="215" alt="Two
+triangles sharing base AB — left shows equilateral-looking triangle
+ABC with apex C high above; right shows a flat, acute triangle ABC
+with internal point F, illustrating how apex position and angle affect
+triangle geometry.">
+</figure>
+
+<p>Let \(AB\) be given base.</p>
+
+<p>At \(A\), in Line \(AB\), make angle \(BAC\) equal to given angle,
+making \(AC = AB\); and join \(BC\).</p>
+
+<p>Then, by hypothesis, \(BC\) is not-greater than \(AC\): i. e. it is
+either equal to it, or less than it.</p>
+
+<p class="space-above2">
+First let \(BC\) be equal to \(AC\). (Fig. 1.)</p>
+
+<p>Then Triangle \(ABC\) is equilateral: i. e. it is an isosceles
+Triangle, on given base \(AB\), and having each base-angle equal to
+given angle.</p>
+
+
+<p class="right">Q.E.F.</p>
+
+
+<p class="space-above2">
+Secondly, let \(BC\) be less than \(AC\). (Fig. 2.)</p>
+
+<p>Then angle \(BAC\) is less than angle \(ABC\).</p>
+
+
+<p class="right">[Euc. I. 18.</p>
+
+
+<p>At \(B\) make angle \(ABF\) equal to angle \(BAC\).</p>
+
+
+<p class="right">[Euc. I. 23.</p>
+
+
+<p>Then Triangle \(ABF\) is isosceles, and is on given base \(AB\), and
+has each base-angle equal to given angle.</p>
+
+
+<p class="right">Q.E.F.</p>
+<p><span class="pagenum" id="Page_7">[Pg 7]</span></p>
+
+
+<p class="nindc"><span class="allsmcap">Corollary.</span></p>
+
+<p><i>The isosceles Triangle, so described, has its vertical angle
+not-less than either of its base-angles.</i></p>
+
+<p>For, in Fig. 1, \(AB = AC\);</p>
+
+<p>\(\therefore \text{angle}\, ACB = \text{angle}\, ABC\).</p>
+
+
+<p class="right">Q.E.D.</p>
+
+
+<p>Again, in Fig. 2, \(AB = AC\);</p>
+
+<p>\(\therefore \text{angle}\, ACB = \text{angle}\, ABC\).</p>
+
+<p>But angle \(AFB\) is greater than angle \(ACB\);</p>
+
+
+<p class="right">[Euc. I. 16.</p>
+
+
+<p>and angle \(ABF\) is less than angle \(ABC\);</p>
+
+<p>\(\therefore \text{angle}\, AFB\, \text{is greater than angle}\, ABF\).</p>
+
+
+<p class="right">Q.E.D.</p>
+<p><span class="pagenum" id="Page_8">[Pg 8]</span></p>
+
+
+<p class="nindc space-above2">PROP. IV. <span class="allsmcap">Theorem.</span></p>
+
+<p><i>Either all Triangles have the same 'amount'; or else, if \(\alpha\),
+\(\beta\) be two 'possible amounts,' that is, 'amounts' belonging to
+existing Triangles, then any 'amount,' intermediate to \(\alpha\) and
+\(\beta\), is also 'possible.'</i></p>
+
+<figure class="figcenter width500" id="i_p008" style="width: 500px;">
+<img src="images/i_p008.jpg" width="500" height="321" alt="Two
+overlapping triangles on baseline AD — triangle ACB with apex C, and
+triangle AEB with vertical height EB, their sides crossing at point F.
+Illustrates a proposition comparing areas or angles of intersecting
+triangles.">
+</figure>
+
+<p>If all Triangles have the same amount, the Proposition is true. If not,
+let \(ABC\), \(ADE\) be two Triangles whose amounts are different. Call
+their 'amounts' '\(\alpha\), \(\beta\).' And let the two Triangles be
+placed so as to have a common vertex at \(A\), and their bases in the
+same straight Line.</p>
+
+<p>Now Triangle \(ABC\) may be converted into Triangle \(ABF\) by making
+the point, where \(AC\) intersects \(BC\), move from \(C\) to \(F\),
+\(BC\) remaining stationary.</p>
+
+<p>And, during this process, the angle at \(A\) will vary continuously,</p>
+
+
+<p class="right">[Ax. 4.</p>
+
+
+<p>and the angle, at the point where the revolving Line intersects \(BC\),
+will vary continuously;</p>
+
+
+<p class="right">[Ax. 5.</p>
+
+
+<p>\(\therefore\) the sum of these angles will, if it vary at all, vary
+continuously.</p>
+
+
+<p class="right">[Ax. 3.</p>
+
+
+<p>Similarly, Triangle \(ABF\) may be converted, first, into Triangle
+\(ABE\), by making the point, where \(BF\) intersects \(AE\), move from
+\(F\) to \(E\), \(AE\) remaining stationary, and then into Triangle
+\(ADE\), by making the point, where \(EB\) intersects \(AD\), move from
+\(B\) to \(D\), \(AD\) remaining stationary.</p>
+
+<p>And, during the whole process, the 'amount' of the changing Triangle
+will, if it vary at all, vary continuously.</p>
+
+<p>Hence, in changing from the value \(\alpha\) to the value \(\beta\), it
+must pass through all intermediate 'amounts': i. e. all intermediate
+'amounts' are 'possible.'</p>
+
+
+<p class="right">[Ax. 2.</p>
+
+
+<p>Therefore either all Triangles have &c.</p>
+
+
+<p class="right">Q.E.D.</p>
+
+
+
+<p class="nindc"><span class="allsmcap">Corollaries.</span></p>
+<p><span class="pagenum" id="Page_9">[Pg 9]</span></p>
+
+<p class="nindc">1.</p>
+
+<p><i>Among angular magnitudes there is one, and only one, 'possible
+region.'</i></p>
+
+
+<p class="nindc">2.</p>
+
+
+<p><i>This 'possible region' either consists of one single angular
+magnitude, such that it, and it alone, is a 'possible amount'; or it
+consists of a continuous series of angular magnitudes, lying between
+2 'limits,' which 2 limits are such that any magnitude, lying between
+them, is a 'possible amount,' and any magnitude, lying outside them, is
+an 'impossible amount.'</i></p>
+
+
+
+<p class="nindc space-above2">PROP. V. <span class="allsmcap">Theorem.</span></p>
+<p><span class="pagenum" id="Page_10">[Pg 10]</span></p>
+
+<p><i>The angles of any Triangle are together less than three right
+angles.</i></p>
+
+<p>Let a right angle be represented by '\(R\).'</p>
+
+<p>Now any 2 of the angles of a Triangle are together less than \(2R\);</p>
+
+
+<p class="right">[Euc. I. 17.</p>
+
+
+<p>\(\therefore\), adding together the 3 pairs which may be taken, the 3
+angles of a Triangle, taken twice over, are together less than \(6R\);</p>
+
+<p>\(\therefore\), taken once only, they are together less than \(3R\).</p>
+
+<p>Therefore the angles of any Triangle &c.</p>
+
+
+<p class="right">Q.E.D.</p>
+
+
+
+<p class="nindc">PROP. VI. <span class="allsmcap">Theorem.</span></p>
+
+<p><i>There is a Triangle whose angles are together not-greater than two
+right angles.</i></p>
+
+<p>If we deny this, we must assert that any 'amount' is greater than
+\(2R\).</p>
+
+<p>Let this be our First Hypothesis.</p>
+
+<p>Now any 'amount' is less than \(3R\).</p>
+
+
+<p class="right">[Prop. 5.</p>
+
+
+<p>Hence the 'possible region' lies below \(3R\); i. e. it has a 'superior
+limit.'</p>
+
+<p>Now let a 'possible amount' be selected, more than half-way from \(2R\)
+to the 'superior limit' of this region; and call it '\((2R + a)\).'
+Then it is evident that \((2R + 2a)\) will lie <i>above</i> this limit,
+and will therefore be an 'impossible amount.'</p>
+
+<p><span class="pagenum" id="Page_11">[Pg 11]</span></p>
+
+<p>Now any Triangle, whose 'amount' is \((2R + a)\), must be either
+obtuse-angled, or right-angled, or acute-angled.</p>
+
+<p>Hence we must assert that there is either an obtuse-angled Triangle, or
+a right-angled Triangle, or an acute-angled Triangle, whose 'amount' is
+\((2R + a)\).</p>
+
+<p>Let these be our Second, our Third, and our Fourth Hypotheses.</p>
+
+<figure class="figright width500" id="i_p011" style="width: 200px;">
+<img src="images/i_p011.jpg" width="200" height="147" alt="Vertical
+line AD on the right, with point B far left and interior point C,
+forming triangles BAC and BDC. Lines radiate from B through C to AD,
+illustrating a proposition about angles or distances from an external
+point.">
+</figure>
+
+<p>Call the Triangle '\(ABC\).'</p>
+
+<p>First, let it be obtuse-angled; and let \(C\) be the obtuse angle.</p>
+
+<p>At Point \(C\), in Line \(BC\), make angle \(BCD\) equal to angle
+\(BCA\), making \(CD\) equal to \(CA\); and join \(BD\) and \(AD\).</p>
+
+<p>Then Triangle \(BCD\) is equal, in all respects, to Triangle \(BCA\);</p>
+
+
+<p class="right">[Euc. I. 4.</p>
+
+
+<p>\(\therefore\) its 'amount' = \((2R + a)\);</p>
+
+<p>also 'amount' of Triangle \(ACD\) is, by our First Hypothesis, greater
+than \(2R\);</p>
+
+<p>\(\therefore\) 'amounts' of the 3 Triangles are together greater than
+\((6R + 2a)\).</p>
+
+<p>But these make up 'amount' of Triangle \(ABD\), plus angles about
+\(C\), which = \(4R\);</p>
+
+
+<p class="right">[Euc. I. 13. Cor.</p>
+
+
+<p>\(\therefore\) 'amount' of Triangle \(ABD\), plus \(4R\), is greater
+than \((6R + 2a)\);</p>
+
+<p>\(\therefore\) this 'amount,' alone, is greater than \((2R + 2a)\);
+which is absurd, since the latter lies above the 'superior limit,' and
+is therefore an 'impossible amount.'</p>
+
+<p>Hence our Second Hypothesis is false; i.e. no obtuse-angled Triangle
+can have the amount \((2R + a)\).</p>
+
+<p><span class="pagenum" id="Page_12">[Pg 12]</span></p>
+
+<p>Secondly, let it be right-angled; and let \(C\) be the right angle.</p>
+
+<figure class="figleft width500" id="i_p012a" style="width: 200px;">
+<img src="images/i_p012a.jpg" width="200" height="176" alt="Vertical
+line AD on the right with midpoint C, and point B to the left, forming
+two symmetric triangles BAC and BDC sharing horizontal base BC,
+illustrating equal angles or distances about a midpoint.">
+</figure>
+
+<p>At Point \(C\), in Line \(BC\), make angle \(BCD\) equal to angle
+\(BCA\), i. e. equal to \(R\); and make \(CD = CA\); and join \(BD\).</p>
+
+<p>Then \(AC\), \(CD\), are in one straight Line.</p>
+
+
+<p class="right">[Euc. I. 14.</p>
+
+
+<p>Also Triangle \(BCD\) is equal, in all respects, to Triangle \(BCA\);</p>
+
+
+<p class="right">[Euc. I. 4.</p>
+
+
+<p>\(\therefore\) its 'amount' = \((2R + a)\);</p>
+
+<p>\(\therefore\) 'amounts' of the 2 Triangles together = \((4R + 2a)\).</p>
+
+<p>But these make up 'amount' of Triangle \(ABD\), plus angles at \(C\),
+which = \(2R\);</p>
+
+<p>\(\therefore\) 'amount' of Triangle \(ABD\), plus \((2R, = (4R + 2a)\);</p>
+
+<p>\(\therefore\) this 'amount,' alone, = \((2R + 2a)\); which is absurd.</p>
+
+<p>Hence our Third Hypothesis is false; i.e. no right-angled Triangle can
+have the amount \((2R + a)\).</p>
+
+<p>Thirdly, let it be acute-angled.</p>
+
+<figure class="figleft width500" id="i_p012b" style="width: 200px;">
+<img src="images/i_p012b.jpg" width="200" height="129" alt="Two
+overlapping triangles, ABC on the left with interior point D, and a
+larger triangle extending to E on the right, their sides crossing at D,
+illustrating a proposition about overlapping or congruent triangles.">
+</figure>
+
+<p>Bisect \(AC\) at \(D\); join \(BD\), and produce it to \(E\), making
+\(DE = BD\); and join \(CE\).</p>
+
+<p>Now angles \(ADB\), \(CDE\), are equal, being vertical;</p>
+
+
+<p class="right">[Euc. I. 15.</p>
+
+
+<p>\(\therefore\) Triangles \(ADB\), \(CDE\), are equal in all respects;</p>
+
+
+<p class="right">[Euc. I. 4.</p>
+
+
+<p>\(\therefore\) angle \(DCE\) = angle \(A\); and angle \(CED\) = angle
+\(ABD\);</p>
+
+<p>\(\therefore\) 'amount' of Triangle \(BCE\) = that of Triangle \(ABC\).</p>
+
+<p>Again, \(\because\) angle \(CED\) = angle \(ABD\);</p>
+
+<p>\(\therefore\) angles \(CED\), \(CBD\) together = angle \(ABC\);</p>
+
+<p>\(\therefore\) they are together less than \(R\);</p>
+
+<p>\(\therefore\) angle \(BCE\) is, by our First Hypothesis, greater than
+\(R\);</p>
+
+<p>i. e. Triangle \(BCE\) is obtuse-angled;</p>
+
+<p>\(\therefore\) its 'amount' cannot be \((2R + a)\);</p>
+
+<p>\(\therefore\) 'amount' of Triangle \(ABC\) cannot be \((2R + a)\).</p>
+
+<p>Hence our Fourth Hypothesis is false; i.e. no acute-angled Triangle can
+have the 'amount' \((2R + a)\).</p>
+
+<p>Hence, no Triangle can have this amount.</p>
+
+<p>Hence our First Hypothesis, that any 'amount' is greater than \(2R\),
+is false.</p>
+
+<p>Therefore there is a Triangle &amp;c.</p>
+
+
+<p class="right">Q.E.D.</p>
+
+
+
+<p class="nindc space-above2"><span class="allsmcap">Corollary.</span></p>
+<p><span class="pagenum" id="Page_13">[Pg 13]</span></p>
+
+
+<p><i>The 'possible region' does not lie wholly above two right angles.</i></p>
+
+
+<hr class="chap x-ebookmaker-drop">
+
+<div class="chapter">
+<h2 class="nobreak" id="Book_IIa"><span class="smcap">Book II.</span></h2>
+</div>
+
+<p><i>Certain universally-true Propositions, not provable from genuine
+Axioms, but provable if the following Axiom be accepted.</i></p>
+
+
+<p>[N.B. This Axiom cannot claim to be more than a 'Quasi-Axiom,' i. e.
