Commentary on Proposition V.18

Back to Proposition V.18.

$$\begin{aligned} \text{Algebraically, if} \qquad a : b &= c : d, \\ \text{then} \qquad (a + b) : b &= (c + d) : d. \end{aligned}$$

In the enunciation of this proposition there is the same special use of διῃρημένα and συντεθέντα as there was of συγκείμενα and διαιρεθέντα in the last enunciation. Practically, as the algebraical form shows, διῃρημένα might have been left out.

The following is the method of proof employed by Euclid.

Given that $a : b = c : d$,

suppose, if possible, that

$(a + b) : b = (c + d) : (d \pm x)$.

Therefore, separando [V. 17],

$$\begin{aligned} a : b &= (c \mp x) : (d \pm x), \\ \text{whence, by V. 11,} \qquad (c \mp x) : (d \pm x) &= c : d. \\ \text{But} \qquad (c - x) < c, \text{ while } (d + x) &> d, \\ \text{and} \qquad (c + x) > c, \text{ while } (d - x) &< d, \end{aligned}$$

which relations respectively contradict V. 14.

Simson pointed out (as Saccheri before him saw) that Euclid’s demonstration is not legitimate, because it assumes without proof that to any three magnitudes, two of which, at least, are of the same kind, there exists a fourth proportional. Clavius and, according to him, other editors made this an axiom. But it is far from axiomatic; it is not till VI. 12 that Euclid shows, by construction, that it is true even in the particular case where the three given magnitudes are all straight lines.

In order to remove the defect it is necessary either (1) to prove beforehand the proposition thus assumed by Euclid or (2) to prove V. 18 independently of it.

Saccheri ingeniously proposed that the assumed proposition should be proved, for areas and straight lines, by means of Euclid VI. 1, 2 and 12. As he says, there was nothing to prevent Euclid from interposing these propositions immediately after V. 17 and then proving V. 18 by means of them. VI. 12 enables us to construct the fourth proportional when the three given magnitudes are straight lines; and VI. 12 depends only on VI. 1 and 2. “Now,” says Saccheri, “when we have once found the means of constructing a straight line which is a fourth proportional to three given straight lines, we obviously have the solution of the general problem ‘To construct a straight line which shall have to a given straight line the same ratio which two polygons have (to one another).’” For it is sufficient to transform the polygons into two triangles of equal height and then to construct a straight line which shall be a fourth proportional to the bases of the triangles and the given straight line.

The method of Saccheri is, as will be seen, similar to that adopted by Smith and Bryant (loc. cit.) in proving the theorems of Euclid V. 16, 17, 18, 22, so far as straight lines and rectilineal areas are concerned, by means of VI. 1.

De Morgan gives a sketch of a general proof of the assumed proposition that, $B$ being any magnitude, and $P$ and $Q$ two magnitudes of the same kind, there does exist a magnitude $A$ which is to $B$ in the same ratio as $P$ to $Q$.

“The right to reason upon any aliquot part of any magnitude is assumed; though, in truth, aliquot parts obtained by continual bisection would suffice: and it is taken as previously proved that the tests of greater and of less ratio are never both presented in any one scale of relation as compared with another” (see note on V. Def. 7 ad fin.).

“(1) If $M$ be to $B$ in a greater ratio than $P$ to $Q$, so is every magnitude greater than $M$, and so are some less magnitudes; and if $M$ be to $B$ in a less ratio than $P$ to $Q$, so is every magnitude less than $M$, and so are some greater magnitudes. Part of this is in every system: the rest is proved thus. If $M$ be to $B$ in a greater ratio than $P$ to $Q$, say, for instance, we find that $15M$ lies between $22B$ and $23B$, while $15P$ lies before $22Q$. Let $15M$ exceed $22B$ by $Z$; then, if $N$ be less than $M$ by anything less than the 15th part of $Z$, $15N$ is between $22B$ and $23B$: or $N$, less than $M$, is in a greater ratio to $B$ than $P$ to $Q$. And similarly for the other case.

  1. $M$ can certainly be taken so small as to be in a less ratio to $B$ than $P$ to $Q$, and so large as to be in a greater; and since we can never pass from the greater ratio back again to the smaller by increasing $M$, it follows that, while we pass from the first designated value to the second, we come upon an intermediate magnitude $A$ such that every smaller is in a less ratio to $B$ than $P$ to $Q$, and every greater in a greater ratio. Now $A$ cannot be in a less ratio to $B$ than $P$ to $Q$, for then some greater magnitudes would also be in a less ratio; nor in a greater ratio, for then some less magnitudes would be in a greater ratio; therefore $A$ is in the same ratio to $B$ as $P$ to $Q$. The previously proved proposition above mentioned shows the three alternatives to be the only ones.”

Alternative proofs of V. 18.

Simson bases his alternative on V. 5, 6. As the 18th proposition is the converse of the 17th, and the latter is proved by means of V. 1 and 2, of which V. 5 and 6 are converses, the proof of V. 18 by V. 5 and 6 would be natural; and Simson holds that Euclid must have proved V. 18 in this way because “the 5th and 6th do not enter into the demonstration of any proposition in this book as we have it, nor can they be of any use in any proposition of the Elements,” and “the 5th and 6th have undoubtedly been put into the 5th book for the sake of some propositions in it, as all the other propositions about equimultiples have been.”

