Commentary on Proposition V.2

Back to Proposition V.2.

[Folio 0146 | p. 140]

To find the corresponding formula for the result of this proposition, we may suppose $a$ to be the “second” magnitude and $b$ the “fourth.” If now the “first” magnitude is $ma$, the “third” is, by hypothesis, $mb$; and, if the “fifth” magnitude is $na$, the “sixth” is $nb$. The proposition then asserts that $ma + na$ is the same multiple of $a$ that $mb + nb$ is of $b$.

More generally, if $pa$, $qa \dots$ and $pb$, $qb \dots$ be any further equimultiples of $a$, $b$ respectively, $ma + na + pa + qa + \dots$ is the same multiple of $a$ that $mb + nb + pb + qb + \dots$ is of $b$. This extension is stated in Simson’s corollary to V. 2 thus:

“From this it is plain that, if any number of magnitudes AB, BG, $GH$ be multiples of another C; and as many DE, $EK$, $KL$ be the same multiples of F, each of each; the whole of the first, viz. $AH$, is the same multiple of C that the whole of the last, viz. $DL$, is of F.”

The course of the proof, which separates $m$ into its units and also $n$ into its units, practically tells us that the multiple of $a$ arrived at by adding the two multiples is the $(m + n)$th multiple; or practically we are shown that

$$ma + na = (m + n)a,$$

or, more generally, that

$$ma + na + pa + \dots = (m + n + p + \dots)a.$$