+one whose <i>self-evident</i> character is disputable.]</p>
+
+<hr class="r5">
+
+<h3><span class="allsmcap">Axioms.</span></h3>
+<p><span class="pagenum" id="Page_14">[Pg 14]</span></p>
+
+<figure class="figleft width500" id="i_p014" style="width: 150px;">
+<img src="images/i_p014.jpg" width="150" height="150" alt="Diagram
+with inscribed square within a circle, divided by two perpendicular
+diameters into eight triangles, here appearing within the text to
+support a specific proposition about the inscribed tetragon.">
+</figure>
+
+<p class="nindc">1.</p>
+
+<p><span class="antiqua">In any Circle, the inscribed equilateral Tetragon is greater than
+any one of the Segments which lie outside it.</span></p>
+
+
+<hr class="chap">
+
+<div class="blockquot">
+
+<p><span class="allsmcap">Note.</span>—If, in any Circle, 2 Diameters be drawn at right
+angles to each other, and their extremities joined, the joining Lines
+will, by Euc. I. 4 be equal to each other. Hence the Figure, thus
+formed, will be an inscribed equilateral Tetragon. (It will also be
+<i>equiangular</i>; but that is of no importance for our present
+purpose.)</p>
+</div>
+
+
+<h3>PROP. I. <span class="smcap">Theorem.</span></h3>
+<p><span class="pagenum" id="Page_15">[Pg 15]</span></p>
+
+<p><i>An isosceles Triangle, whose vertical angle is one-eighth of a right
+angle, has its base less than either of its sides.</i></p>
+
+<figure class="figcenter width500" id="i_p015" style="width: 300px;">
+<img src="images/i_p015.jpg" width="300" height="307" alt="A
+quarter-circle arc from K to B, with multiple lines radiating from
+corner A to labelled points K, J, H, G, F, E, D, C, B along the arc
+and sides, illustrating how angles or chord lengths vary progressively
+around a quadrant.">
+</figure>
+
+<p>Let \(ABC\) be an isosceles Triangle, whose vertical angle at \(A\) is
+one-eighth of a right angle.</p>
+
+<p>It shall be proved that \(BC\) is less than \(AB\).</p>
+
+<p>If we deny this, we must assert that \(BC\) is not-less than \(AB\).</p>
+
+<p>Let this be our Hypothesis.</p>
+
+<p>Construct 7 more Triangles \(ACD\), &amp;c., equal to \(ABC\). With centre
+\(A\), and distance \(AB\), describe quadrant passing through \(C\),
+\(D\), &amp;c. (See Note.) And join \(BK\), \(BF\), \(FK\), \(BD\), \(DF\),
+\(FH\), \(HK\).</p>
+
+<hr class="chap">
+
+<div class="blockquot">
+
+<p><span class="allsmcap">Note.</span>—The Reader is requested to imagine chords drawn to the
+arcs \(BC\), \(CD\), &amp;c.</p>
+</div>
+
+<p><span class="pagenum" id="Page_16">[Pg 16]</span></p>
+
+<p>Then Triangle \(ABK\) is one-fourth of an equilateral Tetragon
+inscribed in the Circle.</p>
+
+<p>Hence, 4 times this Triangle is greater than Segment \(BCK\).</p>
+
+
+<p class="right">[II. Ax. 1.</p>
+
+
+<figure class="figcenter width500" id="i_p016" style="width: 300px;">
+<img src="images/i_p016.jpg" width="300" height="306" alt="A
+quarter-circle arc from K to B, with multiple lines radiating from
+corner A to labelled points K, J, H, G, F, E, D, C, B along the arc
+and sides, illustrating how angles or chord lengths vary progressively
+around a quadrant.">
+</figure>
+
+<p>Because angle \(BCD\) is greater than angle \(ACD\), and that angle
+\(BDC\) is less than angle \(ADC\);</p>
+
+<p>\(\therefore\) angle \(BCD\) is greater than angle \(BDC\);</p>
+
+<p>\(\therefore\) \(BD\) is greater than \(BC\).</p>
+
+
+<p class="right">[Euc. I. 19.</p>
+
+
+<p>Similarly, \(BF\) is greater than \(BC\).</p>
+
+<p>Hence, on our Hypothesis, \(BD\) and \(BF\) are both of them greater
+than \(AB\).</p>
+
+<p>Also, \(because\) \(BF\) is greater than \(AB\);</p>
+
+<p>\(\therefore\) Triangle \(BFK\) is greater than Triangle \(ABK\);</p>
+
+
+<p class="right">[I. Prop. 2.</p>
+
+
+<p>to each of these add Triangle \(ABK\);</p>
+
+<p>\(\therefore\) Figure \(ABFK\) is greater than twice Triangle \(ABK\).</p>
+
+<p>Again, \(\because\) \(BD\) is greater than \(AB\);</p>
+
+<p>\(\therefore\) Triangle \(BDF\) is greater than Triangle \(ABF\);</p>
+
+
+<p class="right">[I. Prop. 2.</p>
+
+
+<p>\(\therefore\) Triangles \(BDF\), \(FHK\) are together greater than
+Figure \(ABFK\);</p>
+
+<p>to each of these add Figure \(ABFK\);</p>
+
+
+<p>\(\therefore\) Figure \(ABDFHK\) is greater than twice Figure \(ABFK\),
+i.e. greater than 4 times Triangle \(ABK\).</p>
+
+<p>Again, \(\because\) \(BC\) is not-less than \(AB\);</p>
+
+<p>\(\therefore\) Triangle \(BCD\) is not-less than Triangle \(ABD\);</p>
+
+
+<p class="right">[I. Prop. 2.</p>
+
+
+<p>\(\therefore\) Triangles \(BCD\), \(DEF\), \(FGH\), \(HJK\) are
+together not-less than Figure \(ABDFHK\); i.e. they are together
+greater than 4 times Triangle \(ABK\);</p>
+
+<p>\(\therefore\), <i>a fortiori</i>, Segment \(BCK\) is greater than 4
+times Triangle \(ABK\).</p>
+
+<p>But this is absurd, since it has been already proved less than 4 times
+this Triangle.</p>
+
+<p>Hence our Hypothesis, that \(BC\) is not-less than \(AB\), is false;
+i.e. \(BC\) is less than \(AB\).</p>
+
+<p>Therefore an isosceles Triangle &amp;c.</p>
+
+
+<p class="right">Q.E.D.</p>
+
+
+
+<p class="nindc space-above2"><span class="allsmcap">Corollary.</span></p>
+<p><span class="pagenum" id="Page_17">[Pg 17]</span></p>
+
+
+<p><i>Hence, by Book I, Prop. III, it is possible to describe, on a given
+base, an isosceles Triangle having each base-angle equal to one-eighth
+of a right angle.</i></p>
+
+
+<h3>PROP. II. <span class="allsmcap">Theorem.</span></h3>
+<p><span class="pagenum" id="Page_18">[Pg 18]</span></p>
+
+<p><i>The angles of any Triangle are together not-less than one-eighth of
+a right angle.</i></p>
+
+<p>Let one-eighth of a right angle be represented by '\(\theta\).'</p>
+
+<figure class="figleft width500" id="i_p018" style="width: 200px;">
+<img src="images/i_p018.jpg" width="200" height="224" alt="Two nested
+triangles sharing base AB, outer triangle ADB with apex D, inner
+triangle ACB with apex C just below, with apexes D and C positioned
+nearer together.">
+</figure>
+
+<p>Let \(ABC\) be a Triangle; it shall be proved that its 'amount' is
+not-less than \(\theta\).</p>
+
+<p>If we deny this, we must assert that its 'amount' is less than
+\(\theta\).</p>
+
+<p>Let this be our Hypothesis.</p>
+
+<p>Hence each of its angles is less than \(\theta\).</p>
+
+<p>On \(AB\) describe an isosceles Triangle \(ABD\) having each base-angle
+equal to \(\theta\);</p>
+
+<p class="right">[II. Prop. 1. Cor.</p>
+
+<p>hence \(AC\), \(BC\), must lie within this Triangle;</p>
+
+<p>i.e. Triangle \(ABC\) must lie within it;</p>
+
+<p>\(\therefore\) angle \(ADB\) is less than angle \(ACB\);</p>
+
+
+<p class="right">[Euc. I. 21.</p>
+
+
+<p>i.e. less than \(\theta\);</p>
+
+<p>but angle \(ADB\) is not-less than angle \(DAB\);</p>
+
+
+<p class="right">[I. Prop. 3. Cor.</p>
+
+
+<p>i.e. not-less than \(\theta\); which is absurd.</p>
+
+<p>Hence our Hypothesis, that 'amount' of Triangle \(ABC\) is less than
+\(\theta\), is false; i.e. it is not less than \(\theta\).</p>
+
+<p>Therefore the angles of any Triangle &amp;c.</p>
+
+
+<p class="right">Q.E.D.</p>
+
+
+<p class="nindc space-above2"><span class="allsmcap">Corollary.</span></p>
+
+
+<p><i>The 'possible region' does not extend below one-eighth of a right
+angle.</i></p>
+
+
+<h3>PROP. III. <span class="allsmcap">Theorem.</span></h3>
+<p><span class="pagenum" id="Page_19">[Pg 19]</span></p>
+
+<p><i>There is a Triangle whose angles are together not-less than two
+right angles.</i></p>
+
+<p>If we deny this, we must assert that any 'amount' is less than \(2R\).</p>
+
+<p>Let this be our First Hypothesis.</p>
+
+<p>It shall be proved that, if we assert this, we must also assert
+that any 'amount' is not-less-than \(2\theta\), where '\(\theta\)'
+represents one-eighth of a right angle.</p>
+
+<p>If we deny this, we must assert that there is an 'amount' less than
+\(2\theta\).</p>
+
+<p>Let this be our Second Hypothesis.</p>
+
+<p>Now we know that the 'possible region' does not extend below \(\theta\);</p>
+
+
+<p class="right">[II. Prop. 2. Cor.</p>
+
+
+<p>i.e. it has an 'inferior limit.'</p>
+
+<p>Let a 'possible amount' be selected, more than half-way from
+\(2\theta\) to this 'inferior limit,' and call it '\((2\theta - a)\).'
+Then it is evident that \((2\theta - 2a)\) will lie <i>below</i> the
+'inferior limit,' and will therefore be an 'impossible amount.'</p>
+
+<p>Let a Triangle be taken, whose 'amount' is \((2\theta - a)\);</p>
+
+<p>\(\therefore\) any one of its angles, which is not-greater than either
+of the others, is not-greater than \(\dfrac{2\theta - a}{3}\); i.e. is
+less than \(\theta\).</p>
+
+<p>Call this angle '\(A\).'</p>
+
+<p>Now one, at least, of the remaining angles must be acute.</p>
+
+
+<p class="right">[Euc. I. 17.</p>
+
+
+<p>Call this '\(B\)'; and call the third angle \(C\).</p>
+
+<p><span class="pagenum" id="Page_20">[Pg 20]</span></p>
+
+<p>Let 2 such Triangles, \(ABC\) and \(A′B′C′\), be taken; and let them be
+so placed that their \(B\)-vertices coincide and their \(BA\)-sides lie
+in one straight line; and join \(CC′\).</p>
+
+<figure class="figcenter width500" id="i_p020" style="width: 300px;">
+<img src="images/i_p020.jpg" width="300" height="172" alt="Symmetric
+diagram with baseline AA', apex D above, and central point B on the
+base. Inner points C and C' form a small diamond shape, with multiple
+lines radiating from D and B, illustrating a proposition about
+symmetric or mirror triangles.">
+</figure>
+
+<p>On \(AA′\) describe an isosceles Triangle \(AA′D\), having each
+base-angle equal to \(\theta\).</p>
+
+
+<p class="right">[II. Prop. 1. Cor.</p>
+
+
+<p>Now each of the angles \(BAC\), \(BA′C′\), is less than \(\theta\).</p>
+
+<p>\(\therefore\) Lines \(AC\), \(A′C′\), will fall within angles \(BAD\),
+\(BA′D\);</p>
+
+<p>i.e. Figure \(AA′C′C\) will fall within Triangle \(AA′D\).</p>
+
+<p>Join \(DC\), <i>DC′</i>.</p>
+
+<p>Now 'amounts' of Triangles \(ABC\), \(A′BC′\), together = \(2(2\theta -
+a)\), and those of the other 4 Triangles are, by our First Hypothesis,
+together less than \(8R\);</p>
+
+<p>\(\therefore\) 'amounts' of all 6 Triangles are together less than
+\((8R + 4\theta - 2a)\);</p>
+
+<p>but these make up 'amount' of Triangle \(AA′D\), plus angles at \(B\),
+\(C\), \(C′\), which together = \(10 R\);</p>
+
+<p>\(\therefore\) 'amount' of Triangle \(AA′D\), plus \(10 R\), is less
+than \((8R + 4\theta - 2a)\).</p>
+
+<p>Now we know that \(2\theta\) is not-greater than \(2R\);</p>
+
+<p>\(\therefore\), adding these inequalities, 'amount' of Triangle
+\(AA′D\), plus \((10R + 2\theta)\), is less than \((10R + 4\theta-2a)\);
+</p>
+
+<p>\(\therefore\) this 'amount,' alone, is less than \((2\theta - 2a)\);
+which is absurd, since the latter lies below the 'inferior limit,' and
+is therefore an 'impossible amount';</p>
+
+
+<p>\(\therefore\) one of our two Hypotheses must be false;</p>
+
+<p>i.e. either there is an 'amount' not-less than \(2R\), or else any
+'amount' is not-less than \(2\theta\).</p>
+
+<p>Suppose we maintain our First Hypothesis: then we must abandon
+our Second; i.e. we must admit that any 'amount' is not-less than
+\(2\theta\).</p>
+
+<p>It shall be proved that, in this case, we must also admit that any
+'amount' is not-less than \(4\theta\).</p>
+
+<p>For, if we deny this, we must assert that there is an 'amount' less
+than \(4\theta\).</p>
+
+<p>Let this be our Third Hypothesis.</p>
+
+<p>Then, in the above proof, \(\theta\) and \(2\theta\) may be replaced by
+\(2\theta\) and \(4\theta\), and a similar absurdity will follow.</p>
+
+<p>Hence either our First or our Third Hypothesis must be false;</p>
+
+<p>i.e. either there is an 'amount' not-less than \(2R\), or else any
+amount is not-less than \(4\theta\).</p>
+
+<p>A similar proof will hold for \(8\theta\); and then for \(16\theta\).</p>
+
+<p>Hence, either there is an 'amount' not-less than \(2R\), or else any
+amount is not-less than \(16\theta\).</p>
+
+<p>But \(16\theta = 2R\).</p>
+
+<p>Hence the second clause of this alternative contains the first.</p>
+
+<p>Hence the first clause must be true.</p>
+
+<p>That is, there is a Triangle &amp;c.</p>
+
+
+<p class="right">Q.E.D.</p>
+
+
+<p class="nindc space-above2"><span class="allsmcap">Corollary.</span></p>
+<p><span class="pagenum" id="Page_21">[Pg 21]</span></p>
+
+<p><i>The 'possible region' does not lie wholly below two right angles.</i></p>
+
+
+<h3>PROP. IV. <span class="smcap">Theorem.</span></h3>
+<p><span class="pagenum" id="Page_22">[Pg 22]</span></p>
+
+<p><i>There is a Triangle whose angles are together equal to two right
+angles.</i></p>
+
+<p>For the 'possible region' does not lie wholly above \(2R\);</p>
+
+
+<p class="right">[I. Prop. 6. Cor.</p>
+
+
+<p>neither does it lie wholly below \(2R\);</p>
+
+
+<p class="right">[II. Prop. 3. Cor.</p>
+
+
+<p>\(\therefore\) it includes \(2R\).</p>
+
+
+<p class="right">[I. 4. Cor. 2.</p>
+
+
+<p>That is, there is a Triangle &amp;c.</p>
+
+
+<p class="right">Q.E.D.</p>
+
+
+
+<h3>PROP. V. <span class="smcap">Theorem.</span></h3>
+
+<p><i>There is a quadrilateral Figure which is 'rectangular,' that is,
+which has all its angles right angles.</i></p>
+
+<figure class="figleft width500" id="i_p022" style="width: 200px;">
+<img src="images/i_p022.jpg" width="200" height="115" alt="Two
+parallel horizontal lines DE and AB, with vertical lines and diagonals
+connecting labelled points D, H, C, L, E above and A, G, K, B below, via
+intermediate points F and J, illustrating a proposition about parallel
+lines and transversals.">
+</figure>
+
+<p>Let \(ABC\) be a Triangle whose 'amount' = \(2R\).</p>
+
+
+<p class="right">[II. Prop. 4.</p>
+
+
+<p>At \(C\) make angle \(ACD\) equal to angle \(CAB\), and angle \(BCE\)
+equal to angle \(CBA\);</p>
+
+<p>hence angles \(ACD\), \(ACB\), \(BCE\), together = \(2R\);</p>
+
+<p>\(\therefore\) \(DC\), \(CE\), are in a straight Line.</p>
+
+
+<p class="right">[Euc. I. 14.</p>
+
+
+<p>Bisect \(AC\) at \(F\); from \(F\) draw \(FG\) perpendicular to \(AB\);
+from \(CD\) cut off \(CH\) equal to \(AG\); and join \(FH\).</p>
+
+<p>\(\because\), in Triangles \(FAG\), \(FCH\), \(FA\), \(AG\), are
+respectively equal to \(FC\), \(CH\), and angle \(A\) to angle
+\(FCH\), [Euc. I. 4.</p>
+
+<p>\(\therefore\) the Triangles are equal in all respects;</p>
+
+<p>\(\therefore\) angle \(FHC\) is a right angle;</p>
+
+<p>and angle \(AFG = \text{angle}\, CFH\);</p>
+
+<p>\(\therefore\) angles \(AFG\), \(AFH\), together = angles \(CFH\),
+\(AFH\); i. e. together = \(2R\);</p>
+
+<p>\(\therefore\) \(GF\), \(FH\) are in a straight line;</p>
+
+
+<p class="right">[Euc. I. 14.</p>
+
+
+<p>\(\therefore\) \(GH\) is a common perpendicular to Lines \(AB\), \(DE\).</p>
+
+<p>Similarly, by bisecting \(BC\) at \(J\), it may be proved that \(KL\)
+is a common perpendicular.</p>
+
+<p>Hence Figure \(HK\) is rectangular.</p>
+
+<p>Therefore there is a quadrilateral Figure &amp;c.</p>
+
+
+<p class="right">Q.E.D.</p>
+
+
+
+<h3><span class="allsmcap">Definition.</span></h3>
+
+<p><span class="pagenum" id="Page_23">[Pg 23]</span></p>
+
+<p>A rectangular quadrilateral Figure may be called a 'Rectangle.'</p>
+
+
+<h3>PROP. VI. <span class="allsmcap">Theorem.</span></h3>
+<p><span class="pagenum" id="Page_24">[Pg 24]</span></p>
+
+
+<p><i>The opposite sides of a Rectangle are equal.</i></p>
+
+<figure class="figleft width500" id="i_p024" style="width: 200px;">
+<img src="images/i_p024.jpg" width="200" height="189" alt="A rectangle
+ABCD with two crossing diagonals inside, plus point C' above D and
+D' on the right side, illustrating a proposition about diagonals,
+rectangles, and points displaced from the corners.">
+</figure>
+
+<p>Let \(ABCD\) be a Rectangle; and let it be reversed so that \(A\),
+\(B\), may change places.</p>
+
+<p>Then \(AD\) will lie along \(BC\), and \(BC\) along \(AD\).</p>
+
+<p>Now, if \(AD\) were not equal to \(BC\), \(D\), \(C\), would not change
+places, but would take new positions, as \(D′C′\);</p>
+
+<p>hence exterior angle \(ADC\) would be greater than interior opposite
+angle \(AC′D′\);</p>
+
+
+<p class="right">[Euc. I. 18.</p>
+
+
+<p>but they are also equal, being right angles; which is absurd;</p>
+
+<p>\(\therefore\) \(AD = BC\).</p>
+
+<p>Similarly it may be proved that \(AB = DC\).</p>
+
+<p>Therefore the opposite sides &amp;c.</p>
+
+
+<p class="right">Q.E.D.</p>
+
+
+
+<h3>PROP. VII. <span class="smcap">Theorem.</span></h3>
+
+<p><i>There is a Pair of Lines, each of which is 'equidistant' from the
+other, that is, is such that all Points on it are equally distant from
+the other Line.</i></p>
+
+<p>Let \(ABCD\) be a Rectangle; and let a vertical Line be supposed, first
+to coincide with \(AD\), and then to move along \(AB\), continuing
+always at right angles to it, till it reaches some intermediate
+position \(A′D′\).</p>
+
+<figure class="figcenter width500" id="i_p025" style="width: 300px;">
+<img src="images/i_p025.jpg" width="300" height="200" alt="Rectangle
+ABCD with interior point D' connected by lines to corners D and C
+above, and to point A' on the base, illustrating a proposition about an
+internal point and its distances to the sides or corners.">
+</figure>
