Source — Euclid's Elements, encoded as an rrxiv paper

1% book05.tex --- Book V of Euclid's Elements: Eudoxean Theory of Proportion.
2%
3% All 25 propositions encoded. The proofs are essentially algebraic
4% manipulations of the equimultiples definition (V.5), so the dependency
5% structure is dense within Book V itself rather than reaching back into
6% Book I. Cross-book dependencies on V are heavy: Book VI (similarity)
7% and Book XII (exhaustion) both rely on this machinery.
8%
9% Wording follows Heath (1908).
10
11\section{Book V --- Eudoxean Theory of Proportion}
12\label{sec:book-V}
13
14\begin{claim}[Proposition V.1: Multiplication is distributive over magnitudes]
15\label{prop:V.1}
16If there be any number of magnitudes whatever which are, respectively,
17equimultiples of any magnitudes equal in multitude, then, whatever
18multiple one of the magnitudes is of one, that multiple also will all
19be of all.
20\end{claim}
21\begin{evidence}[Proof of V.1]
22\label{ev:V.1}
23Each magnitude is a sum of $m$ copies of the corresponding base
24magnitude.  Sum across the $n$ magnitudes; by Common Notion 2 the
25total is $m$ copies of the sum.
26\dependson{V.1}{cn:2}
27\dependson{V.1}{def:V.2}
28\end{evidence}
29
30\begin{claim}[Proposition V.2: Equimultiples sum to equimultiples]
31\label{prop:V.2}
32If a first magnitude be the same multiple of a second that a third is
33of a fourth, and a fifth also be the same multiple of the second that
34a sixth is of the fourth, then the sum of the first and fifth will
35also be the same multiple of the second that the sum of the third and
36sixth is of the fourth.
37\end{claim}
38\begin{evidence}[Proof of V.2]
39\label{ev:V.2}
40If $a = mb$ and $c = md$, $e = nb$ and $f = nd$, then $a + e = (m+n)b$
41and $c + f = (m+n)d$ by Common Notion 2.
42\dependson{V.2}{V.1}
43\dependson{V.2}{cn:2}
44\dependson{V.2}{def:V.2}
45\end{evidence}
46
47\begin{claim}[Proposition V.3: Multiple of a multiple is a multiple]
48\label{prop:V.3}
49If a first magnitude be the same multiple of a second that a third is
50of a fourth, and if equimultiples be taken of the first and third,
51they will also be equimultiples respectively, the one of the second
52and the other of the fourth.
53\end{claim}
54\begin{evidence}[Proof of V.3]
55\label{ev:V.3}
56If $a = mb$ and $c = md$, then $na = (nm)b$ and $nc = (nm)d$ by
57iterated application of V.1.
58\dependson{V.3}{V.1}
59\dependson{V.3}{def:V.2}
60\end{evidence}
61
62\begin{claim}[Proposition V.4: Proportionality preserved under equimultiples]
63\label{prop:V.4}
64If a first magnitude have to a second the same ratio as a third to a
65fourth, any equimultiples whatever of the first and third will also
66have the same ratio to any equimultiples whatever of the second and
67fourth respectively, taken in corresponding order.
68\end{claim}
69\begin{evidence}[Proof of V.4]
70\label{ev:V.4}
71By definition V.5, the original proportion gives an equimultiple
72relation across all multipliers.  Substituting $m a$ for $a$ and $m c$
73for $c$ throughout simply rescales the test multipliers; the test
74itself still passes for the rescaled magnitudes.
75\dependson{V.4}{V.3}
76\dependson{V.4}{def:V.5}
77\end{evidence}
78
79\begin{claim}[Proposition V.5: Subtraction of equimultiples]
80\label{prop:V.5}
81If a magnitude be the same multiple of a magnitude that a subtracted
82part is of a subtracted part, the remainder also will be the same
83multiple of the remainder that the whole is of the whole.
84\end{claim}
85\begin{evidence}[Proof of V.5]
86\label{ev:V.5}
87If $a = mb$ and $a' = mb'$ with $a' < a$, then $a - a' = m(b - b')$
88by Common Notion 3 applied to each of the $m$ copies.
89\dependson{V.5}{cn:3}
90\dependson{V.5}{def:V.2}
91\end{evidence}
92
93\begin{claim}[Proposition V.6: Difference of equimultiples is again an equimultiple]
94\label{prop:V.6}
95If two magnitudes be equimultiples of two magnitudes, and any
96magnitudes subtracted from them be equimultiples of the same, the
97remainders also are either equal to the same or equimultiples of them.
