Commentary on Proposition V.8

Back to Proposition V.8.

The two separate cases found in the Greek text of the demonstration can practically be compressed into one. Also the expositor of the two cases makes them differ more than they need. It is necessary in each case to select the smaller of the two segments AE, EB of AB with a view to taking a multiple of it which is greater than D; in the first case therefore AE is taken, in the second EB. But, while in the first case successive multiples of D are taken in order to find the first multiple that is greater than GH (or K), in the second case the multiple is taken which is the first that is greater than FG. This difference is not necessary; the first multiple of D that is greater than GH would equally serve in the second case. Lastly, the use of the magnitude K might have been dispensed with in both cases; it is of no practical use and only lengthens the proofs. For these reasons Simson considers that Theon, or some other unskilful editor, has vitiated the proposition. This however seems an unsafe assumption; for, while it was not the habit of the great Greek geometers to discuss separately a number of different cases (e.g. in I. 7 and I. 35 Euclid proves one case and leaves the others to the reader), there are many exceptions to prove the rule, e.g. Eucl. III. 25 and 33; and we know that many fundamental propositions, afterwards proved generally, were first discovered in relation to particular cases and then generalised, so that Book V., presenting a comparatively new theory, might fairly be expected to exhibit more instances than the earlier books do of unnecessary subdivision. The use of the K is no more conclusive against the genuineness of the proofs.

Nevertheless Simson’s version of the proof is certainly shorter, and moreover it takes account of the case in which AE is equal to EB, and of the case in which AE, EB are both greater than D (though these cases are scarcely worth separate mention).

“If the magnitude which is not the greater of the two AE, EB be (1) not less than D, take FG, GH the doubles of AE, EB.

But if that which is not the greater of the two AE, EB be (2) less than D, this magnitude can be multiplied so as to become greater than D whether it be AE or EB.

Let it be multiplied until it becomes greater than D, and let the other be multiplied as often; let FG be the multiple thus taken of AE and GH the same multiple of EB,

therefore FG and GH are each of them greater than D.

And, in every one of the cases, take L the double of D, M its triple and so on, till the multiple of D be that which first becomes greater than GH.

Let N be that multiple of D which is first greater than GH, and M the multiple of D which is next less than N.

Then, because N is the multiple of D which is the first that becomes greater than GH, the next preceding multiple is not greater than GH ;

that is, GH is not less than M.

And, since FG is the same multiple of AE that GH is of EB, GH is the same multiple of EB that FH is of AB ; [V. 1] wherefore FH, GH are equimultiples of AB, EB.

And it was shown that GH was not less than M ;

and, by the construction, FG is greater than D ;
therefore the whole FH is greater than M, D together.

But M, D together are equal to N ;

therefore FH is greater than N.

But GH is not greater than N ; and FH, GH are equimultiples of AB, BE,

and N is a multiple of D ;
[V. Def. 7]therefore AB has to D a greater ratio than BE (or C) has to D.

Also D has to BE a greater ratio than it has to AB.

For, having made the same construction, it may be shown, in like manner, that N is greater than GH but that it is not greater than FH ; and N is a multiple of D, and GH, FH are equimultiples of EB, AB ;

[V. Def. 7]Therefore D has to EB a greater ratio than it has to AB.”

The proof may perhaps be more readily grasped in the more symbolical form thus.

Take the $m$th equimultiples of C, and of the excess of AB over C (that is, of AE), such that each is greater than D ; and, of the multiples of D, let $pD$ be the first that is greater than $mC$, and $nD$ the next less multiple of D.

Then, since $mC$ is not less than $nD$, and, by the construction, $m(AE)$ is greater than D, the sum of $mC$ and $m(AE)$ is greater than the sum of $nD$ and D.

That is, m(AB) is greater than $pD$.

And, by the construction, $mC$ is less than $pD$.

Therefore [V. Def. 7] $AB$ has to D a greater ratio than C has to D.

Again, since $pD$ is less than $m(AB)$, and $pD$ is greater than $mC$,

D has to C a greater ratio than D has to AB.