+
+<p>Now, if its top be not now on \(DC\), it must have either dropped below
+it or risen above it.</p>
+
+<p>First, let it be supposed to have dropped below it: and join \(DD′\),
+\(D′C\).</p>
+
+<p>Then, if the Figure \(AA′D′D\) be reversed, and applied to the same
+base, it is evident that \(D\) and \(D′\) will exchange places;</p>
+
+<p>\(\therefore\) angle \(A′D′D = \text{angle}\, ADD′\), i.e. it is less
+than \(R\);</p>
+
+<p>Similarly angle \(A′D′C\) is less than \(R\);</p>
+
+<p>but angle \(DD′C\) is less than \(2R\).</p>
+
+<p>\(\therefore\) angles at \(D′\) are together less than \(4R\); which is
+absurd;</p>
+
+
+<p class="right">[Euc. I. 13. Cor.</p>
+
+
+<p>\(\therefore\) top of vertical Line has not dropped below \(DC\).</p>
+
+<p>Similarly it may be proved that it has not risen above \(DC\).</p>
+
+<p>Hence it moves along \(DC\); i. e. it describes a straight Line, and
+will evidently continue to do so, however far the vertical Line move,
+either way, along \(AB\).</p>
+
+<p>Therefore there is a Pair of Lines &amp;c.</p>
+
+
+<p class="right">Q.E.D.</p>
+
+
+<p class="nindc space-above2"><span class="allsmcap">Corollaries.</span></p>
+<p><span class="pagenum" id="Page_25">[Pg 25]</span></p>
+
+<p class="nindc">1.</p>
+
+<p><i>If a Pair of Lines have a common perpendicular: each of them is
+equidistant from the other.</i></p>
+
+
+<p class="nindc space-above2"><span class="allsmcap">Corollary 2.</span></p>
+<p><span class="pagenum" id="Page_26">[Pg 26]</span></p>
+
+<p><i>It is possible to form a Rectangle of any given width and height.</i></p>
+
+<p>For, in the above Pair of horizontal equidistant Lines, 2 common
+perpendiculars may be drawn, at a given width apart; and the Figure, so
+formed, will be a Rectangle; and its sides will be a Pair of vertical
+equidistant Lines, which may be treated in the same way.</p>
+
+
+<h3>PROP. VIII. <span class="allsmcap">Theorem.</span></h3>
+
+<p><i>The angles of any Triangle are together equal to two right
+angles.</i></p>
+
+<figure class="figcenter width500" id="i_p026" style="width: 300px;">
+<img src="images/i_p026.jpg" width="300" height="211" alt="Rectangle
+with points F, C, E along the top and A, D, B along the base. Point
+C connects by lines to A, D, and B, forming triangles within,
+illustrating a proposition about a point on the top side and its
+distances to the base.">
+</figure>
+
+<p>Let \(ABC\) be a Triangle, so placed that each base-angle is acute;
+from \(C\) draw \(CD\) perpendicular to \(AB\); and make Rectangles
+\(ADCF\), \(BDCE\).</p>
+
+
+<p class="right">[II. Prop. 7. Cor. 2.</p>
+
+
+<p>\(\because\) \(FD\) has its opposite sides equal,</p>
+
+
+<p class="right">[II. Prop. 6.</p>
+
+
+<p>\(\therefore\), in Triangles \(ADC\), \(CFA\), the sides of the one are
+respectively equal to the sides of the other;</p>
+
+<p>\(\therefore \text{angle}\, CAD = \text{angle}\, ACF\);</p>
+
+
+<p class="right">[Euc. I. 8.</p>
+
+
+<p>Similarly angle \(CBD = \text{angle}\, BCE\);</p>
+
+<p>\(\therefore\) the angles of the Triangle \(ABC\) together = the angles
+\(FCA\), \(ACB\), \(BCE\); i. e. together = \(2R\).</p>
+
+<p>Therefore the angles of any Triangle &amp;c.</p>
+
+
+<p class="right">Q.E.D.</p>
+
+
+<h3>PROP. IX. <span class="smcap">Theorem.</span></h3>
+<p><span class="pagenum" id="Page_27">[Pg 27]</span></p>
+
+<p><i>A Pair of Lines, which are equally inclined to a certain
+transversal, are so to any transversal.</i></p>
+
+<figure class="figcenter width500" id="i_p027" style="width: 300px;">
+<img src="images/i_p027.jpg" width="300" height="103" alt="Two
+parallel horizontal lines AB and CD, with a small triangle formed
+between them by points E, G above and F, H below, illustrating a
+proposition about transversals or angles between parallel lines.">
+</figure>
+
+<p>Let \(AB\), \(CD\), be equally inclined to transversal \(EF\); and let
+\(GH\) be any other transversal.</p>
+
+<p>Join \(EH\).</p>
+
+<p>Now 'amounts' of Triangles \(EFH\), \(EGH\), together = \(4R\).</p>
+
+
+<p class="right">[II. Prop. 8.</p>
+
+
+<p>But these make up angles of Figure \(FG\);</p>
+
+<p>\(\therefore\) angles of Figure \(FG\) together = \(4R\);</p>
+
+<p>but angles \(GEF\), \(EFH\) together = \(2R\);</p>
+
+
+<p class="right">[I. Prop. 1.</p>
+
+
+<p>\(\therefore\) angles \(EGH\), \(GHF\), together = \(2R\);</p>
+
+<p>\(\therefore\) \(AB\), \(CD\), are equally inclined to \(GH\).</p>
+
+
+<p class="right">[I. Prop. 1.</p>
+
+
+<p>Therefore a Pair of Lines &amp;c.</p>
+
+
+<p class="right">Q.E.D.</p>
+
+
+<h3><span class="allsmcap">Axioms</span> (<i>continued</i>).</h3>
+
+<p class="nindc">2.</p>
+
+<p>If two homogeneous magnitudes be both of them finite: the lesser may be
+so multiplied, by a finite number, as to exceed the greater.</p>
+
+
+
+<h3>PROP. X. <span class="smcap">Theorem.</span></h3>
+<p><span class="pagenum" id="Page_28">[Pg 28]</span></p>
+
+<p><i>If a Pair of Lines make, with a certain transversal, two interior
+angles, on the same side of it, which are together less than two right
+angles, the defect being a finite angle: these Lines are intersectional
+on that side of the transversal.</i></p>
+
+<figure class="figcenter width500" id="i_p028" style="width: 300px;">
+<img src="images/i_p028.jpg" width="300" height="287" alt="A diagram
+showing multiple lines radiating from central point E toward labelled
+points Z, K, V, Y, T, S, R, N, H, M, B, D, F, L, C, G, A,a dense fan of
+rays and intersecting lines illustrating a proposition about angles,
+parallels, or convergence.">
+</figure>
+
+<p>Let \(AB\), \(CD\) make with \(EF\) the interior angles \(BEF\),
+\(EFD\) together less than \(2R\).</p>
+
+<p>Make angle \(FEG\) equal to angle \(EFD\); produce \(GE\) to \(H\);
+from \(E\) draw \(EK\) at right angles to \(AB\), and \(EL\) at right
+angles to \(CD\).</p>
+
+<p>Hence \(EL\) is also at right angles to \(GH\);</p>
+
+
+<p class="right">[II. Prop. 9.</p>
+
+
+<p>i. e. \(CD\), \(GH\) have a common perpendicular;</p>
+
+<p>\(\therefore\) each is equidistant from the other;</p>
+
+
+<p class="right">[II. Prop. 7, Cor. 1.</p>
+
+
+<p>also \(EL\) = the common distance between them.</p>
+
+<p><span class="pagenum" id="Page_29">[Pg 29]</span></p>
+
+<p>Now angle \(BEH\) is the defect, from \(2R\), of the sum of the 2
+interior angles \(BEF\), \(EFD\);</p>
+
+<p>hence, by hypothesis, it is finite;</p>
+
+<p>\(\therefore\) it may be so multiplied, by a finite number, as to
+exceed angle \(BEK\).</p>
+
+
+<p class="right">[II. Ax. 2.</p>
+
+
+<p>Call this finite number '\(n\).'</p>
+
+<p>In \(EB\), produced if necessary, take \(EM\) \(n\)-times \(EL\); from
+\(M\) draw \(MN\) at right angles to \(EH\); turn Triangle \(ENM\)
+about \(EN\) into position \(ENR\); then, about \(ER\), into position
+\(ERS\), and so on, till there are \(n\) such Triangles altogether; and
+let its final position be \(EYZ\).</p>
+
+<p>Then angle \(MEZ\) is \(n\)-times angle \(MEH\); i. e. it is greater
+than angle \(MEK\).</p>
+
+<p>Let \(EK\) cut broken-Line \(MRTZ\) at \(V\); and join \(MV\).</p>
+
+<p>Then \(MRTZ\) is greater than \(MRTV\), which is greater than \(MV\),
+which is greater than \(EM\);</p>
+
+
+<p class="right">[Euc. I. 20, 17, 19.</p>
+
+
+<p>\(\therefore\) \(MRTZ\) is greater than \(EM\);</p>
+
+<p>but \(MN\) is the same fraction of \(MRTZ\) which \(EL\) is of \(EM\);</p>
+
+<p>\(\therefore\) \(MN\) is greater than \(EL\); i. e. the distance of
+\(M\), from \(GH\), is greater than the common distance between \(CD\)
+and \(GH\).</p>
+
+<p>\(\therefore\) \(AB\), \(CD\) are intersectional towards \(B\), \(D\).</p>
+
+<p>Therefore, if a Pair of Lines &c.</p>
+
+
+<p class="right">Q.E.D.</p>
+
+
+<hr class="chap x-ebookmaker-drop">
+
+<div class="chapter">
+<p><span class="pagenum" id="Page_31">[Pg 31]</span></p>
+<h2 class="nobreak" id="APPENDICES">APPENDICES<br>
+<span class="allsmcap">TO</span>
+<br>
+PART I.</h2>
+</div>
+
+
+<hr class="chap x-ebookmaker-drop">
+
+<div class="chapter">
+<p><span class="pagenum" id="Page_32">[Pg 32]</span></p>
+
+<h2 class="nobreak" id="APPENDIX_I">APPENDIX I.</h2>
+</div>
+
+<p class="nindc"><i>Containing an alternative Axiom, which may be substituted for Axiom
+1 at p. 14.</i></p>
+
+
+<h3><span class="allsmcap">Definition.</span></h3>
+
+
+<p>The Segment, cut off, from any Sector of a Circle, by its Chord, may be
+called its '<b>outer Segment</b>'; and the Triangle, contained by its
+Chord and its two Radii, may be called its '<b>central Triangle</b>.'
+And, if its Arc be bisected and the point of bisection joined to the
+ends of its Chord, the isosceles Triangle, so formed, may be called its
+'<b>inscribed isosceles Triangle</b>.'</p>
+
+
+<h3>PROPOSITIONS.<br>
+<br>
+PROP. A. <span class="allsmcap">Theorem.</span></h3>
+<p><span class="pagenum" id="Page_33">[Pg 33]</span></p>
+
+<p><i>If, in any Sector of a Circle, its Chord be not-less than its
+Radius: then, in a Sector whose vertical angle is twice as great, its
+outer Segment is greater than its central Triangle.</i></p>
+
+<figure class="figcenter width500" id="i_p033" style="width: 300px;">
+<img src="images/i_p033.jpg" width="300" height="264"
+alt="Quadrilateral CABD with crossing diagonals, plus two arcs between
+C and D and between D and B, illustrating a proposition about a
+quadrilateral, its diagonals, and the circular arcs connecting its
+vertices.">
+</figure>
+
+<p>Let \(ABD\) be a Sector whose Chord \(BD\) is not-less than its Radius
+\(AB\). Make angle \(DAC\) equal to angle \(BAD\); and join \(BC\),
+\(CD\).</p>
+
+<p>Then vertical angle of Sector \(ABDC\) is twice as great as that of
+Sector \(ABD\).</p>
+
+<p>It shall be proved that its outer Segment \(BDC\) is greater than its
+central Triangle \(ABC\).</p>
+
+<p>Because \(BD\) is not-less than \(AB\);</p>
+
+<p>\(\therefore\) Triangle \(BDC\) is not-less than Triangle \(ABC\);</p>
+
+
+<p class="right">[I. Prop. 2.</p>
+
+
+<p>\(\therefore\), <i>a fortiori</i>, Segment \(BDC\) is greater than
+Triangle \(ABC\).</p>
+
+<p>Hence if, in any Sector &amp;c.</p>
+
+
+<p class="right">Q.E.D.</p>
+
+
+<h3>PROP. B. <span class="allsmcap">Theorem.</span></h3>
+<p><span class="pagenum" id="Page_34">[Pg 34]</span></p>
+
+<p><i>If, in any Sector of a Circle, each of the equal Sides of its
+inscribed isosceles Triangle be not-less than its Radius; and if
+its outer Segment be greater than a certain multiple of its central
+Triangle: then, in a Sector, whose vertical angle is twice as great,
+its outer Segment is greater than twice that multiple of its central
+Triangle.</i></p>
+
+<figure class="figcenter width500" id="i_p034" style="width: 300px;">
+<img src="images/i_p034.jpg" width="300" height="272" alt="An
+elaboration of the previous diagram, adding points F on the top arc
+and E on the right arc, with additional crossing lines from A and
+C, illustrating a more complex proposition about arcs, chords, and
+intersecting diagonals.">
+</figure>
+
+<p>Let \(ABED\) be a Sector such that each of the equal sides of its
+inscribed isosceles Triangle \(BED\) is not-less than its Radius
+\(AB\), and such that its outer Segment \(BED\) is greater than \(m\)
+times its central Triangle \(ABD\). Make angle \(DAC\) equal to angle
+\(BAD\); bisect angles \(BAD\), \(DAC\) by Lines \(AE\), \(AF\); and
+join \(BC\), \(CD\), \(CF\), \(FD\).</p>
+
+<p>Then vertical angle of Sector \(ABDC\) is twice as great as that of
+Sector \(ABED\).</p>
+
+<p>It shall be proved that its outer Segment \(BDC\) is greater than
+\(2m\) times its central Triangle \(ABC\).</p>
+
+<p>Because angle \(BED\) is greater than angle \(AED\), and that angle
+\(BDE\) is less than angle \(ADE\);</p>
+
+<p>\(\therefore\) angle \(BED\) is greater than angle \(BDE\);</p>
+
+<p>\(\therefore\) \(BD\) is greater than \(BE\);</p>
+
+
+<p class="right">[Euc. I. 25.</p>
+
+
+<p>but \(BE\) is not-less than \(AB\);</p>
+
+<p>\(\therefore\) \(BD\) is greater than \(AB\);</p>
+
+<p>\(\therefore\) Triangle \(BDC\) is greater than Triangle \(ABC\);</p>
+
+
+<p class="right">[I. Prop. 2.</p>
+
+
+<p>to each of these add Triangle \(ABC\);</p>
+
+<p>\(\therefore\) Figure \(ABDC\) is greater than twice Triangle \(ABC\).</p>
+
+<p>Again, \(\because\) Segment \(BED\) is given to be greater than \(m\)
+times Triangle \(ABD\);</p>
+
+<p>\(\therefore\) Segments \(BED\), \(DFC\) are together greater than
+\(m\) times Figure \(ABDC\); i. e. are together greater than \(2m\)
+times Triangle \(ABC\);</p>
+
+<p>\(\therefore\), <i>a fortiori</i>, Segment \(BDC\) is greater than
+\(2m\) times Triangle \(ABC\).</p>
+
+<p>Hence if, in any Sector &amp;c.</p>
+
+
+<p class="right">Q.E.D.</p>
+
+
+<h3><span class="allsmcap">Axiom.</span></h3>
+<p><span class="pagenum" id="Page_35">[Pg 35]</span></p>
+
+<figure class="figright width500" id="i_p035" style="width: 200px;">
+<img src="images/i_p035.jpg" width="200" height="209" alt="The
+recurring core diagram — inscribed square within a circle divided by
+two perpendicular diameters into eight triangles, appearing again to
+support the book's central theorem about the inscribed equilateral
+tetragon.">
+</figure>
+
+<div class="blockquot">
+
+<p>[An alternative Axiom, to be substituted for Axiom 1, at p. 14, if
+the Reader feel any difficulty in granting that Axiom. In this case,
+certain portions of the foregoing Propositions will also need to be
+replaced by new matter, which is hereto appended.]</p>
+</div>
+
+<p><span class="antiqua">In every Circle, the inscribed equilateral tetragon, multiplied by
+\(2^{a}\) ('\(a\)' being a certain selected finite number), is greater
+than any one of the Segments which lie outside it.</span></p>
+
+<hr class="chap">
+
+<div class="blockquot">
+
+<p><span class="allsmcap">Note.</span>—The reader can assign to '\(a\)' any finite value which
+he finds large enough to induce him to accept this Axiom. For example,
+if he be willing to grant that 1024 times the Tetragon is greater than
+the Segment, he can assign to it the value '10.'</p>
+</div>
+
+
+<h3>PROP. C. <span class="smcap">Theorem.</span></h3>
+<p><span class="pagenum" id="Page_36">[Pg 36]</span></p>
+
+<p class="nindc">[To be substituted for Prop. I, at p. 15.]</p>
+
+<p><i>An isosceles Triangle, whose vertical angle is \(\dfrac{1}{2^{a +3}}\) 
+of a right angle, has its base less than either of its sides.</i></p>
+
+<figure class="figcenter width500" id="i_p036" style="width: 300px;">
+<img src="images/i_p036.jpg" width="300" height="327" alt="A square
+ADBC with a quarter-circle arc from D to B, point C on the arc, and
+diagonal DB crossing the figure, illustrating a proposition about a
+square, its inscribed quarter-circle, and a point on the arc.">
+</figure>
+
+<p>Let \(\dfrac{1}{2^{a + 3}}\) of a right angle be represented by
+'\(\phi\).'</p>
+
+<p>Let \(ABC\) be an isosceles Triangle, whose vertical angle at \(A\) is
+\(\phi\).</p>
+
+<p>It shall be proved that \(BC\) is less than \(AB\).</p>
+
+<p>Draw \(AD\) at right angles to \(AB\), and, with centre \(A\), and
+distance \(AB\), describe Quadrant. (See Note.) And join \(BD\).</p>
+
+<p>Then Triangle \(ABD\) is one-fourth of an equilateral Tetragon
+inscribed in the Circle.</p>
+
+<p>Hence \(2^{a + 2}\) times this Triangle is greater than Segment \(BCD\).</p>
+
+
+<p class="right">[Alternative Axiom.</p>
+
+
+<p>Now, if we deny that \(BC\) is less than \(AB\), we must assert that
+\(BC\) is not-less than \(AB\).</p>
+
+<p>Let this be our Hypothesis.</p>
+
+<hr class="chap">
+
+<div class="blockquot">
+
+<p><span class="allsmcap">Note.</span>—The Reader is requested to imagine a chord drawn to the
+arc \(BC\).</p>
+</div>
+
+<p><span class="pagenum" id="Page_37">[Pg 37]</span></p>
+
+<p>Hence it may be proved, as in Book II, Prop. I, that any Chord, drawn
+from \(B\) to any Point on the Arc \(CD\), is greater than \(BC\), and
+therefore not-less than \(AB\).</p>
+
+<p>Now, on our Hypothesis, \(ABC\) is a Sector whose Chord is not-less
+than its Radius;</p>
+
+<p>\(\therefore\), in a Sector whose vertical angle is \(2\phi\), its
+outer Segment is greater than its central Triangle;</p>
+
+
+<p class="right">[Prop. A.</p>
+
+
+<p>i.e., in a Sector whose vertical angle is \(2\phi\), each of the equal
+sides of its inscribed isosceles Triangle is not-less than its Radius,
+and its outer Segment is greater than once its central Triangle;</p>
+
+<p>\(\therefore\), in a Sector, whose vertical angle is \(4\phi\), its
+outer Segment is greater than twice its central Triangle;</p>
+
+
+<p class="right">[Prop. B.</p>
+
+
+<p>\(\therefore\), similarly, in a Sector, whose vertical angle is
+\(8\phi\), its outer Segment is greater than 4 times its central
+Triangle;</p>
+
+<p>and so on;</p>
+
+<p>\(\therefore\), ultimately, in a Sector, whose vertical angle is \(2^{{}a
++ 3}\phi\), its outer Segment is greater than \(2^{a + 2}\) times its
+central Triangle;</p>
+
+<p>but \(2^{a + 3}\phi = R\);</p>
+
+<p>\(\therefore\), in Sector \(ABCD\), Segment \(BCD\) is greater than
+\(2^{a + 2}\) times Triangle \(ABD\).</p>
+
+<p>But this is absurd, since it has been already proved less than \(2^{{}a +
+2}\) times this Triangle.</p>
+
+<p>Hence our Hypothesis, that \(BC\) is not-less than \(AB\), is false; i.