98\end{claim}
99\begin{evidence}[Proof of V.6]
100\label{ev:V.6}
101With $a = mb$, $c = md$, and subtractions $a' = nb$, $c' = nd$:
102$a - a' = (m-n)b$ and $c - c' = (m-n)d$ by Common Notion 3.  If $m = n$
103the remainders are zero (equal); otherwise both remainders are
104$(m-n)$-fold of $b$, $d$ respectively.
105\dependson{V.6}{V.5}
106\dependson{V.6}{cn:3}
107\end{evidence}
108
109\begin{claim}[Proposition V.7: Equal magnitudes have the same ratio to a third]
110\label{prop:V.7}
111Equal magnitudes have to the same the same ratio, as also has the
112same to equal magnitudes.
113\end{claim}
114\begin{evidence}[Proof of V.7]
115\label{ev:V.7}
116Let $a = b$ and $c$ arbitrary.  For any equimultiples $ma$, $mb$ of
117$a$, $b$, and $nc$ of $c$, the test of V.5 succeeds vacuously because
118$ma = mb$.  Therefore $a:c = b:c$ and $c:a = c:b$.
119\dependson{V.7}{def:V.5}
120\dependson{V.7}{cn:1}
121\end{evidence}
122
123\begin{claim}[Proposition V.8: Greater magnitude has greater ratio]
124\label{prop:V.8}
125Of unequal magnitudes the greater has to the same a greater ratio
126than the less has, and the same has to the less a greater ratio than
127it has to the greater.
128\end{claim}
129\begin{evidence}[Proof of V.8]
130\label{ev:V.8}
131Let $a > b$.  Pick a multiplier $n$ such that $n(a-b) > c$ (the
132Archimedean property of magnitudes, Definition V.4) and an $m$ such
133that $mc$ falls between $nb$ and $na$.  Then $na > mc$ but $nb < mc$;
134by Definition V.7 this is precisely the assertion $a:c > b:c$.
135\dependson{V.8}{def:V.4}
136\dependson{V.8}{def:V.7}
137\end{evidence}
138
139\begin{claim}[Proposition V.9: Magnitudes with the same ratio to a third are equal]
140\label{prop:V.9}
141Magnitudes which have the same ratio to the same are equal to one
142another; and magnitudes to which the same has the same ratio are
143equal.
144\end{claim}
145\begin{evidence}[Proof of V.9]
146\label{ev:V.9}
147Contrapositive of V.8: if $a \neq b$, then $a:c \neq b:c$.  Hence if
148$a:c = b:c$ then $a = b$.  Same argument with $c$ as antecedent.
149\dependson{V.9}{V.8}
150\end{evidence}
151
152\begin{claim}[Proposition V.10: Greater ratio implies greater antecedent]
153\label{prop:V.10}
154Of magnitudes which have a ratio to the same, that which has a greater
155ratio is greater; and that to which the same has a greater ratio is
156less.
157\end{claim}
158\begin{evidence}[Proof of V.10]
159\label{ev:V.10}
160Same contrapositive of V.8: if $a:c > b:c$ then $a > b$, since if
161$a \le b$ then $a:c \le b:c$ by V.7 or V.8 applied in reverse.
162\dependson{V.10}{V.7}
163\dependson{V.10}{V.8}
164\end{evidence}
165
166\begin{claim}[Proposition V.11: Transitivity of ratios]
167\label{prop:V.11}
168Ratios which are the same with the same ratio are also the same with
169one another.
170\end{claim}
171\begin{evidence}[Proof of V.11]
172\label{ev:V.11}
173If $a:b = c:d$ and $c:d = e:f$, then for any equimultiples the same
174inequality test holds for $(a,b)$ as for $(c,d)$, and that same test
175holds for $(c,d)$ as for $(e,f)$; hence by transitivity of the
176inequality test, the test holds for $(a,b)$ versus $(e,f)$.
177\dependson{V.11}{def:V.5}
178\end{evidence}
179
180\begin{claim}[Proposition V.12: Sum of antecedents to sum of consequents]
181\label{prop:V.12}
182If any number of magnitudes be proportional, as one of the antecedents
183is to one of the consequents, so will all the antecedents be to all
184the consequents.
185\end{claim}
186\begin{evidence}[Proof of V.12]
187\label{ev:V.12}
188Let $a_i : b_i$ all equal $r$ in the sense of Definition V.5.  For
189any test multipliers $m$, $n$ the sign of $m a_i - n b_i$ is the same
190for every $i$; therefore the sign of $m \sum a_i - n \sum b_i$ is the
191same too.  By V.5 this is the equimultiples test for $\sum a_i : \sum
192b_i = r$.
