Commentary on Proposition V.10

Back to Proposition V.10.

No better example can, I think, be found of the acuteness which Simson brought to bear in his critical examination of the $Elements$, and of his great services to the study of Euclid, than is furnished by the admirable note on this proposition where he points out a serious flaw in the proof as given in the text.

For the first time Euclid is arguing about greater and less ratios, and it will be found by an examination of the steps of the proof that he assumes more with regard to the meaning of the terms than he is entitled to assume, having regard to the fact that the definition of greater ratio (Def. 7) is all that, as yet, he has to go upon. That we cannot argue, at present, about greater and less as applied to ratios in the same way as about the same terms in relation to magnitudes is indeed sufficiently indicated by the fact that Euclid does not assume for ratios what is in Book I. an axiom, viz. that things which are equal to the same thing are equal to one another; on the contrary, he proves, in Prop. 11, that ratios which are the same with the same ratio are the same with one another.

Let us now examine the steps of the proof in the text. First we are told that

“A is greater than B. For, if not, it is either equal to B or less than it. Now A is not equal to B; for in that case each of the two magnitudes A, B would have had the same ratio to C: [V. 7] but they have not:

therefore A is not equal to B.”

As Simson remarks, the force of this reasoning is as follows. If A has to C the same ratio as B has to C, then—supposing any equimultiples of A, B to be taken and any multiple of C— by Def. 5, if the multiple of A be greater than the multiple of C, the multiple of B is also greater than that of C.

But it follows from the hypothesis (that A has a greater ratio to C than B has to C) that, by Def. 7, there must be some equimultiples of A, B and some multiple of C such that the multiple of A is greater than the multiple of C, but the multiple of B is not greater than the same multiple of C.

And this directly contradicts the preceding deduction from the supposition that A has to C the same ratio as B has to C;

therefore that supposition is impossible.

The proof now goes on thus :

“Nor again is A less than B ; for, in that case, A would have had to C a less ratio than B has to C ; [V. 8] but it has not ;

therefore A is not less than B.”

It is here that the difficulty arises. As before, we must use Def. 7. “A would have had to C a less ratio than B has to C,” or the equivalent statement that B would have had to C a greater ratio than A has to C, means that there would have been some equimultiples of B, A and some multiple of C such that

  1. the multiple of B is greater than the multiple of C, but

  2. the multiple of A is not greater than the multiple of C,

and it ought to have been proved that this can never happen if the hypothesis of the proposition is true, viz. that A has to C a greater ratio than B has to C: that is, it should have been proved that, in the latter case, the multiple of A is always greater than the multiple of C whenever the multiple of B is greater than the multiple of C (for, when this is demonstrated, it will be evident that B cannot have a greater ratio to C than A has to C). But this is not proved (cf. the remark of De Morgan quoted in the note on V. Def. 7, p. 130), and hence it is not proved that the above inference from the supposition that A is less than B is inconsistent with the hypothesis in the enunciation. The proof therefore fails.

Simson suggests that the proof is not Euclid’s, but the work of some one who apparently “has been deceived in applying what is manifest, when understood of magnitudes, unto ratios, viz. that a magnitude cannot be both greater and less than another.”

The proof substituted by Simson is satisfactory and simple.

“Let A have to C a greater ratio than B has to C ; A is greater than B.

For, because A has a greater ratio to C than B has to C, there are some equimultiples of A, B and some multiple of C such that the multiple of A is greater than the multiple of C, but the multiple of B is not greater than it. [V. Def. 7]

Let them be taken, and let $D$, $E$ be equimultiples of A, B, and $F$ a multiple of C, such that

$D$ is greater than $F$,
but $E$ is not greater than $F$.
Therefore $D$ is greater than $E$.

And, because $D$ and $E$ are equimultiples of A and B, and $D$ is greater than $E$,

[Simson's 4th Ax.]therefore A is greater than B.

Next, let C have a greater ratio to B than it has to A ; B is less than A.

For there is some multiple $F$ of C and some equimultiples $E$ and $D$ of B and A such that

[V. Def. 7]$F$ is greater than $E$ but not greater than $D$.
Therefore $E$ is less than $D$ ;

and, because $E$ and $D$ are equimultiples of B and A,

therefore B is less than A.”