+e. \(BC\) is less than \(AB\).</p>
+
+<p>Therefore an isosceles Triangle &amp;c.</p>
+
+
+<p class="right">Q.E.D.</p>
+
+
+<p class="nindc"><span class="allsmcap">Corollary to Prop. C.</span></p>
+<p><span class="pagenum" id="Page_38">[Pg 38]</span></p>
+
+<p><i>Hence, by Book I, Prop. III, it is possible to describe, on a
+given base, an isosceles Triangle having each base-angle equal to
+\(\dfrac{1}{2^{a + 3}}\) of a right angle.</i></p>
+
+<div class="blockquot">
+
+<p>[N.B. If the Reader be willing to grant, as axiomatic, that 1024
+times the Tetragon is greater than the Segment, he must now admit, as
+logically proved, that an isosceles Triangle, whose vertical angle is
+\(\dfrac{1}{8192}\) of a right angle, has its base less than either of
+its sides. If such a Triangle were actually drawn, having each side 140
+yards long, its base would be found to be less than an inch!]</p>
+</div>
+
+
+<h3>PROP. D. <span class="smcap">Theorem.</span></h3>
+
+<div class="blockquot">
+
+<p>[To be substituted for Prop. II, at p. 18.]</p>
+</div>
+
+<p><i>The angles of any Triangle are together not-less than
+\(\dfrac{1}{2^{a + 3}}\) of a right angle.</i></p>
+
+<div class="blockquot">
+
+<p>[The proof of Prop. II will serve here, without any change, except the
+substitution of '\(\phi\)' for '\(\theta\)'.]</p>
+</div>
+
+<div class="blockquot">
+
+<p>[N.B. If the Reader be willing to grant, as axiomatic, that 1024
+times the Tetragon is greater than the Segment, he must now admit, as
+logically proved, that the angles of any Triangle are together not-less
+than \(\dfrac{1}{8192}\) of a right angle.]</p>
+</div>
+
+
+<h3>PROP. E. <span class="smcap">Theorem.</span></h3>
+<p><span class="pagenum" id="Page_39">[Pg 39]</span></p>
+
+<div class="blockquot">
+
+<p>[To be substituted for Prop. III, at p. 19.]</p>
+</div>
+
+<p><i>There is a Triangle whose angles are together not-less than two
+right angles.</i></p>
+
+<div class="blockquot">
+
+<p>[The proof of Prop. III will serve here, without any change, except the
+substitution of '\(\phi\)' for '\(\theta\),' and '\(\dfrac{1}{2^{a+3}}\)' 
+for 'one-eighth,' down to the words 'A similar proof &amp;c.' at
+foot of p. 21; for which the following is to be substituted.]</p>
+</div>
+
+<p>A similar proof will hold for \(8\phi\), \(16\phi\), and so on; and
+ultimately for \(2^{a + 4}\phi\).</p>
+
+<p>Hence, either there is an 'amount' not-less than \(2R\), or else every
+'amount' is not-less than \(2^{a + 4}\phi\).</p>
+
+<p>But \(2^{a + 4}\phi = 2R\).</p>
+
+<p>Hence the second clause of this alternative contains the first.</p>
+
+<p>Hence the first clause must be true.</p>
+
+<p>That is, there is a Triangle &amp;c.</p>
+
+
+<p class="right">Q.E.D.</p>
+
+<hr class="chap x-ebookmaker-drop">
+
+<div class="chapter">
+<h2 class="nobreak" id="APPENDIX_II">APPENDIX II.</h2>
+</div>
+
+<p class="nindc"><i>Is Euclid's Axiom true?</i></p>
+
+<hr class="r5">
+
+<p class="nindc space-above2">§ 1.</p>
+<p><span class="pagenum" id="Page_40">[Pg 40]</span></p>
+
+<p class="nindc"><i>Infinite and Finite Magnitudes.</i></p>
+
+<p>The answer I propose to give to this alarming question is that, though
+true for Finite Magnitudes—the sense in which, as I believe, Euclid
+<i>meant it to be taken</i>—it is <i>not</i> universally true.</p>
+
+<p>Will the gentle Reader be so kind as to join me in contemplating,
+for a few minutes, the Infinite Space which surrounds our tiny
+planet? We believe—those of us, at least, who answer <i>fully</i>
+to the ancient definition of Man, '<i>animal rationale</i>'—that
+it <i>is</i> infinite. And that, not because we profess to have
+grasped the conception of Infinity, but because the <i>contrary</i>
+hypothesis <i>contradicts</i> Reason: and what contradicts Reason
+we feel ourselves authorised to deny. <i>Both</i> conceptions—that
+Space has a limit, and that it has none—are <i>beyond</i> our Reason:
+but the former is also <i>against</i> our Reason: for we may fairly
+say "When we have reached the limit, what then? What do we come to?
+There <i>must</i> be either Something, or Nothing. If Something, it
+is <i>full</i> Space, '<i>plenum</i>': if Nothing, it is <i>empty</i>
+Space, '<i>vacuum</i>.' That there should be <i>neither</i> of these is
+a logical impossibility. Such an hypothesis would be—in the<span class="pagenum" id="Page_41">[Pg 41]</span> words of
+Master Constable Dogberry—'most tolerable and not to be endured.'"</p>
+
+<p>I propose to show, by certain considerations which begin with Infinite
+Space, but will speedily condescend to Finite Magnitudes, that it is
+possible for two homogeneous Magnitudes to be so related to each other
+that <i>no</i> multiple of the lesser will exceed the greater. (It
+is of course assumed that a 'multiple' of a Magnitude is the result
+produced by the use of a 'multiplier,' and that a 'multiplier' is a
+<i>nameable</i>—and therefore a <i>finite</i>—number.)</p>
+
+<p>"Yet surely," the gentle Reader will protest, "Euclid has assumed the
+exact <i>contrary</i> of this? Does he not, in Book X, Prop. 1, tacitly
+assume the Axiom that the lesser of two Magnitudes may be so multiplied
+as to exceed the greater?"</p>
+
+<p>Gentle Reader, he <i>does</i>! But <i>my</i> contention is that, in so
+doing, he excludes from his view both Infinities and Infinitesimals,
+and is contemplating <i>Finite Magnitudes only</i>.</p>
+
+<p>For consider Euclid's Definitions of the word 'Ratio' and of the phrase
+'to have a Ratio to.' (Book V. Def. 3, 4.)</p>
+
+<p>(3) λόγος ἐστὶ δύο μεγεθῶν ὁμογενῶν ἡ κατὰ πηλικότητα πρὸς ἄλληλα ποιὰ
+σχέσις.</p>
+
+<p>(4) λόγον ἔχειν πρὸς ἄλληλα μεγέθη λέγεται, ἅ δύναται πολλαπλασιαζόμενα
+ἀλλήλων ὑπερέχειν.</p>
+
+<p>Quite literally, these are:—</p>
+
+<p>(3) "Ratio is a certain relationship, as to size, of two homogeneous
+Magnitudes, each to the other."</p>
+
+<p>(4) "Magnitudes, which can, (on) being multiplied, exceed each the
+other, are said to have a Ratio, each to the other."</p>
+
+<p>But they become more intelligible when less literally translated:—</p>
+
+<p>(3) "Ratio is a certain relationship, as to size, borne, by a
+Magnitude, to another Magnitude homogeneous with it."</p>
+
+<p><span class="pagenum" id="Page_42">[Pg 42]</span></p>
+
+<p>(4) "A Magnitude is said 'to bear a Ratio to' another Magnitude,
+homogeneous with it, when either of them, that is not greater than the
+other, can be made so by multiplication."</p>
+
+<p>Some translators introduce the word 'mutual' into No. (3), and tell
+us that Ratio is 'a <i>mutual</i> relation of two Magnitudes': but
+this seems to me incorrect, as seeming to imply that the Ratio, borne
+by \(A\) to \(B\), is <i>identical</i> with that borne by \(B\) to
+\(A\). But, if \(A\) were 3-4ths of \(B\), \(B\) would <i>not</i> be
+3-4ths, but 4-3ds, of \(A\): and the Ratio '4-3ds,' though of the same
+<i>nature</i> as the Ratio '3-4ths,' is not identical with it.</p>
+
+<p>Now it seems to me clear that Euclid does <i>not</i> mean to imply that
+<i>any</i> two homogeneous Magnitudes bear 'Ratios' to each other:
+for in No. (4) he gives us a test, by which to know in what cases two
+such Magnitudes <i>do</i>, and in what cases they do <i>not</i>, bear
+'Ratios' to each other. This test would be wholly superfluous if it
+were true, of <i>any</i> two homogeneous Magnitudes, that one could be
+said 'to bear a Ratio to' the other.</p>
+
+<p>What cases then, does Euclid mean to <i>exclude</i> by this
+test? My answer is "all cases in which one of the Magnitudes is
+<i>infinitely greater than the other</i>." Take, as an example, these
+two Magnitudes—a Cubic Inch, and Infinite Space. It is <i>not</i>
+possible, by multiplying a Cubic Inch by any finite number, however
+great, to make it exceed, or even equal, Infinite Space: hence
+Euclid's test <i>fails</i> in this case, and Euclid would, no doubt,
+<i>decline</i> to say that either of these Magnitudes, though they are
+strictly <i>homogeneous</i>, bears a 'Ratio' to the other.</p>
+
+<p>My conclusion, then, is that, in Book X. Prop. 1, Euclid is limiting
+his view to the case of two homogeneous Magnitudes <i>which are
+such that neither of them is infinitely greater than the other</i>:
+nay, more—for such a limitation would not exclude the case of two
+Infinities of the same order—that he is contemplating <i>Finite
+Magnitudes only</i>.</p>
+
+
+<p class="nindc space-above2">§ 2.</p>
+<p><span class="pagenum" id="Page_43">[Pg 43]</span></p>
+
+<p class="nindc"><i>Infinitesimal Lines and Strips.</i></p>
+
+<p>We have already seen that, in the case of two homogeneous Magnitudes,
+one Finite and the other Infinite, <i>no</i> multiple of the lesser
+will exceed the greater: and I now propose to show that it is possible
+for the same thing to happen in the case of two homogeneous Magnitudes,
+<i>neither of them being Infinite</i>.</p>
+
+<p>Let us imagine an Infinite Plane, placed upright, and facing us—as
+if we were standing in front of a wall, which extended to infinity,
+upwards, downwards, to the right-hand, and to the left. If we divide
+this Plane by a single horizontal straight Line, extended to infinity
+to the right-hand and to the left, we get <i>half</i> of the whole
+Plane <i>above</i> the Line, and <i>half below</i> it, wherever
+we choose to place the Line. This may be deduced from the logical
+principle that we have no reason for believing <i>either</i> to be
+greater than the other; or we may adopt M. Bertrand's Axiom, "Two
+spaces, whether finite or infinite, are equal, when one can be placed
+upon the other so that any point whatsoever of either coincides with a
+point of the other"—a condition which seems, theoretically, readily
+attainable, by making one portion of the Plane revolve round the
+boundary-line, as a hinge, till it coincides with the other portion.
+Each portion is, of course, an Infinity of the <i>same</i> order as the
+whole Plane.</p>
+
+<p>Now let us imagine <i>two</i> such infinite horizontal straight Lines,
+'separational' from each other (i. e. never intersecting), placed at
+a finite distance apart, and therefore having between them a Strip,
+finite in width, infinite in length, and therefore infinite in area.</p>
+
+<p>Now it clearly is <i>not</i> possible, by multiplying this Infinite
+Strip by any finite number, however great, to make it exceed, or even
+equal, the whole Infinite-Plane. Here again, then, Euclid's test fails,
+and neither of these Magnitudes, though they are <i>homogeneous</i> can
+be said to have a 'Ratio' to the other.<span class="pagenum" id="Page_44">[Pg 44]</span> In fact, the Infinite-Strip is
+an Infinity <i>of a lower order</i> than the Infinite-Plane.</p>
+
+<p>Comparing this Strip with a single square-inch, you will, I fancy,
+be willing to grant at once that <i>no</i> multiple of the latter
+can possibly reach—much less exceed—the former. We have, in fact,
+established the existence of <i>three</i> kinds of Area, viz. Finite
+Areas (e.g. a square inch), Infinities of the first order (e.g. the
+Infinite-Strip we have been considering), and Infinities of the second
+order (e.g. the upper half of the whole Infinite-Plane).</p>
+
+<p>Now let us go a step further. Let the two sides of this Strip be
+supposed to gradually approach each other—still maintaining their
+'separational' character—and let us consider the effect of this
+process on the intervening area.</p>
+
+<p>When the width of the Strip has been reduced to half-an-inch, you
+will grant, I suppose, that the area is exactly half what it was at
+first—since <i>two</i> such Strips, laid side by side, evidently make
+up the original Strip. And so, by reducing the width to a quarter-inch,
+&amp;c., we obtain a number of different areas, all Infinite alike, and
+yet having finite ratios to one another: in fact, so long as the width
+continues to be a <i>finite fraction</i> (i. e. a fraction with a
+finite numerator and denominator) of an inch, the area continues to be
+an Infinity <i>of the same order</i> as the original Strip. (It seems
+obvious that Infinities <i>of the same order</i> have finite ratios to
+one another, and that any one of them can be so multiplied, by a finite
+number, as to exceed any other.)</p>
+
+<p>But this narrowing process may be continued until the two Lines
+absolutely <i>coincide</i>: and what <i>then</i> becomes of the area?
+It cannot be denied that it is then <i>Zero</i>.</p>
+
+<p>Now I have ventured (see p. 2) to lay it down, as an Axiom, that, when
+a Magnitude varies <i>continuously</i>, and has changed from a certain
+value to a certain other value, it must have passed through <i>every
+intermediate value</i>. And what intermediate values do we find between
+an Infinite Area and Zero?<span class="pagenum" id="Page_45">[Pg 45]</span> Surely every <i>Finite</i> Area, that can
+be named, lies between them? The Reader can, if he likes, part company
+with me at this point: but to <i>my</i> Reason it seems absolutely
+clear—first, that the Strip does diminish <i>continuously</i>, and not
+'<i>per saltum</i>'; and secondly, that its area has, at some time or
+other during the process, every conceivable <i>finite</i> value. At one
+time, for instance, it contains a square-mile: and, rather later in its
+career, it is reduced to a single <i>square-inch</i>.</p>
+
+<p>Let us contemplate it in this last-named condition. Its length? As
+Infinite, clearly, as it was at first. Its Area? One square-inch,
+undoubtedly. <i>And what is its width?</i></p>
+
+<p>Can you, oh gentle Reader, find any reasonable answer to this question,
+except the following? "Its width is <i>infinitely small</i>—having, in
+fact, exactly the same relation to a linear inch which that linear inch
+has to an infinite Line."</p>
+
+<p>If this be so (and I see no way out of it), we have found two
+Magnitudes, both linear, neither of them infinite, and yet <i>such that
+no finite multiple of the lesser can possibly exceed the greater</i>.