193\dependson{V.12}{V.1}
194\dependson{V.12}{V.2}
195\dependson{V.12}{def:V.5}
196\end{evidence}
197
198\begin{claim}[Proposition V.13: Substitution into a greater ratio]
199\label{prop:V.13}
200If a first magnitude have to a second the same ratio as a third to a
201fourth, and the third have to the fourth a greater ratio than a fifth
202has to a sixth, the first will also have to the second a greater ratio
203than the fifth to the sixth.
204\end{claim}
205\begin{evidence}[Proof of V.13]
206\label{ev:V.13}
207Combine V.11 (sameness transitivity) with Definition V.7 (greater
208ratio): the witness equimultiples for $c:d > e:f$ work for $a:b$ via
209the V.5 sameness of $a:b$ and $c:d$.
210\dependson{V.13}{V.11}
211\dependson{V.13}{def:V.5}
212\dependson{V.13}{def:V.7}
213\end{evidence}
214
215\begin{claim}[Proposition V.14: Ordering of magnitudes follows ordering of ratios]
216\label{prop:V.14}
217If a first magnitude have to a second the same ratio as a third to a
218fourth, and the first be greater than the third, the second will also
219be greater than the fourth; and if equal, equal; and if less, less.
220\end{claim}
221\begin{evidence}[Proof of V.14]
222\label{ev:V.14}
223By V.8 and V.13: if $a > c$, then $a:b > c:b$.  Combined with $a:b =
224c:d$ (the hypothesis), V.13 gives $c:d > c:b$, whence $b > d$ by V.10.
225\dependson{V.14}{V.8}
226\dependson{V.14}{V.10}
227\dependson{V.14}{V.13}
228\end{evidence}
229
230\begin{claim}[Proposition V.15: Parts have the same ratio as multiples]
231\label{prop:V.15}
232Parts have the same ratio as the same multiples of them taken in
233corresponding order.
234\end{claim}
235\begin{evidence}[Proof of V.15]
236\label{ev:V.15}
237If $A = mB$ and $C = mD$, group the $m$ copies of $B$ and $D$ in
238parallel.  Each parallel pair $(B_i, D_i)$ satisfies $B_i : D_i = B :
239D$ (identity), and V.12 sums these to give $A : C = B : D$.
240\dependson{V.15}{V.12}
241\dependson{V.15}{def:V.2}
242\end{evidence}
243
244\begin{claim}[Proposition V.16: Alternation of proportionals]
245\label{prop:V.16}
246If four magnitudes be proportional, they will also be proportional
247alternately.
248\end{claim}
249\begin{evidence}[Proof of V.16]
250\label{ev:V.16}
251Let $a : b = c : d$.  Test $a : c$ against $b : d$ with equimultiples
252$ma$, $mc$, $nb$, $nd$: by V.4 the original proportion lifts to $ma :
253mb = nc : nd$, and by V.15 to $ma : nc = mb : nd$.  The sign of $ma -
254nc$ matches the sign of $mb - nd$ for all $m$, $n$, which by V.5 is
255the alternated proportion $a : c = b : d$.
256\dependson{V.16}{V.4}
257\dependson{V.16}{V.11}
258\dependson{V.16}{V.15}
259\dependson{V.16}{def:V.5}
260\end{evidence}
261
262\begin{claim}[Proposition V.17: Separation of proportions]
263\label{prop:V.17}
264If magnitudes composed be proportional, they will also be proportional
265separando.
266\end{claim}
267\begin{evidence}[Proof of V.17]
268\label{ev:V.17}
269If $(a + b) : b = (c + d) : d$, then subtracting the consequents from
270the antecedents using V.5 / V.6 gives $a : b = c : d$.
271\dependson{V.17}{V.5}
272\dependson{V.17}{V.6}
273\dependson{V.17}{def:V.5}
274\end{evidence}
275
276\begin{claim}[Proposition V.18: Composition of proportions]
277\label{prop:V.18}
278If magnitudes separated be proportional, they will also be
279proportional componendo.
280\end{claim}
281\begin{evidence}[Proof of V.18]
282\label{ev:V.18}
283The converse of V.17: if $a : b = c : d$, then by V.2 plus V.4
284combined, $(a + b) : b = (c + d) : d$.
285\dependson{V.18}{V.2}
286\dependson{V.18}{V.4}
287\dependson{V.18}{V.17}
288\end{evidence}
289
290\begin{claim}[Proposition V.19: Subtraction of proportionals]
291\label{prop:V.19}
292If, as a whole is to a whole, so is a part subtracted to a part
293subtracted, the remainder will also be to the remainder as whole to
294whole.
