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Curiosa Mathematica: Part I | Project Gutenberg
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<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>
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<p>Simple typographical errors have been silently corrected; unbalanced
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