+They are, in fact, <i>not of the same order</i>; one of them being
+Finite, and the other an Infinitesimal of the first order.</p>
+
+<p>But is not the gentle Reader saying to himself, all this time, "I
+<i>cannot</i> believe in the existence of a Strip, Infinite in length,
+and yet Finite in Area! No doubt, if such a Strip <i>could</i> exist,
+its width must be what you call 'Infinitesimal'; since a Finite width
+must give an Infinite area. But I don't believe in the existence of an
+Infinitesimal width! My belief is that, if you make the edges of the
+Infinite-Strip gradually approach till they coincide, its width will
+continue <i>Finite</i> till the last moment, and will then suddenly
+become Zero; and that its area will continue an <i>Infinity of the
+first order</i> till the last moment, and will then suddenly become
+Zero."</p>
+
+<p><span class="pagenum" id="Page_46">[Pg 46]</span></p>
+
+<p>Very good. My gentle Reader has formulated his views, very clearly and
+definitely. Permit me now to offer to his consideration a Strip, which
+I will <i>prove</i> to be at once Infinite in length and Finite in area.</p>
+
+<figure class="figcenter width500" id="i_p046" style="width: 300px;">
+<img src="images/i_p046.jpg" width="300" height="173" alt="Coordinate
+axes OX and OY with a descending curve from C through F and K to M, and
+stepped rectangular constructions at intervals A, D, G, illustrating
+convergence or limits as successive rectangles diminish toward the
+axis.">
+</figure>
+
+<p>Let \(OX\), \(OY\), be rectangular Axes of Reference; and let us
+trace the Curve \(y = 2^{-x}\), where \(OC = OA = AD = DG\) =
+unit-line. Hence, when \(x = 0\), \(y = 1\); when \(x = 1\), 
+\(y = \tfrac{1}{2}\); when \(x = 2\), \(y = \tfrac{1}{4}\); when \(x = 3\),
+\(y = \tfrac{1}{8}\); and so on. Hence the Curve is \(LCFKMN\). Also,
+however large \(x\) becomes, \(y\) can <i>never</i> be Zero; i. e. the
+Strip, intercepted between the Curve and the \(x\)-Axis, and bounded
+at the left-hand end by \(CO\), is Infinite in length. And now let
+us estimate its area. Its first portion, \(COAF\), is less than the
+rectangle \(OB\); its second portion is less than \(AE\); and so on.
+Hence its area is less than \((OB + AE + \&c. \text{for ever})\); i.
+e., is less than \((1 + \tfrac{1}{2} + \tfrac{1}{4}\) + &c. for
+ever; therefore, <i>a fortiori</i>, it is less than 2.</p>
+
+<p>Now an area, which is less than '2' is surely <i>Finite</i>? Does the
+gentle Reader see any escape from admitting this? And, if he admits
+this, does he <i>still</i> maintain that the <i>width</i> of this
+Infinite-Strip (which obviously dwindles, as you go along the Strip,
+but never becomes Zero) <i>never ceases to be Finite</i>? Yet surely a
+Strip, Infinite in <i>length</i>, and <i>nowhere less than Finite in
+width</i>, must be <i>Infinite</i> in area?</p>
+
+<p>In brief, I place before my gentle Reader that savouriest of Logical
+dishes, a <i>Trilemma</i>! Either he must assert that a Strip,
+Infinite in length, and nowhere less than Finite in width, is only
+<i>Finite</i> in area; or he must assert that the <i>length</i> of
+this Strip is <i>Finite</i>, i.e. he must assert that <i>the Curve
+meets the x-Axis</i>; or else he must admit that its <i>width</i>
+ceases to be Finite<span class="pagenum" id="Page_47">[Pg 47]</span> without becoming Zero, i.e. he must admit that its
+width becomes <i>Infinitesimal</i>! Let him take his choice, and help
+himself. 'May good digestion wait on appetite, And health on both!'</p>
+
+<p>Now let us cut off, from the infinitely-long Strip named in p. 45,
+whose area is a square-inch, a piece just an inch long. What will its
+area be? It is evident that no finite multiple of this short Strip can
+ever make up the infinitely-long Strip; that is, no finite multiple
+of its area can make up a square-inch. Hence its area must be an
+Infinitesimal of the first order.</p>
+
+<p>But this Infinitesimal area, inconceivably small as it
+is, is nevertheless <i>greater than Zero</i>. Hence our
+continuously-diminishing Strip is bound, before reaching Zero, to
+pass through this singularly unassuming value. And, at that moment,
+<i>what will be its width</i>? Its length will be Infinite, as usual:
+its area will be an Infinitesimal of the first order: but its width
+cannot be an Infinitesimal of the same order as the previous width;
+for <i>that</i> would yield a finite area, as we have seen. What else,
+then, can it be but <i>an Infinitesimal of the second order</i>? A
+Line of such stupendous brevity that no finite multiple of it can even
+make up an Infinitesimal of the first order. Evidently we might repeat
+this process <i>ad libitum</i>, and so get Infinitesimals of the third
+order, the fourth order, and so on.</p>
+
+<p>To sum up our results, so far. We see that a Line may be either Finite,
+or may extend to an Infinity of the first order, or may dwindle to an
+Infinitesimal of the first, second, or any order we choose: and that an
+Area may be either Finite, or may extend to an Infinity of the first
+order (an Infinite-Strip), or of the second order (e.g. the whole
+Infinite-Plane), or may dwindle to an Infinitesimal of any order we
+choose.</p>
+
+<p>Now we may reasonably expect to find that all, that has been here
+said, is equally applicable to <i>any</i> kind of Magnitude that is
+capable of <i>continuous</i> increase and decrease. Let us consider,
+then, whether it is possible to have Infinitesimal <i>Angles</i> of the
+various orders.</p>
+
+
+<p class="nindc space-above2">§ 3.</p>
+<p><span class="pagenum" id="Page_48">[Pg 48]</span></p>
+
+<p class="nindc"><i>Infinitesimal Angles and Sectors.</i></p>
+
+<p>For this purpose, let us return to our upright Infinite-Plane, and,
+taking some Point at random as a centre, let us imagine two Lines
+radiating from it (say at an angle of 45°), and both of them extended
+to infinity, and therefore having between them a Sector, finite in
+angular magnitude, infinite in length, and therefore infinite in area.</p>
+
+<p>I named 45° as my specimen-angle, because it is the <i>one single
+angle</i>, other than a right angle, with which Society is acquainted.
+Enquire of some chatty traveller, who is relating his experiences of
+an Alpine Pass, what <i>slope</i> it was that he had to climb. "The
+ground sloped at an angle of forty-five," he is sure to reply. Nay, I
+once met a gentleman who, on hearing it mentioned that the 'dip' of a
+certain river-bed was, in one place, "one in forty-five," cautiously
+remarked "I suppose that means an angle of forty-five degrees?" Imagine
+a <i>river</i> sloping at that angle! And then imagine the labour
+of rowing <i>up</i> it, and the headlong, wild delight of rowing
+<i>down</i> it! But this is a digression.</p>
+
+<p>Now it clearly <i>is</i> possible, this time, by multiplying this
+Infinite-Sector by a certain finite number, namely '8,' to make it
+equal to the whole Infinite-Plane. Hence this Infinite-Sector is an
+Infinity <i>of the same order</i> as the whole Infinite-Plane; i.e. it
+is of the <i>second</i> order.</p>
+
+<p>Now let us suppose that the two sides of this Infinite Sector gradually
+approach each other until they coincide. It cannot be denied that the
+area then becomes Zero. It has dwindled then, from an Infinity of the
+second order (when its angular magnitude was <i>finite</i>), down to
+absolute Zero. And there seems no room to doubt that it has done this
+<i>continuously</i>, and not 'per saltum.' What values, then, has it
+gone through on the way? Can we reasonably doubt that it has gone
+through, first, all Infinities of the <i>first</i> order (during
+which process its angular magnitude would be an Infinitesimal of the
+<i>first</i> order), secondly, all <i>Finite</i> values (its angular
+magnitude being an Infinitesimal of the <i>second</i> order), thirdly,
+all Infinitesimals of the <i>first</i> order (its angular magnitude
+being an Infinitesimal of the <i>third</i> order), and so on—the
+angular magnitude being always <i>two</i> degrees ahead of the area, in
+this long and fatiguing competition in the Dwindling-Race.</p>
+
+
+<p class="nindc space-above2">§ 4.</p>
+<p><span class="pagenum" id="Page_49">[Pg 49]</span></p>
+
+<p class="nindc"><i>Pairs of Lines.</i></p>
+
+<p>We are now in a position to examine the phenomena of intersection, or
+non-intersection, with regard to a Pair of Lines, by imagining one of
+them to revolve about a fixed Point.</p>
+
+<figure class="figcenter width500" id="i_p049" style="width: 300px;">
+<img src="images/i_p049.jpg" width="300" height="323" alt="Four axes
+meeting at centre V, vertical X1 Y1, horizontal CD, and diagonal rays
+X2, X3 upper-left and Y2, Y3 lower-right, dividing the plane into four
+numbered sectors, illustrating angular regions around a point.">
+</figure>
+
+<p>Let \(AB\) be one of the 2 Lines: and let \(X_{1}VY_{1}\) be, at first
+at right angles to it: and let it then revolve, about \(V\), so as
+to take the successive positions \(X_{2}VY_{2}\), \(X_{3}VY_{3}\),
+\(CVD\), its final position being at right angles to its first
+position, and therefore parallel to \(AB\).</p>
+
+<p>Let us further suppose that the angle, contained between \(CV\) and
+the upper part of the revolving Line, dwindles, from a right angle,
+through all possible <i>finite</i> lesser values, while its<span class="pagenum" id="Page_50">[Pg 50]</span> upper
+edge revolves from \(VX_{1}\), to \(VX_{2}\); that, the moment the
+upper edge goes <i>below</i> \(VX_{2}\), this angle becomes an
+<i>Infinitesimal of the first degree</i>, and so dwindles, through
+all such values, while its upper edge revolves from \(VX_{2}\)
+to \(VX_{3}\); that, the moment the upper edge goes <i>below</i>
+\(VX_{3}\), this angle becomes an <i>Infinitesimal of the second
+degree</i>; and so on.</p>
+
+<figure class="figcenter width500" id="i_p050" style="width: 300px;">
+<img src="images/i_p050.jpg" width="300" height="315" alt="Nearly
+identical to previous figure but without the horizontal AB line
+intersecting at V, the same four sectors and radiating rays X2 ,X3 and
+Y2, Y3, showing a variant case of angular regions around centre point
+V."> 
+</figure>
+
+<p>Now let us suppose all the Lines in the diagram to extend to infinity
+both ways: and let us call the Infinite-Sector, lying between
+\(VC\)-produced and the upper part of the revolving-Line, 'No. 1'; the
+semi-Infinite-Strip (I mean, by 'semi-Infinite,' that it is terminated
+at <i>one</i> end), lying between the Lines \(VC\)-produced and
+\(Y_{1}A\)-produced, and bounded at the right-hand end by \(VY_{1}\),
+'No. 2'; the surface, lying between the Line \(Y_{1}B\)-produced
+and the lower part of the revolving-Line, and bounded at the
+left-hand end by \(VY_{1}\), (which will be a Triangle, so long as
+\(Y_{1}B\)-produced and the lower part of the revolving-Line continue
+to intersect, and will become a semi-Infinite-Strip when they cease
+to intersect,) 'No. 3'; and, with regard to the Infinite-Sector
+lying between the Line \(VD\)-produced and the lower part of the
+revolving-Line, let us call that portion of it, which lies <i>above</i>
+\(Y_{1}B\)-produced (which portion will be a semi-Infinite-Strip, so
+<span class="pagenum" id="Page_51">[Pg 51]</span>long as \(Y_{1}B\)-produced and the lower part of the revolving-Line
+continue to intersect, and will become the whole Infinite-Sector if
+they should cease to intersect before the revolving-Line reaches
+the position \(VD\)), 'No. 4'; and that portion of it, which lies
+<i>below</i> \(Y_{1}B\)-produced (which portion will cease to exist as
+soon as these Lines cease to intersect) 'No. 5.'</p>
+
+<p>Let us now investigate the changes, in the <i>areas</i> of these 5
+surfaces, caused by the changes in the position of the revolving-Line.</p>
+
+<p>First as to No. 1. This is clearly, throughout its history, an
+Infinite-Sector, whose vertical-angle is at first a right angle, and
+ultimately Zero. Also its area is at first one-quarter of the whole
+Infinite-Plane, and ultimately Zero. Also, so long as the upper part of
+the revolving-Line ranges between \(VX_{1}\) and \(VX_{2}\), the area
+continues to be an Infinity of the <i>second</i> order; and, as the
+revolving-Line crosses the position \(VX_{2}\), the area changes, from
+a very small (!) <i>Infinity of the second order</i>, to a very large
+<i>Infinity of the first order</i>. Similarly, as the revolving-Line
+crosses the position \(VX_{3}\), the area changes, from a very small
+<i>Infinity of the first order</i>, to a very large <i>Finite value</i>.</p>
+
+<p>A little further on, it will of course become an <i>Infinitesimal of
+the first order</i>; and so on, through the other orders, till it
+finally reaches the value <i>Zero</i>.</p>
+
+<p>Next, as to No. 2. This is clearly, throughout its history, one-half of
+the Infinite-Strip lying between \(AB\) and \(CD\), and is therefore an
+<i>Infinity of the first order</i>. Its area is a constant quantity,
+being unaffected by the revolving-Line.</p>
+
+<p>Next, as to Nos. 4, 5. (I take these next, because they will help
+us to investigate the properties of No. 3.) It is evident that, so
+long as No. 5 continues to exist, the two together constitute—and
+that, when No. 5 has ceased to exist, No. 4 by itself constitutes—an
+Infinite-Sector, which is the exact counterpart of No. 1—the two
+having what Euclid calls 'opposite vertical angles.' Hence, while
+the lower part of the revolving-Line ranges between \(VY_{1}\) and
+<span class="pagenum" id="Page_52">[Pg 52]</span>\(VY_{2}\), the area of No. 4 continues to be an <i>Infinity of the
+second order</i>, and entirely declines to be 'cribb'd, cabin'd,
+and confined' within such narrow limits as our Infinite-Strip!