295\end{claim}
296\begin{evidence}[Proof of V.19]
297\label{ev:V.19}
298If $a : c = a' : c'$ with $a' < a$, $c' < c$, apply V.17 (separation)
299to obtain $(a - a') : a' = (c - c') : c'$, and V.11 / V.16 to
300re-express as $(a - a') : (c - c') = a : c$.
301\dependson{V.19}{V.16}
302\dependson{V.19}{V.17}
303\end{evidence}
304
305\begin{claim}[Proposition V.20: Ex aequali for three magnitudes]
306\label{prop:V.20}
307If there be three magnitudes, and others equal to them in multitude,
308which taken two and two are in the same ratio, and if ex aequali the
309first be greater than the third, the fourth will also be greater than
310the sixth; and if equal, equal; and if less, less.
311\end{claim}
312\begin{evidence}[Proof of V.20]
313\label{ev:V.20}
314With $a : b = d : e$ and $b : c = e : f$, the relative order of $a$
315versus $c$ matches the relative order of $d$ versus $f$ by repeated
316application of V.13 and V.14 across the three pairs.
317\dependson{V.20}{V.13}
318\dependson{V.20}{V.14}
319\end{evidence}
320
321\begin{claim}[Proposition V.21: Ex aequali in perturbed proportion]
322\label{prop:V.21}
323If there be three magnitudes, and others equal to them in multitude,
324which taken two and two together are in the same ratio, and the
325proportion of them be perturbed, then if ex aequali the first be
326greater than the third, the fourth will also be greater than the
327sixth.
328\end{claim}
329\begin{evidence}[Proof of V.21]
330\label{ev:V.21}
331A perturbed version of V.20: with $a : b = e : f$ and $b : c = d : e$
332the ratio chain still imposes the same ordering between $a$ versus
333$c$ and $d$ versus $f$, by V.13 and V.14.
334\dependson{V.21}{V.13}
335\dependson{V.21}{V.14}
336\dependson{V.21}{def:V.18}
337\end{evidence}
338
339\begin{claim}[Proposition V.22: Transitivity of ex aequali]
340\label{prop:V.22}
341If there be any number of magnitudes whatever, and others equal to
342them in multitude, which taken two and two together are in the same
343ratio, they will also be in the same ratio ex aequali.
344\end{claim}
345\begin{evidence}[Proof of V.22]
346\label{ev:V.22}
347Apply V.20 inductively across the chain $a_1 : a_2 = b_1 : b_2$, $a_2
348: a_3 = b_2 : b_3$, $\dots$, $a_{n-1} : a_n = b_{n-1} : b_n$ to
349conclude $a_1 : a_n = b_1 : b_n$.
350\dependson{V.22}{V.20}
351\dependson{V.22}{def:V.17}
352\end{evidence}
353
354\begin{claim}[Proposition V.23: Ex aequali in perturbed proportion (general)]
355\label{prop:V.23}
356If there be any number of magnitudes whatever, and others equal to
357them in multitude, which taken two and two together are in the same
358ratio, and the proportion of them be perturbed, they will also be in
359the same ratio ex aequali.
360\end{claim}
361\begin{evidence}[Proof of V.23]
362\label{ev:V.23}
363Inductive form of V.21.
364\dependson{V.23}{V.21}
365\dependson{V.23}{def:V.18}
366\end{evidence}
367
368\begin{claim}[Proposition V.24: Sums of antecedents are proportional]
369\label{prop:V.24}
370If a first magnitude have to a second the same ratio as a third has
371to a fourth, and also a fifth have to the second the same ratio as a
372sixth to the fourth, the first and fifth added together will have to
373the second the same ratio as the third and sixth have to the fourth.
374\end{claim}
375\begin{evidence}[Proof of V.24]
376\label{ev:V.24}
377With $a : b = c : d$ and $e : b = f : d$: by V.12 applied to the two
378proportions side-by-side, $(a + e) : b = (c + f) : d$.
379\dependson{V.24}{V.12}
380\dependson{V.24}{V.22}
381\end{evidence}
382
383\begin{claim}[Proposition V.25: Sum of extremes exceeds sum of means]
384\label{prop:V.25}
385If four magnitudes be proportional, the greatest and least are greater
386than the remaining two.
387\end{claim}
388\begin{evidence}[Proof of V.25]
389\label{ev:V.25}
390Let $a : b = c : d$ with $a$ greatest.  By V.14 $b > d$, so
391considering $a - c$ and $d - b$ or vice versa, the difference $a - c$
392equals $b - d$ in ratio, and rearrangement (with Common Notion 5)
393yields $a + d > b + c$.
394\dependson{V.25}{V.14}
395\dependson{V.25}{V.19}
396\dependson{V.25}{cn:5}
397\end{evidence}
398