Commentary on Proposition V.5

Back to Proposition V.5.

This proposition corresponds to V. 1, with subtraction taking the place of addition. It proves the formula

$$ma - mb = m(a - b).$$

Euclid’s construction assumes that, if AE is any multiple of CF, and EB is any other magnitude, a fourth straight line can be found such that EB is the same multiple of it that AE is of CF, or in other words that, given any magnitude, we can divide it into any number of equal parts. This is however not proved, even of straight lines, much less other magnitudes, until VI. 9. Peletarius had already seen this objection to the construction. The difficulty is not got over by regarding it merely as a hypothetical construction ; for hypothetical constructions are not in Euclid’s manner. The remedy is to substitute the alternative construction given by Simson, after Peletarius and Campanus’ translation from the Arabic, which only requires us to add a magnitude to itself a certain number of times. The demonstration follows Euclid’s line exactly.

G A E B C F D

“Take AG the same multiple of FD that AE is of CF; therefore AE is the same multiple of CF that EG is of CD. [V. 1]

But AE, by hypothesis, is the same multiple of CF that AB is of CD; therefore EG is the same multiple of CD that AB is of CD;

wherefore EG is equal to AB.

Take from them the common magnitude AE; the remainder AG is equal to the remainder EB.

Wherefore, since AE is the same multiple of CF that AG is of FD, and since AG is equal to EB,

therefore AE is the same multiple of CF that EB is of FD.

But AE is the same multiple of CF that AB is of CD; therefore EB is the same multiple of FD that AB is of CD.”

Euclid’s proof amounts to this.

Suppose a magnitude $x$ taken such that

$$ma - mb = mx, \text{ say.}$$

Add $mb$ to each side, whence (by V. 1)

$$ma = m(x + b).$$

Therefore

$$a = x + b, \text{ or } x = a - b,$$

so that

$$ma - mb = m(a - b).$$

Simson’s proof, on the other hand, argues thus.

Take $x = m(a - b)$, the same multiple of $(a - b)$ that $mb$ is of $b$.

Then, by addition of $mb$ to both sides, we have [V. 1]

$$x + mb = ma,$$

or

$$x = ma - mb.$$

That is,

$$ma - mb = m(a - b).$$