+Hence the revolving-Line continues, all this time, to intersect
+\(Y_{1}B\)-produced. (N.B. Here we have a proof of the truth of
+Euclid's Axiom, when amended, as I have done at p. 28, by inserting the
+words 'the defect being a finite angle.')</p>
+
+<figure class="figcenter width500" id="i_p052" style="width: 300px;">
+<img src="images/i_p052.jpg" width="300" height="319" alt="A third
+variant of the four-sector diagram, rays X2, X3 and Y2, Y3 now spread
+slightly wider apart from centre V, with sectors 1, 2, 3, 4 labelled,
+showing another angular case in the same progressive series.">
+</figure>
+
+<p>Let us now, reserving for future consideration the phenomena of the
+period while the upper part of the revolving-Line is crossing from
+\(VX_{2}\) to \(VX_{3}\), suppose it to have passed \(VX_{3}\), so that
+the angle, which it makes with \(VC\), has become an Infinitesimal
+of the <i>second</i> order. Will its lower part continue to
+intersect \(VD\)-produced? It may easily be shown, by a <i>reductio
+ad absurdum</i>, that it will <i>not</i>: for, if it did, Nos. 1,
+2, 3 would then make up an Infinite Sector, whose vertical angle
+would be an Infinitesimal of the <i>second</i> order, and whose area
+would therefore be <i>finite</i>. But the area of No. 2 is always an
+<i>Infinity</i> of the first order. Which is absurd. Hence, after the
+upper part of the revolving-Line has passed \(VX_{3}\), its lower part
+does <i>not</i> intersect \(Y_{1}B\)-produced.</p>
+
+<p>We have now to answer a far more puzzling question, namely, what
+happens while the upper part of the revolving-Line is crossing
+from \(VX_{2}\) to \(VX_{3}\)? Does its lower part intersect
+<span class="pagenum" id="Page_53">[Pg 53]</span>\(Y_{1}B\)-produced, all the while? Or does it fall clear of it, all
+the while? Or does it at first intersect it, and afterwards cease to do
+so?</p>
+
+<p>First, suppose it to intersect \(Y_{1}B\)-produced. In this case No. 3
+is clearly a closed Triangle, whose vertical-angle is an Infinitesimal
+of the first order. This looks as if its <i>area</i> must also be an
+Infinitesimal of the first order: but this, we know, <i>cannot</i> be,
+since it contains within it the <i>finite</i> area possessed by No. 3
+while the upper part of the revolving-Line was passing from \(VX_{1}\)
+to \(VX_{2}\). Hence its area must be, at least, <i>finite</i>. The
+only way I can see out of this difficulty is to assume that the
+<i>sides</i> of this Triangle have become <i>infinite</i>; i.e. that
+the revolving-Line intersects \(Y_{1}B\)-produced <i>at an infinite
+distance</i>.</p>
+
+<p>Next, suppose it to fall clear of \(Y_{1}B\)-produced. In this
+case No. 3 is a semi-Infinite-Strip, but we cannot be certain that
+its <i>area</i> is, like such Strips when of a uniform width,
+<i>infinite</i>; for its width dwindles so much towards the
+right-hand that it may possibly be, in the early part of the period,
+<i>finite</i>. I see no way of settling this question; but, luckily, it
+is not relevant to the question of <i>intersection</i>.</p>
+
+<p>My own inclination is to believe that, during this second period of
+revolution, the lower part of the revolving-Line at first intersects
+\(Y_{1}B\)-produced, at an infinite distance, and then ceases to
+intersect it.</p>
+
+<p>After the revolving-Line has once ceased to intersect
+\(Y_{1}B\)-produced, No. 4 is of course equal to No. 1. That is to
+say, the surface, lying between the whole revolving-Line and \(AB\),
+is, from that moment until the revolving-Line coincides with \(CD\),
+of a constant area, since it is the sum-total of Nos. 1, 2, 3,
+which is equal to the sum-total of Nos. 2, 3, 4, i.e. to the whole
+Infinite-Strip. And this will continue true, while No. 1 and No. 4
+dwindle down, through all finite and infinitesimal values, till they
+finally reach Zero.</p>
+
+<p><span class="pagenum" id="Page_54">[Pg 54]</span></p>
+
+<p>The results we have arrived at will perhaps be more easily understood
+by examining the following Table of values. The symbols used in it are
+as follows:—</p>
+
+
+<table class="autotable">
+<thead><tr>
+<th class="tdc bb bl bt2 br">Symbols.</th>
+<th class="tdc bb bt2 br">Meanings.</th>
+</tr></thead>
+<tbody><tr>
+<td class="tdc bl br">\(R\)</td>
+<td class="tdl br">a right angle.</td>
+</tr><tr>
+<td class="tdc bl br">\(F\)</td>
+<td class="tdl br">a large finite magnitude</td>
+</tr><tr>
+<td class="tdc bl br">\(f\)</td>
+<td class="tdl br">a small&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;"&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;"</td>
+</tr><tr>
+<td class="tdc bl br">\(M\)</td>
+<td class="tdl br">a large Infinitesimal of 1st order.</td>
+</tr><tr>
+<td class="tdc bl br">\(M^{2}\)</td>
+<td class="tdl br">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;"&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;"&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;2nd&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;"</td>
+</tr><tr>
+<td class="tdc bl br">\(J\)</td>
+<td class="tdl br">a large Infinity of 1st order.</td>
+</tr><tr>
+<td class="tdc bl br">\(S\)</td>
+<td class="tdl br">area of the semi-Infinite-Strip No. 2.</td>
+</tr><tr>
+<td class="tdc bl br"></td>
+<td class="tdc br">[an Infinity of 1st order.]</td>
+</tr><tr>
+<td class="tdc bl br">\(P\)</td>
+<td class="tdl br">area of one-fourth of the Infinite-Plane.</td>
+</tr><tr>
+<td class="tdc bl bb br"></td>
+<td class="tdc bb br">[an Infinity of 2nd order.]</td>
+</tr>
+</tbody>
+</table>
+
+<p>&nbsp;</p>
+
+<table class="autotable">
+<thead><tr>
+<th class="tdc bb bl bt2 br" rowspan="2">Position of<br>
+upper part<br>
+of revolving-<br>
+Line.</th>
+<th class="tdc bb bl bt2 br" rowspan="2">Vertical<br>
+angle of<br>
+Space<br> 
+No. 1.</th>
+<th class="tdc bb bl bt2 br" colspan="5">Areas of Spaces.</th>
+</tr><tr>
+<th class="tdc bb bl bt2 br">No. 1.</th>
+<th class="tdc bb bl bt2 br">No. 2.</th>
+<th class="tdc bb bl bt2 br">No. 3.</th>
+<th class="tdc bb bl bt2 br">No. 4.</th>
+<th class="tdc bb bl bt2 br">No. 5.</th>
+</tr>
+</thead>
+<tbody><tr>
+<td class="tdc bl br">\(X_{1}V\)</td>
+<td class="tdc br">\(R\)</td>
+<td class="tdc br">\(P\)</td> 
+<td class="tdc br">\(S\)</td>
+<td class="tdc br">Zero</td>
+<td class="tdc br">\(S\)</td>
+<td class="tdc br">\(P\)</td>
+</tr><tr>
+<td class="tdc bl br"></td>
+<td class="tdc br">\(F\)</td>
+<td class="tdc br">\(J^2\)</td> 
+<td class="tdc br">"</td>
+<td class="tdc br">\(f\)</td>
+<td class="tdc br">"</td>
+<td class="tdc br">\(J^2\)</td>
+</tr><tr>
+<td class="tdc bl br"></td>
+<td class="tdc bb br">\(f\)</td>
+<td class="tdc bb br">\(j^2\)</td> 
+<td class="tdc bb br">"</td>
+<td class="tdc bb br">\(F\)</td>
+<td class="tdc bb br">"</td>
+<td class="tdc bb br">\(j^2\)</td>
+</tr><tr>
+<td class="tdc bl br">\(X_{2}V\)</td>
+<td class="tdc br">\(M\)</td>
+<td class="tdc br">\(J\)</td> 
+<td class="tdc br">"</td>
+<td class="tdc br">?</td>
+<td class="tdc br">?</td>
+<td class="tdc br">?</td>
+</tr><tr>
+<td class="tdc bl br"></td>
+<td class="tdc bb br">\(m\)</td>
+<td class="tdc bb br">\(j\)</td> 
+<td class="tdc bb br">"</td>
+<td class="tdc bb br"></td>
+<td class="tdc bb br"></td>
+<td class="tdc bb br"></td>
+</tr><tr>
+<td class="tdc bl br">\(X_{3}V\)</td>
+<td class="tdc br">\(M^2\)</td>
+<td class="tdc br">\(F\)</td> 
+<td class="tdc br">"</td>
+<td class="tdc br">\(S\)</td>
+<td class="tdc br">\(F\)</td>
+<td class="tdc br">does not</td>
+</tr><tr>
+<td class="tdc bl br"></td>
+<td class="tdc bb br">\(m^2\)</td>
+<td class="tdc bb br">\(f\)</td> 
+<td class="tdc bb br">"</td>
+<td class="tdc bb br">"</td>
+<td class="tdc bb br">\(M\)</td>
+<td class="tdc bb br">exist</td>
+</tr><tr>
+<td class="tdc bl br"></td>
+<td class="tdc br">\(M^3\)</td>
+<td class="tdc br">\(M\)</td> 
+<td class="tdc br">"</td>
+<td class="tdc br">"</td>
+<td class="tdc br">\(M\)</td>
+<td class="tdc br">"</td>
+</tr><tr>
+<td class="tdc bl br"></td>
+<td class="tdc bb br">\(m^3\)</td>
+<td class="tdc bb br">\(m\)</td> 
+<td class="tdc bb br">"</td>
+<td class="tdc bb br">"</td>
+<td class="tdc bb br">\(m\)</td>
+<td class="tdc bb br">"</td>
+</tr><tr>
+<td class="tdc bl br">\(CV\)</td>
+<td class="tdc br">&c.</td>
+<td class="tdc br">&c.</td> 
+<td class="tdc br">"</td>
+<td class="tdc br">"</td>
+<td class="tdc br">&c.</td>
+<td class="tdc br">"</td>
+</tr><tr>
+<td class="tdc bl bb br"></td>
+<td class="tdc bb br">Zero.</td>
+<td class="tdc bb br">Zero.</td> 
+<td class="tdc bb br">"</td>
+<td class="tdc bb br">"</td>
+<td class="tdc bb br">Zero.</td>
+<td class="tdc bb br">"</td>
+</tr>
+</tbody>
+</table>
+
+
+<p><span class="pagenum" id="Page_55">[Pg 55]</span></p>
+
+<figure class="figcenter width500" id="i_p055" style="width: 300px;">
+<img src="images/i_p055.jpg" width="300" height="312" alt="Fourth
+variant of the sector diagram, rays X2, X3 upper-left and Y2, Y3
+lower-right now spread wider still, approaching the horizontal CD,
+showing the limiting case as the parallel postulate is approached.">
+</figure>
+
+<p>The sum of the whole matter appears to be this. If a Pair of Lines
+make, with a certain transversal, two interior angles on the same side
+of it together less than two right angles, then, so long as the defect
+is <i>finite</i>, there is no doubt that the Lines intersect: also,
+if the theory be true, that the area of an Infinite-Sector, whose
+vertical-angle is finite, and whose area is therefore undoubtedly an
+Infinity of the second order, passes, as its vertical angle dwindles
+to Zero, through infinite values of the first order, and then
+through finite values, its vertical angle meanwhile passing through
+infinitesimal values of the first and second order—if all this be
+true, it follows that, when the 'defect from two right angles' becomes
+an Infinitesimal of the <i>first</i> order, the Lines may possibly
+intersect, but can only do so at an infinite distance; and that, when
+the defect has become an Infinitesimal of the <i>second</i> order, the
+Lines have ceased to intersect.</p>
+
+<p>The theory, here discussed, may be bewildering; but it is at least
+consistent with itself: and it seems to me to be <i>quite</i> as
+credible as the theory that 'Infinitesimals' are mythical, and that a
+Finite Magnitude, dwindling down to Zero, continues Finite to its last
+gasp.</p>
+
+<p><span class="pagenum" id="Page_56">[Pg 56]</span></p>
+
+<p>Another process—a <i>negative</i> one—has occurred to me, for
+disproving the <i>absolute</i> truth of Euclid's Axiom. It is a very
+simple process, and has the great recommendation of not requiring any
+belief in the existence of Infinitesimals.</p>
+
+<p>On an upright Infinite-Plane let us suppose 2 horizontal
+Infinite-Lines, having a common perpendicular, and therefore of course
+never intersecting. Let us call them 'No. 1' and 'No. 2,' No. 1 being
+above No. 2: and let us suppose the common perpendicular to be an inch
+long, so that the 2 Lines are the edges of an Infinite-Strip, whose
+uniform width is an inch.</p>
+
+<p>Now let us suppose Line No. 1 to begin to revolve about the upper end
+of the common perpendicular. The believer in the <i>absolute</i> truth
+of Euclid's Axiom is bound to believe that, <i>in the very act of
+beginning to revolve</i>, it also begins to intersect No. 2: he cannot
+allow the one process any such <i>start</i> of the other as might
+enable him to say "No. 1 has begun to revolve, but has <i>not</i> yet
+begun to intersect No. 2": such a state of things <i>must</i> be, in
+<i>his</i> view, <i>absolutely</i> impossible. 'And what's impossible
+ca'n't be, And never never comes to pass!'</p>
+
+<p>Very good. The actual process of <i>beginning</i> to intersect No. 2 is
+a deep mystery, no doubt: but that the thing <i>happens</i> is quite
+undeniable. We are able to say "Now it <i>isn't</i> intersecting No.
+2—and now it <i>is</i>!" So, though we cannot conceive <i>how</i> it
+managed to begin, it has certainly <i>done</i> it.</p>
+
+<p>Now let us suppose <i>another</i> horizontal Line, 'No. 3,' lying an
+inch below No. 2. The believer in Euclid's Axiom is bound to assert, as
+to No. 3, <i>exactly</i> what he asserted as to No. 2. Has he, then,
+any logical escape from the conclusion that the revolving Line begins
+to intersect Nos. 2 and 3 <i>together</i>? I see none, myself. And yet
+how <i>can</i> it get at No. 3, without <i>first</i> going through
+No. 2? <i>Any</i> point on No. 1 (I am careful not to say 'every': I
+know how gleefully the logical Reader would swoop down upon me with
+the crushing sarcasm "What does 'every' mean, when there is no limit
+to the number of points?": but 'any' is a safe epithet) is amenable to
+the simple rule of<span class="pagenum" id="Page_57">[Pg 57]</span> "First come, first served. Cross No. 2 first: and
+<i>afterwards</i> (not by any means <i>simultaneously</i>) you have our
+gracious permission to cross No. 3." Now, what is true of <i>any</i>
+point on the Infinite-Line No. 1 is surely true of that Infinite-Line
+itself? To say "No. 1 intersects No. 3" is tantamount to saying "A
+certain <i>point</i> of No. 1 has reached No. 3." And <i>how did that
+Point get below No. 2</i>?</p>
+
+<p>Of two things, one. Either <i>some</i> point of No. 1 has crossed Nos.
+2 and 3 <i>at the same moment</i>: or else <i>no</i> point of No. 1
+has crossed No. 3 until <i>after</i> it had crossed No. 2. That is a
+logical Dilemma. Which of its two horns does the Reader prefer?</p>
+
+<p>The choice of the <i>first</i> horn involves the Reader's acceptance
+of <i>ubiquitous points</i>! In the event of his choosing the
+<i>second</i> horn, he seems logically bound to admit it as a
+<i>possible</i> state of things, that No. 1 should have <i>begun</i> to
+revolve, and yet should <i>not</i> have begun to intersect No. 3—which
+is a surrender of his belief in the <i>universal</i> truth of Euclid's
+Axiom.</p>
+
+<p>My final answer, then, to the question "<i>Is Euclid's Axiom true?</i>"
+is as follows:—</p>
+
+<p>"If the defect, from the sum of two right angles, be <i>finite</i>,
+the Lines <i>certainly meet</i>: if it be an Infinitesimal of the
+<i>first</i> order, they may meet, or not, according to circumstances:
+if it be an Infinitesimal of the <i>second</i>, third, or any higher
+order, they <i>certainly do not meet</i>."</p>
+
+<p>The question, with which this Appendix is headed, was sufficiently
+startling: but a more startling one remains to be answered. "If
+Euclid's Axiom be not <i>universally</i> true, what becomes of all the
+Propositions which he has made to depend upon it, such as I. 29? Has he
+done no more than prove them to be <i>partially</i> true?"</p>
+
+<p>To this disheartening question I will give as re-assuring an answer
+as the case seems to admit of. It must be admitted that Euc. I. 29
+requires—if we would make it <i>strictly</i> true—the same limitation
+as the Axiom: it should run as follows:—"Two<span class="pagenum" id="Page_58">[Pg 58]</span> Lines, which do not
+meet, make, with all transversals, angles which are equal <i>so far
+as finite differences are concerned</i>" (i. e. angles so nearly
+equal that the difference, if any, is <i>infinitesimal</i>). And of
+course Euc. I. 32, as proved from Euc. 1. 29, would need a similar
+qualification. It seems to me very doubtful whether Euclid ever noticed
+this defect in his Axiom. If he did, it is possible that he may have
+thought fit to ignore it, on the ground that, when Finite Magnitudes
+differ only by an Infinitesimal, they are, for all <i>practical</i>
+purposes, equal.</p>
+
+<p>But I feel bound to admit that, for the purpose of proving Euc. I.
+32 to be, as it really is, <i>universally</i> true, neither Euclid's
+Axiom, nor any other that deals with <i>intersection</i> of Lines, will
+suffice: and that some Axiom, not involving that principle, must be
+substituted for it.</p>
+
+<p class="space-above2">
+Let me say in conclusion that, though I assert the <i>absolute</i>
+truth of Euclid's Axiom—with the limiting clause I have introduced,
+'<i>the defect being a finite angle</i>'—it still remains, in my
+opinion, a 'disputable' Axiom; i.e. it is not properly admissible as an
+<i>Axiom</i>, but ought to be, if possible, proved as a <i>Theorem</i>.</p>
+
+
+<hr class="chap x-ebookmaker-drop">
+
+<div class="chapter">
+<p><span class="pagenum" id="Page_59">[Pg 59]</span></p>
+<h2 class="nobreak" id="APPENDIX_III">APPENDIX III.</h2>
+</div>
+
+<p class="nindc"><i>How should Parallels be defined?</i></p>
+
+
+<p>We know that, if a Pair of Lines has either of the following
+properties</p>
+
+<div class="blockquot">
+
+<p>(1) they are equally inclined to a certain transversal,</p>
+
+<p>(2) one of them contains 2 Points on the same side of, and equidistant
+from, the other,</p>
+</div>
+
+<p class="nind">it has all the following properties</p>
+
+<div class="blockquot">
+
+<p>(3) they are equally inclined to all transversals,</p>
+
+<p>(4) any 2 Points, on either of them, are on the same side of, and
+equidistant from, the other,</p>
+
+<p>(5) they do not meet, however far produced.</p>
+</div>
+
+<p>Any one of these properties may be used as a Definition. Let us take
+them one by one.</p>
+
+<p>No. (1). This has the advantage that we need not begin by proving that
+such Lines <i>exist</i>, but may assume it as axiomatic. It has been
+used as a Definition by Varignon, Bezout, Cooley, &amp;c.: at least, they
+say "which make equal angles with<span class="pagenum" id="Page_60">[Pg 60]</span> <i>a</i> transversal," leaving
+it uncertain whether they mean "a <i>certain</i> transversal" or
+"<i>any</i> transversal": if the latter, it is of course No. (3) they
+are proposing to use. From No. (1) we can prove No. (5) without using any
+disputable Axiom (see Euc. I. 27, 28), but not No. (3) or No. (4).</p>
+
+<p>No. (2). This has the same advantage as No. (1). It has been used as a
+Definition by D'Alembert. From <i>it</i>, also, we can prove No. (5)
+without using any disputable Axiom; and it, also, fails to prove No. (3)
+or No. (4).</p>
+
+<p>No. (3). This cannot be used, as a Definition, till we have proved that
+such Lines <i>exist</i>—which has not yet been done without employing
+some disputable Axiom. If Varignon, &amp;c. mean <i>this</i> to be their
+Definition, they are assuming the <i>existence</i> of such Lines,—a
+<i>very</i> un-axiomatic Axiom. But, when once their <i>existence</i>
+has been granted, or proved, No. (4) can be deduced.</p>
+
+<p>No. (4). This cannot be used, as a Definition, till we have proved
+that such Lines <i>exist</i>—which has not yet been done without
+employing some disputable axiom. When once their <i>existence</i> has
+been granted, or proved, No. (3) can be deduced. No. (4) has been used as a
+Definition by Wolf, Boscovich, T. Simpson, and Bonnycastle.</p>
+
+<p>No. (5). This has the advantage that it is easy to prove (as in
+Euc. I. 27, 28) that such Lines <i>exist</i>. It has been used as a
+Definition by Euclid and a host of other geometers. It has, however,
+the enormous <i>dis</i>advantage that, whereas Nos. (3), (4), give
+us a <i>unique</i> Pair of Lines (e.g. given a Line and a Point not
+on it, we can prove that only <i>one</i> Line can be drawn, through
+the Point, such that the Pair shall have property No. (3)), No. (5)
+does <i>not</i>: on the contrary, given a Line and a Point not on
+it, a whole 'pencil' of Lines may be drawn, through the Point, and
+not meeting the given Line: all we need to do is to take care, after
+drawing <i>one</i> such Line, that the others shall make with it angles
+which are <i>Infinitesimals of the second order with regard to a right
+angle</i>.</p>
+
+<p><span class="pagenum" id="Page_61">[Pg 61]</span></p>
+
+<p>We see, then, that the word "Parallels" has been already used with
+<i>four</i>, and possibly with <i>five</i>, different meanings: so that
+any fresh writer, who uses the word, is liable to be misunderstood
+unless he first defines it. The derivation of the word would seem to
+suggest No. (4) as its Definition; but Euclid's adoption of No. (5) has
+led to that being the popular meaning attached to the word.</p>
+
+<p>It is easy, however, to avoid all this ambiguity by the use of new
+terms. Rejecting Nos. (1) and (2), as useless for the purpose of
+definition, we may call a Pair of Lines, which possesses</p>
+
+<div class="blockquot">
+
+<p class="nindc">No. (3), "equiangulated";</p>
+
+<p class="nindc">No. (4), "equidistantial";</p>
+
+<p class="nindc">No. (5), "separational";</p>
+</div>
+
+<p class="nind">and thus avoid the dangerous word "parallel" altogether.</p>
+
+<p>My own belief is that No. (3) is, on the whole, the property best
+adapted for practical use as a Definition.</p>
+
+
+<hr class="chap x-ebookmaker-drop">
+
+<div class="chapter">
+<h2 class="nobreak" id="APPENDIX_IV">APPENDIX IV.</h2>
+</div>
+
+<p class="nindc"><i>How the Question stands to-day.</i></p>
+
+
+<p>I will first enumerate certain Theorems connected with Pairs of Lines.</p>
+
+<p>I will then notice three other methods which have been suggested for
+superseding Euclid's Axiom.</p>
+
+<p>And, in conclusion, I will indicate what seem to me the most hopeful
+directions for future efforts at exploring this fascinating, but very
+obscure, region of mathematical research.</p>
+
+
+<p class="nindc space-above2">§ 1.</p>
+<p><span class="pagenum" id="Page_62">[Pg 62]</span></p>
+
+<div class="blockquot">
+
+<p><i>Certain universally-true Theorems, provable from genuine Axioms (i.
