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justify-content: center;} +.poetry-container {text-align: center;} +.poetry {text-align: left; margin-left: 5%; margin-right: 5%;} +.poetry .stanza {margin: 1em auto;} +.poetry .verse {text-indent: -3em; padding-left: 3em;} + +/* Transcriber's notes */ +.transnote {background-color: #E6E6FA; + color: black; + font-size:small; + padding:0.5em; + margin-bottom:5em; + font-family:sans-serif, serif; +} + +/* Poetry indents */ +.poetry .indent0 {text-indent: -3em;} +.poetry .indent4 {text-indent: -1em;} + +/* css needed in m2svg output: displayed equations and prevention of bad breaks*/ + .align-center { + display: block; + text-align: center; + text-indent: 0; + margin-top: 1em; + margin-bottom: 1em; + } + .nowrap { + white-space: nowrap; + } + + + </style> +</head> + +<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, &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 '&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 &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 &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 \(ACD\), &c., equal to \(ABC\). With centre +\(A\), and distance \(AB\), describe quadrant passing through \(C\), +\(D\), &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 \(BC\), \(CD\), &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 &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 &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 &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 &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 &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 &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 &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 &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 &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 &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 &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 &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 &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 &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, +&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 " "</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"> " " 2nd "</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> </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, &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, &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, &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 & 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. |