Simson’s proof is however, as it seems to me, intolerably long and difficult to follow unless it be put in the symbolical form as follows.

Suppose that $a$ is to $b$ as $c$ is to $d$; it is required to prove that $a + b$ is to $b$ as $c + d$ is to $d$.

Take any equimultiples of the last four magnitudes, say

$m(a + b), \quad mb, \quad m(c + d), \quad md,$

and any equimultiples of $b, d$, as

$nb, \quad nd.$

Clearly, if

$nb$ is greater than $mb$, $nd$ is greater than $md$; if equal, equal; and if less, less.

  1. Suppose $nb$ not greater than $mb$, so that $nd$ is also not greater than $md$.

Now $m(a + b)$ is greater than $mb$: therefore $m(a + b)$ is greater than $nb$.
Similarly $m(c + d)$ is greater than $nd$.

  1. Suppose $nb$ greater than $mb$.

Since $m(a + b), mb, m(c + d), md$ are equimultiples of $(a + b), b, (c + d), d$,

$ma$ is the same multiple of $a$ that $m(a + b)$ is of $(a + b)$, and $mc$ is the same multiple of $c$ that $m(c + d)$ is of $(c + d)$, so that $ma, mc$ are equimultiples of $a, c$. [V. 5]

Again $nb, nd$ are equimultiples of $b, d$, and so are $mb, md$; therefore $(n - m)b, (n - m)d$ are equimultiples of $b, d$ and, whether $n - m$ is equal to unity or to any other integer [V. 6], it follows, by Def. 5, that, since $a, b, c, d$ are proportionals,

if $ma$ is greater than $(n - m)b$, then $mc$ is greater than $(n - m)d$; if equal, equal; and if less, less.

  1. If now $m(a + b)$ is greater than $nb$, subtracting $mb$ from each, we have

$ma$ is greater than $(n - m)b$; therefore $mc$ is greater than $(n - m)d$,

and, if we add $md$ to each,

$m(c + d)$ is greater than $nd$.

  1. Similarly it may be proved that,

if $m(a + b)$ is equal to $nb$, then $m(c + d)$ is equal to $nd$, and (3) that, if $m(a + b)$ is less than $nb$, then $m(c + d)$ is less than $nd$.

But (under I. above) it was proved that, in the case where $nb$ is not greater than $mb$,

$m(a + b)$ is always greater than $nb$, and $m(c + d)$ is always greater than $nd$.

Hence, whatever be the values of $m$ and $n$, $m(c + d)$ is always greater than, equal to, or less than $nd$ according as $m(a + b)$ is greater than, equal to, or less than $nb$.

Therefore, by Def. 5,

$a + b$ is to $b$ as $c + d$ is to $d$.

Todhunter gives the following short demonstration from Austin (Examination of the first six books of Euclid’s Elements).

“Let AE be to EB as CF is to FD:

AB shall be to BE as CD is to DF.

For, because AE is to EB as CF is to FD, therefore, alternately,

[V. 16]$AE$ is to $CF$ as EB is to FD.

And, as one of the antecedents is to its consequent, so is the sum of the antecedents to the sum of the consequents: [V. 12]

therefore, as $EB$ is to $FD$, so are AE, EB together to CF, FD together; that is,

$AB$ is to $CD$ as $EB$ is to $FD$.

Therefore, alternately,

$AB$ is to $BE$ as $CD$ is to $FD$.”
A E B C F D

The objection to this proof is that it is only valid in the case where the proposition V. 16 used in it is valid, i.e. where all four magnitudes are of the same kind.

Smith and Bryant’s proof avails where all four magnitudes are straight lines, where all four magnitudes are rectilineal areas, or where one antecedent and its consequent are straight lines and the other antecedent and its consequent rectilineal areas.

Suppose that

$A : B = C : D$.

$First$, let all the magnitudes be areas.

Construct a rectangle abcd equal to A, and to bc apply the rectangle bcef equal to B. Also to ab, bf apply the rectangles ag, bk equal to C, D respectively.

d c e a b f h g k

Then, since the rectangles ac, be have equal heights bc, they are to one another as their bases. [VI. 1]

Hence

ab : bf = rect. ac : rect. be = A : B = C : D = rect. ag : rect. bk.

Therefore [VI. 1, converse] the rectangles $ag, bk$ have the same height, so that k is on the straight line hg.

Hence

A + B : B = rect. ae : rect. be = af : bf = rect. ak : rect. bk = C + D : D.

$Secondly$, let the magnitudes $A$, $B$ be straight lines and the magnitudes $C$, $D$ areas.

Let ab, bf be equal to the straight lines A, B, and to ab, bf apply the rectangles ag, bk equal to C, D respectively. Then, as before, the rectangles ag, bk have the same height. Now

$$\begin{aligned} A + B : B &= af : bf \\ &= \text{rect. } ak : \text{rect. } bk \\ &= C + D : D. \end{aligned}$$

$Thirdly$, let all the magnitudes be straight lines. Apply to the straight lines $C$, $D$ rectangles $P$, $Q$ having the same height.

Then

$$P : Q = C : D. \quad [\text{VI. } 1]$$

Hence, by the second case,

$$A + B : B = P + Q : Q.$$

Also

$$P + Q : Q = C + D : D.$$

Therefore

$$A + B : B = C + D : D.$$