+e. from Axioms whose self-evident character is indisputable).</i></p>
+</div>
+
+<p>(1) A Pair of Lines, which are equally inclined to a certain
+transversal, are non-intersectional (i. e. never intersect, however far
+produced). [Euc. I. 27, 28.]</p>
+
+<p>(2) [<i>contranominal</i> of (1)] A Pair of intersectional Lines (i.
+e. which either intersect or would do so if produced) are unequally
+inclined to any transversal. [Euc. I. 16, 17.]</p>
+
+<p>(3) A Pair of Lines, such that two Points on one of them are on the
+same side of, and equidistant from, the other, are non-intersectional.</p>
+
+<p>(4) [<i>contranominal of</i> (3)] A Pair of intersectional Lines are
+non-equidistantial (i. e. are such that any two Points on either of
+them, which are on the same side of the other, are non-equidistant from
+it, that which is further from the Point of intersection being also
+further from the other Line).</p>
+
+<p>(5) If there be a Triangle whose angles are together equal to two right
+angles: the angles of any Triangle are together equal to two right
+angles.</p>
+
+<p>(6) There is a Triangle whose angles are together not-greater than two
+right angles.</p>
+
+
+<p class="nindc space-above2">§ 2.</p>
+<p><span class="pagenum" id="Page_63">[Pg 63]</span></p>
+
+<div class="blockquot">
+
+<p><i>Certain universally-true Theorems, not provable from genuine Axioms,
+but provable if any one of them be accepted as an Axiom.</i></p>
+</div>
+
+<div class="blockquot">
+
+<p>[N.B. These will be hereafter referred to as the 'Nine Quasi-Axioms,'
+their self-evident character being disputable.]</p>
+</div>
+
+<p>(1) Through a given Point, outside a given Line, a Line may be drawn,
+such that the Pair shall be equally inclined to any transversal.</p>
+
+<p>(2) A Pair of Lines, which are equally inclined to a certain
+transversal, are so to any transversal. [Deducible from Euc. I. 27, 28,
+29.]</p>
+
+<p>(3) [<i>contranominal of (2)</i>] A pair of Lines, which are unequally
+inclined to a certain transversal, are so to any transversal.</p>
+
+<p><span class="pagenum" id="Page_64">[Pg 64]</span></p>
+
+<p>(4) If a Point move so as to be at a constant distance from a given
+Line, its path shall be a straight Line.</p>
+
+<p>*(5) Through a given Point, outside a given Line, a Line may be drawn
+equidistantial from the given Line (i. e. such that any two Points on
+it shall be equidistant from the given Line).</p>
+
+<p>(6) A Pair of Lines, such that two Points on one of them are on the
+same side of, and equidistant from, the other, are equidistantial (i.
+e. are such that any two Points on either of them are equidistant from
+the other).</p>
+
+<p>(7) [<i>contranominal of</i> (6)] A Pair of Lines, such that two
+Points on one of them are non-equidistant from the other, are
+non-equidistantial (i. e. are such that any two Points on either of
+them, which are on the same side of the other, are non-equidistant from
+the other).</p>
+
+<p>*(8) A Line cannot recede from and then approach another; nor can it
+approach and then recede from another while on the same side of it.</p>
+
+<p>*(9) In any Circle, the inscribed equilateral Tetragon is greater than
+any one of the Segments which lie outside it.</p>
+
+<p class="space-above2">
+If any one of these 9 Theorems be granted as an Axiom, the rest can
+be proved from it. But only 3 of them, so far as I know, have been
+used as Axioms—No. 5 by Clavius, No. 8 by Dr. R. Simson, and No. 9
+by myself. Clavius' Axiom requires us to assure ourselves that it
+will continue true when the Lines are produced <i>without limit</i>;
+and the strain on the imagination, caused by the effort of following
+them into Infinite Space, is one to be, if possible, avoided. Dr. R.
+Simson's Axiom gains a certain plausibility from the fact that, for
+<i>intersecting</i> Lines, it admits of actual <i>proof</i> (see § 1.
+(4)): where, however, the Lines are <i>not</i> known to intersect, it
+is an appeal to the eye, of <i>very</i> doubtful force.</p>
+
+
+<p class="nindc space-above2">§ 3.</p>
+<p><span class="pagenum" id="Page_65">[Pg 65]</span></p>
+
+<div class="blockquot">
+
+<p><i>Certain universally-true Theorems, not provable from genuine Axioms,
+but provable if any one of the 'Nine Quasi-Axioms,' given in § 2, be
+accepted.</i></p>
+</div>
+
+<p>(1) There is a finite angular magnitude such that the angles of any
+Triangle are together not-less than it.</p>
+
+<p>(2) There is a Triangle whose angles are together not-less than two
+right angles.</p>
+
+<p>(3) There is a Triangle whose angles are together equal to two right
+angles.</p>
+
+<p>(4) The angles of any Triangle are together equal to two right angles.
+[Euc. I. 32.]</p>
+
+
+<p class="nindc space-above2">§ 4.</p>
+
+<div class="blockquot">
+
+<p><i>Certain partially-true Theorems, not provable from any
+universally-true Axioms, whether genuine or 'quasi,' but provable if
+any one of themselves be accepted as an Axiom.</i></p>
+</div>
+
+<div class="blockquot">
+
+<p>[N.B. By 'partially-true' is meant 'true for <i>finite</i>
+magnitudes.' They become universally-true, if 'magnitude' be taken
+to mean '<i>finite</i> magnitude'; 'equal' to mean 'not differing
+by a <i>finite</i> difference'; 'unequal' to mean 'differing by a
+<i>finite</i> difference'; 'multiplied' to mean 'multiplied by a
+<i>finite</i> number'; and 'intersectional' to mean 'intersectional at
+a <i>finite</i> angle.'</p>
+
+<p>These Theorems will be hereafter referred to as the 'Nine
+Pseudo-Axioms,' the name being chosen to indicate that they are not
+even universally <i>true</i>—far less <i>self-evident</i>.]</p>
+
+<p><span class="pagenum" id="Page_66">[Pg 66]</span></p>
+
+<p>(1) The lesser of two homogeneous Magnitudes may be so multiplied as to
+exceed the greater.</p>
+
+<p>(2) A Pair of Lines, which are unequally inclined to a certain
+transversal, are intersectional. [Euclid's Axiom.]</p>
+
+<p>(3) [<i>contranominal of</i> (2)] A Pair of non-intersectional Lines
+are equally inclined to any transversal. [Euc. I. 29.]</p>
+
+<p>(4) A Pair of Lines, such that two Points on one of them are
+non-equidistant from the other, are intersectional.</p>
+
+<p>(5) [<i>contranominal of</i> (4)] A Pair of non-intersectional Lines
+are equidistantial.</p>
+
+<p>(6) On either of two intersectional Lines a Point may be found, whose
+distance from the other Line shall exceed any assigned length.</p>
+
+<p>(7) On either of two intersectional Lines a Point may be found, such
+that the distance, from its projection on the other Line to the point
+of intersection of the two Lines, shall exceed any assigned length.</p>
+
+<p>(8) A Pair of intersectional Lines cannot be, both of them,
+non-intersectional with a third Line.</p>
+
+<p>(9) [<i>contranominal of</i> (8)] A Pair of Lines, which are, both of
+them, non-intersectional with a third Line, are non-intersectional.</p>
+</div>
+
+<p>Six of these have been used as Axioms, viz. (2), which is Euclid's
+celebrated Axiom; (4), by T. Simpson; (6), by Proclus; (7), by
+Franceschini; (8), by Ludlam, Playfair, &amp;c.; (1), by Legendre, in
+the 12th Volume of the 'Memoirs of the Institute,' being his "latest
+attempt" (I quote from De Morgan's article on Parallels in Knight's
+Cyclopædia) "at the solution of the problem." He only succeeds,
+however, in proving Prop. 6 at p. 10 of this book, and in proving
+that Prop. 8, at p. 26, follows logically from Prop. 4, at p. 22.
+In order to prove Euc. I. 32, he introduces the principle of Limits
+and Vanishing Quantities, which takes us at once into the region of
+Infinities and Infinitesimals. But a greater success than this has, I
+understand, rewarded some recent investigations made by Professor J.
+Cook Wilson, of Oriel College, Oxford, who has deduced Euclid's 12th
+Axiom from No. (1) of this Section.</p>
+
+<p>In all these systems, however, including Euclid's, the deductions, from
+the proposed Axiom, labour under the same defect as the Axiom itself,
+that is, they are only <i>partially</i>, and not <i>universally</i>,
+true.</p>
+
+
+<p class="nindc space-above2">§ 5.</p>
+<p><span class="pagenum" id="Page_67">[Pg 67]</span></p>
+
+<p class="nindc"><i>Other methods of treatment.</i></p>
+
+<p>Three other methods of treating this subject call for notice.</p>
+
+<p>Playfair and (more recently) Wilson have tried to deduce the properties
+of Parallels (and thence Euc. I. 32) from the idea of <i>sameness of
+direction</i>, as predicated of two non-coincidental Lines (i. e. which
+do not lie in one and the same straight Line).</p>
+
+<p>The foundation-stones of this Theory—without which it has no <i>raison
+d'être</i> whatever—are the two Axioms, which must necessarily be
+<i>somewhere</i> assumed, whether expressly or tacitly, first, that it
+is possible for non-coincident Lines to have 'the same direction'; and
+secondly, that Lines, which have the same direction, make equal angles
+with all transversals.</p>
+
+<p>Before discussing the first of these two Axioms, permit me to remind
+the Reader that the question before us is not "is it <i>true</i>?" but
+"is it <i>axiomatic</i>?", that is, 'does the average human intellect
+<i>accept</i> it as true, <i>without proof</i>?' It is a question in
+Mental Physiology rather than in Geometry. The 47th Proposition of
+Euclid is quite as <i>true</i> as the Axiom 'things that are equal to
+the same are equal to one another'; but the average human intellect,
+while accepting the latter without proof, does most certainly demand
+a good deal of proof before it will accept the former. Intellectual
+beings may conceivably exist, to whom Euc. I. 47 is axiomatic; but our
+books are not written for <i>them</i>.</p>
+
+<p>Now there is one preliminary step, that is absolutely indispensable
+before the human intellect can accept any Axiom<span class="pagenum" id="Page_68">[Pg 68]</span> whatever: and that is,
+it must attach some <i>meaning</i> to it. We cannot, rationally, either
+assent to, or deny, any Proposition the words of which convey to us no
+idea.</p>
+
+<p>We have, then, <i>two</i> questions to answer: first, "what <i>idea</i>
+is conveyed to the average human intellect by the phrase '<i>in the
+same direction</i>,' when applied to non-coincident Lines?", secondly,
+"in accepting the Axiom, that 'Lines, which have the same direction,
+make equal angles with all transversals,' to what other assertions are
+we committing ourselves?"</p>
+
+<p>If we contemplate a fixed Point in a Plane, and imagine one or more
+Lines passing through it, it is not difficult to grasp the following
+ideas—that the 'direction' of any such Line is that property of it
+which determines its <i>position</i>, now that one Point in it is
+already fixed—that any <i>two</i> such Lines form 4 angles, whose
+common vertex is the fixed Point—that, if one of those 4 angles be
+zero, the 2 Lines <i>coincide</i>; if not, they <i>intersect</i>—that,
+in the first case, they have <i>the same direction</i>, in the second,
+<i>different directions</i>—and that the difference of the directions
+of such Lines is measured by <i>the angle between them</i>.</p>
+
+<p>But these ideas are of little use to us, when confronted with a
+given Line and a given Point <i>outside</i> it, and when told to
+imagine a new Line drawn, through the given Point, and '<i>in the
+same direction</i>' as the given Line, the 2 Lines having no common
+Point. For the directions of the Lines are no longer <i>directly
+comparable</i>. There is no use in asking "do they contain a
+zero-angle?" when they contain no angle <i>whatever</i>. An angle
+cannot exist without a <i>vertex</i>: and <i>where is the vertex</i>?</p>
+
+<p>What idea, then, is conveyed to the mind by the phrase "these Lines
+have <i>the same direction</i>"? If they were finite Lines, and the
+question concerned sameness of <i>length</i>, the process, of grasping
+this idea, would be a very simple one. We could either imagine one of
+the 2 Lines laid upon the other, and then apply the Axiom 'magnitudes
+which coincide are equal': or we could imagine a movable third Line,
+first applied to one of the 2 Lines and ascertained to have '<i>the
+same length</i>' with it,<span class="pagenum" id="Page_69">[Pg 69]</span> and then carried across the intervening
+space and applied to the other Line; and, on finding it to have 'the
+same length' with <i>that</i> also, we should pronounce the 2 Lines to
+have 'the same length,' in full confidence that our movable Line had
+preserved its length <i>unchanged during the journey</i>.</p>
+
+<p>Can we, then, use a similar process in grasping the idea of sameness
+of <i>direction</i>? That is, can we imagine a movable third
+Line, first applied to one of the 2 Lines and ascertained to have
+'<i>the same direction</i>' with it, and then carried across the
+intervening space and applied to the second Line, to see if it has
+'<i>the same direction</i>' with <i>it</i> also? But this process
+would convey to the mind no idea of '<i>sameness of direction</i>,'
+unless we had some guarantee that the movable Line had preserved its
+direction <i>unchanged during the journey</i>. The only process,
+for securing this, that presents itself to my mind, is to imagine a
+<i>transversal</i>, cutting the 2 Lines, and thus bridging over the
+intervening space, and then to imagine the movable Line shifted along
+it, so as always to cut it <i>at a constant angle</i>.</p>
+
+<p>I have thought this matter out very carefully, and I feel convinced
+that <i>this</i> is the mental process by which we grasp the idea of
+'<i>sameness of direction</i>,' when predicated of Lines that have no
+common Point, and therefore cannot be said to contain a zero-angle.</p>
+
+<p>Now this 'constant angle' is of course the angle at which the
+transversal cuts the first Line. Hence, the mental picture of a
+Line moving away from another, and yet maintaining '<i>sameness of
+direction</i>' with it, is the picture of its so moving as that a
+certain transversal shall cut the two Lines <i>at the same angle</i>.
+And this is my answer to our <i>first</i> question, namely, "what idea
+is conveyed to the average human intellect by the phrase '<i>in the
+same direction</i>,' when applied to non-coincident Lines?"</p>
+
+<p>If this be granted, our <i>second</i> question, "in accepting the
+Axiom that 'Lines, which have the same direction, make equal angles
+with all transversals,' to what other assertions are we<span class="pagenum" id="Page_70">[Pg 70]</span> committing
+ourselves?", must be answered "we are consciously committing ourselves
+to the assertion that Lines, which are equally inclined to a certain
+transversal, <i>are so to any transversal</i>."</p>
+
+<p>We see, then, that, if this be so, the advocates of the
+Direction-Theory have not escaped the necessity of assuming, as
+axiomatic, the second Theorem enunciated in § 2. (See p. 63.)</p>
+
+<p>I must ask the Reader's pardon for this long digression: but the
+fallacy, which (as I believe) lies at the root of the Direction-Theory,
+is a very subtle one, and has cost me a great many hours of hard
+thinking to un-earth it.</p>
+
+<p class="space-above2">
+Yet another process has been invented—quite fascinating in its
+brevity and its elegance—which, though involving the same fallacy as
+the Direction-Theory, proves Euc. I. 32 without even mentioning the
+dangerous word 'Direction.'</p>
+
+<figure class="figcenter width500" id="i_p070" style="width: 300px;">
+<img src="images/i_p070.jpg" width="300" height="213" alt="Lines EF
+and CAD intersecting, with small triangle ABG formed at their crossing
+point, and short ray AH below, illustrating a proposition about angles
+formed when a line crosses another with a small angular deviation.">
+</figure>
+
+<p>We are told to take any Triangle \(ABC\); to produce \(CA\) to \(D\);
+to make part of \(CD\), viz. \(AD\), revolve, about \(A\), into the
+position \(ABE\); then to make part of this Line, viz. \(BE\), revolve,
+about \(B\), into the position \(BCF\); and lastly to make part of this
+Line, viz. \(CF\), revolve, about \(C\), till it lies along \(CD\), of
+which it originally formed a part. We are then assured that it must
+have revolved through 4 right angles: from which it easily follows
+that the interior angles of the Triangle are together equal to 2 right
+angles.</p>
+
+<p><span class="pagenum" id="Page_71">[Pg 71]</span></p>
+
+<p>The disproof of this fallacy is almost as brief and elegant as the
+fallacy itself. We first quote the general principle that we cannot
+reasonably be told to make a Line fulfil <i>two</i> conditions, either
+of which is enough by itself to fix its position: e.g. given 3 Points,
+\(X\), \(Y\), \(Z\), we cannot reasonably be told to draw a Line, from
+\(X\), which shall pass through \(Y\) <i>and</i> \(Z\): we can make it
+pass through \(Y\), but it must then take its chance of passing through
+\(Z\); and <i>vice versâ</i>.</p>
+
+<p>Now let us suppose that, while one part of \(AE\), viz. \(BE\),
+revolves into the position \(BF\), another little bit of it, viz.
+\(AG\), revolves, through an equal angle, into the position \(AH\);
+and that, while \(CF\) revolves into the position of lying along
+\(CD\), \(AH\) revolves—and here comes the fallacy. You must not say
+"revolves, through an equal angle, into the position of lying along
+\(AD\)," for this would be to make \(AH\) <i>fulfil two conditions at
+once</i>. If you say that the one condition involves the other, you are
+virtually asserting that the Lines \(CF\), \(AH\) are equally inclined
+to \(CD\)—and this in <i>consequence</i> of \(AH\) having been so
+drawn that these same Lines are equally inclined to \(AE\). That is,
+you are asserting § 2. (2). (See p. 63.)</p>
+
+<p class="space-above2">
+One other proof—a very beautiful one, though largely dealing with
+Infinities and Infinitesimals—may here be mentioned, that of M.
+Bertrand. It rests on the principle that an infinite <i>Sector</i>
+(with a vertical angle which has a finite ratio to a right angle)
+is an Infinity of the <i>same</i> order as an infinite Plane,
+whereas an infinite <i>Strip</i> (i. e. the area contained between 2
+'separational' Lines) is an Infinity of a <i>lower</i> order. Hence he
+concludes that no such Sector, however small its vertical angle, will
+lie wholly between 2 such 'separational' Lines, however far apart:
+hence, if a Line intersect one of 2 'separational' Lines, it must, if
+produced, intersect the other. He thus proves the Theorem numbered (8)
+in § 4: but his results are, of course, only <i>partially</i>, and not
+<i>absolutely</i>, true.</p>
+
+<p><span class="pagenum" id="Page_72">[Pg 72]</span></p>
+
+<p>As a rather interesting example of the ease with which an unwary
+explorer may tumble into a pit-fall, I may refer to a "Note on Euclid's
+12th Axiom" by a Mr. W. Hanna, which will be found at p. 27 of Vol.
+XIII of "Mathematical Questions" reprinted from the "Educational
+Times." Mr. Hanna takes two Lines, situated as in Euclid's Axiom,
+and drops a perpendicular, from a point on one, upon the other: from
+the foot of this he drops a second perpendicular back upon the first
+line: and so on, backwards and forwards, till the diagram slightly
+resembles the side of a laced-up boot. He then easily proves that these
+perpendiculars continually decrease in length. From this he infers that
+"the perpendicular will ultimately become less than any assignable
+line"! The fallacy is really too obvious to be worth pointing out.</p>
+
+<p class="space-above2">
+Another writer in the "Educational Times" (Mr. J. Walmsley, B.A.: his
+article will be found at p. 103 of Vol. XVII of the Reprint) has fallen
+into a rather less obvious trap. He endeavours to prove that "any
+straight line, perpendicular to one of two parallel straight lines,
+will meet the other." (This, if it could be proved without assuming
+any disputable Axiom, would indeed be a splendid success! Mr. J.
+Walmsley has hardly realised, as yet, the <i>fearful</i> difficulty of
+persuading two Lines, under any conceivable circumstances, to do such
+a thing as "meet." Whether, at the outset of geometrical discovery,
+Lines were not properly introduced to each other—or whether some
+mischief-making Point has been insinuating that one of them went and
+intersected another when it was looking the other way—certain it is,
+that Lines will do almost <i>anything</i> you like to propose, rather
+than "meet" one another!) Mr. Walmsley assumes, as axiomatic, that when
+one Line lies between two others (whatever "between" may mean), those
+two others lie on <i>opposite</i> sides of it. Now let Mr. Walmsley
+(or any other champion of his theory) draw three Lines diverging
+from a Point at equal angles (of 120° each), and thus dividing the
+infinite Plane into three equal Sectors. In each of these Sectors let
+him draw a branch of a Hyperbola, having the sides of the Sector as
+its Asymptotes. Now, each of these Hyperbolæ lies (in a way) "between"
+the other two: and yet no two can be said to lie on opposite sides of
+the other one! "But," says Mr. Walmsley (or the champion aforesaid)
+"these are <i>Curves</i>, not <i>straight Lines</i>!" Most true: but
+how are you to know that straight Lines will not behave just like
+Hyperbolæ, if only they are put far enough apart? Produce those three
+radiating Lines, that we began with, until each is a million miles
+long (paper, pen, and ink, provided regardless of expense): then,
+across their extremities, draw three Lines perpendicular to them.
+Why shouldn't these three Lines (of course shunning a "meeting," as
+all Lines do) perpetually face inwards, so that no one of them ever
+commits the discourtesy of turning its back upon either of the other
+two? "It cannot be," say Messrs. Walmsley and Co.: "these Lines
+<i>must</i> intersect, if produced far enough." What? In consequence
+of their relative situation? That relative situation being, for each
+pair of them, that they make with a certain transversal two interior
+angles together less than two right angles? In assuming <i>this</i>,
+I very much fear that Messrs. Walmsley and Co. are performing the
+not-wholly-unprecedented feat of assuming Euclid's 12th Axiom!</p>
+
+
+<p class="nindc space-above2">§ 6.</p>
+<p><span class="pagenum" id="Page_73">[Pg 73]</span></p>
+
+<p class="nindc"><i>The Outlook.</i></p>
+
+<p>In conclusion let me address myself to the young and eager explorer
+who, Alpine-staff in hand, and duly furnished with all the necessaries
+for his perilous quest—pick-axe, theodolite, paper-collars, Brown's
+Sticking-Plaster, Jones's Cough-Pills, and Robinson's Insect-Powder,
+"the only known remedy for Phlebitis"—is preparing to sally forth, to
+do or die!</p>
+
+<p><span class="pagenum" id="Page_74">[Pg 74]</span></p>
+
+<p>To him let me address myself, as being, perchance, an older and a more
+experienced traveller—one who has wandered much, and pondered long,
+and who can best describe himself in the words of a lady well-known in
+the literary world (I am glad to have this opportunity of recording
+them, as they have never been printed. They were written "for music,"
+for which purpose, I imagine, the amount of <i>sense</i> required is
+not excessive).</p>
+
+<div class="poetry-container">
+<div class="poetry">
+  <div class="stanza">
+    <div class="verse indent0">"<i>I have wandered,</i></div>
+    <div class="verse indent0"><i>I have pondered,</i></div>
+    <div class="verse indent0"><i>I have squandered</i></div>
+    <div class="verse indent4"><i>Many a boon:</i></div>
+    <div class="verse indent0"><i>In the sadness,</i></div>
+    <div class="verse indent0"><i>In the gladness,</i></div>
+    <div class="verse indent0"><i>In the madness</i></div>
+    <div class="verse indent4"><i>Of the moon.</i></div>
+  </div>
+  <div class="stanza">
+    <div class="verse indent0">"<i>Seek thy pillow</i></div>
+    <div class="verse indent0"><i>By the billow,</i></div>
+    <div class="verse indent0"><i>Where the willow</i></div>
+    <div class="verse indent4"><i>Doth not weep:</i></div>
+    <div class="verse indent0"><i>Few will wonder</i></div>
+    <div class="verse indent0"><i>Who lies under,</i></div>
+    <div class="verse indent0"><i>Hearing thunder,</i></div>
+    <div class="verse indent4"><i>Fast asleep!</i>"</div>
+  </div>
+</div>
+</div>
+
+<p>Poetry like this speaks for itself: vain were it to hope that any poor
+words of mine could serve to illuminate, or even elucidate, its almost
+ethereal beauty!</p>
+
+<p>To what point of the compass, then, should this young and eager
+explorer be advised to direct his steps?</p>
+
+<p>I think his <i>best</i> chance—and that only a slender one—is to find
+some elementary proof for my Axiom, or for one of the many Theorems
+which will serve the same purpose, a few of which I will enumerate.
+In the first place, <i>any</i> Polygon will do, and <i>any</i> ratio,
+between it and the out-lying Segment, so long as<span class="pagenum" id="Page_75">[Pg 75]</span> it is a <i>finite</i>
+ratio. What I want it for is to prove that there is <i>some</i>
+isosceles Triangle, with a definite vertical angle (i. e. some named
+fraction of a right angle), whose base is less than one of its sides.
+And that, again, is wanted in order to prove it possible to draw, on a
+given base, an isosceles Triangle, whose base-angles shall have some
+nameable value. And that, again, is wanted in order to prove that there
+is <i>some</i> finite minimum value for the sum of the angles of a
+Triangle. And that, again, is wanted in order to prove Prop. 3, at p.
+19. So, if any one of these propositions could be either assumed as an
+Axiom or proved as a Theorem, it would suffice for the proof of Euc. I.
+32 and Co.</p>
+
+<p>If, for example, you will grant me, as an Axiom, that there is
+<i>some</i> finite minimum value for the sum of the angles of a
+Triangle, no matter how small you make it, all is easy at once: grant
+me, for instance, that no Triangle can possibly have the sum of its
+angles less than a millionth of a right angle, and I am happy!</p>
+
+<p>Finally, I am inclined to believe that, if ever Euc. I. 32 is proved
+without a new Axiom, it will be by some new and ampler definition
+of <i>the Right Line</i>—some definition which shall connote that
+peculiar and mysterious property, which it must somehow possess, which
+causes Euc. I. 32 to be true. Try <i>that</i> track, my gentle Reader!
+It is not much trodden as yet. And may success attend your search!</p>
+
+
+<p class="nindc space-above2 space-below2">THE END.</p>
+
+
+<p class="right space-above2 space-below2">[TURN OVER.</p>
+
+
+<hr class="chap x-ebookmaker-drop">
+
+<div class="chapter">
+<h2 class="nobreak" id="WORKS_BY_C_L_DODGSON">WORKS BY C. L. DODGSON.</h2>
+</div>
+
+<p class="nindc space-above2">PUBLISHED BY</p>
+
+<p class="nindc">MACMILLAN &amp; CO., LONDON.</p>
+
+<hr class="r5">
+
+<p class="hanging2 space-below2">
+<b>THE FORMULÆ OF PLANE TRIGONOMETRY</b>, printed with symbols, instead
+of words, to express the goniometrical ratios. (First published in
+1861.) Crown 4to, sewed. Price 1<i>s.</i></p>
+
+<p class="hanging2 space-below2">
+<b>EUCLID AND HIS MODERN RIVALS.</b> (First published in 1879.) Second
+Edition, 1885. Crown 8vo, cloth. Price 6<i>s.</i></p>
+
+<p class="hanging2 space-below2">
+<b>SUPPLEMENT TO FIRST EDITION OF "EUCLID AND HIS MODERN RIVALS</b>,"
+containing a Notice of Henrici's Geometry, together with Selections
+from the Reviews. (First published in 1885.) Crown 8vo, sewed. Price
+1<i>s.</i></p>
+
+<p class="hanging2 space-below2">
+<b>EUCLID, BOOKS I, II</b>, with Notes. (First published in 1882.)
+Sixth Edition, 1888. Crown 8vo, cloth. Price 2<i>s.</i></p>
+
+<p class="hanging2">
+<b>CURIOSA MATHEMATICA.</b></p>
+
+<p>Part I. A New Theory of Parallels. (First published in 1888.) Fourth
+Edition, 1895. Crown 8vo, cloth. Price 2<i>s.</i></p>
+
+<p>Part II. Pillow-Problems, thought out during Wakeful Hours. (First
+published in 1893.) Fourth Edition, 1895. Crown 8vo, cloth. Price
+2<i>s.</i></p>
+
+
+
+<hr class="chap x-ebookmaker-drop">
+
+<div class="chapter">
+<div class="transnote spa1">
+<p class="nindc"><b>TRANSCRIBER’S NOTES</b></p>
+
+<p>Simple typographical errors have been silently corrected; unbalanced
+quotation marks were remedied when the change was obvious, and
+otherwise left unbalanced.</p>
+
+<p>Punctuation, hyphenation, and spelling were made consistent when a
+predominant preference was found in the original book; otherwise they
+were not changed.</p>
+</div></div>
+<div style='text-align:center'>*** END OF THE PROJECT GUTENBERG EBOOK 78586 ***</div>
+</body>
+</html>
diff --git a/78586-src/README-math.txt b/78586-src/README-math.txt
new file mode 100644
index 0000000..ba31115
--- /dev/null
+++ b/78586-src/README-math.txt
@@ -0,0 +1,115 @@
+MathJax HTML source file instructions
+====================
+This project has math with high complexity equations (integrals, partial
+derivatives, gradients, matrices, etc.), grouped equations, etc. The source is
+a preliminary HTML file that uses MathJax to define mathematical expressions,
+which is processed to generate a final HTML file, along with SVG images with
+file names consisting of a number and “.svg”, such as “4.svg”.
+
+This source file is kept for the purpose of applying errata fixes. Although the
+MathJax takes some learning, it is clearer than the generated final. This also
+allows the SVG images to be regenerated with changes.
+
+Owing to the complexity of this type of book, it is not required to submit a
+text file.
+
+Inline svg note
+====================
+It is possible to use inline svg in the html file rather than images. It is
+essential in this case also to keep the source file.
+
+
+MathML note
+====================
+For books with less complex math, this may be a good choice. It does not
+require extra processing.
+
+
+Tools
+====================
+See the ppmath GitHub repository:
+    https://github.com/DistributedProofreaders/ppmath.
+Follow the instructions to install m2svg. Don't forget to run "npm update -g"
+before installing m2svg.
+
+Command line:
+    m2svg -i input.htm -o output.htm
+
+- The SVG files will be placed in a subdirectory of the working directory
+  called "images".
+
+- In the converted file, the maths expressions, delimited by the tags `\[`
+  and `\]` for *display* expressions or `\(` and `\)` for *inline*
+  expressions, are replaced by `<img>` links.
+
+- The "data-tex" attribute will contain the original maths expression.
+
+
+Inline code example
+====================
+For the expression AB^2 = AG × BD
+the input
+    `\(\mathrm{AB}^{2} = \mathrm{AG} \times \mathrm{BD}\)`
+
+becomes
+    `<span class="nowrap"><img style="vertical-align: -0.186ex; width: 16.872ex;
+    height: 2.253ex;" src="images/4.svg" alt="" data-tex="\mathrm{AB}^{2}
+    = \mathrm{AG} \times \mathrm{BD}">,</span>`
+
+The file images/4.svg displays the desired expression.
+
+
+Source files structure on Project Gutenberg
+====================
+(eBook 75107 is used as an example)
+
+- 75107/
+  - 75107-0.txt           (optional)
+  - 75107-h/
+    - 75107-h.htm         (final HTML file)
+    - images/             (generated SVGs + other images)
+      - cover.jpg
+      - 1.svg
+      - (etc.)
+  - 75107-source/
+    - 75107-source.htm    (source HTML file with MathJax)
+    - README-math.txt     (this file)
+
+
+Submission process
+====================
+- Generated final HTML and images should be submitted as normal.
+- In addition, the source HTML will be included in a "source/" subdirectory.
+- This readme will also be included in the "source/" subdirectory.
+  - Having it with the eBook makes it obvious, and avoids issues with
+    procedures changing in the future.
+- The submitter should add a note that this is a MathJax project, and whether a
+  text file is included.
+- Workflow will rename the source/ subdirectory and HTML file.
+
+
+Errata process
+====================
+(eBook 75107 is used as an example)
+
+1. Download the project files using Errata Workbench, and unzip.
+2. Install m2svg if not already done.
+3. Make the desired changes to 75107-source.htm.
+4. Execute command line "m2svg -i 75107-source.htm -o 75107-h.htm".
+   - The image files will be placed in a subdirectory of the working directory
+     called "images".
+   - Check error report: filename_svgerr.err (as generated by Guiguts 2)
+   - Preview 75107-h.htm in a browser to make sure that all changes are properly
+     added.
+5. Move 75107-h.htm to the 75107-h directory.
+6. Move the contents of the images directory to the 75107-h/images directory.
+   - The generated images are just a number with .svg extension. There could be
+     other svg files which are not generated by m2svg. It is best to remove the
+     existing generated files from 75107-h/images first.
+   - Any non-generated images should be kept. Otherwise, if the edit results in
+     there being fewer generated images than before, some old ones could be left
+     behind.
+7. Remove the generated images subdirectory and any other temporary files.
+8. Run PPhtml from Guiguts 2 or Post-Processing Workbench to verify all images
+   are used, and none are missing.
+9. Zip the project directory and upload to Errata Workbench.
author	pgww <pgww@lists.pglaf.org>	2026-05-22 04:26:47 -0700
committer	pgww <pgww@lists.pglaf.org>	2026-05-22 04:26:47 -0700
commit	910d9b48123eaeb874a8b5ef170bfa7461528c0e (patch)
tree	788cfc2c53ce9592463e4f578d6aa54fb6accd3e
parent	2d7780f2b7089667375296b0fcbd54a379917ea3 (diff)