Commentary on Proposition X.5

Back to Proposition X.5.

The argument is as follows. If $a, b$ be commensurable magnitudes, they have some common measure $c$, and

$$a = mc,$$

$$b = nc,$$

where $m, n$ are integers.

It follows that

$$c : a = 1 : m \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(1),$$

or, inversely,

$$a : c = m : 1;$$

and also that

$$c : b = 1 : n,$$

so that, ex aequali,

$$a : b = m : n.$$

It will be observed that, in stating the proportion (1), Euclid is merely expressing the fact that $a$ is the same multiple of $c$ that $m$ is of $1$. In other words, he rests the statement on the definition of proportion in VII. Def. 20. This, however, is applicable only to four numbers, and $c, a$ are not numbers but magnitudes. Hence the statement of the proportion is not legitimate unless it is proved that it is true in the sense of V. Def. 5 with regard to magnitudes in general, the numbers $1, m$ being magnitudes. Similarly with regard to the other proportions in the proposition.

There is, therefore, a hiatus. Euclid ought to have proved that magnitudes which are proportional in the sense of VII. Def. 20 are also proportional in the sense of V. Def. 5, or that the proportion of numbers is included in the proportion of magnitudes as a particular case. Simson has proved this in his Proposition C inserted in Book V. (see Vol. II. pp. 126—8). The portion of that proposition which is required here is the proof that, if

$$ \left.\begin{aligned} a &= mb \\ c &= md \end{aligned}\right\}, $$

then

$$a : b = c : d\text{, in the sense of V. Def. 5.}$$

Take any equimultiples $pa, pc$ of $a, c$ and any equimultiples $qb, qd$ of $b, d$.

Now

$$ \left.\begin{aligned} pa &= pmb \\ pc &= pmd \end{aligned}\right\}. $$

But, according as $pmb \gtreqless qb$, $pmd \gtreqless qd$.

Therefore, according as $pa \gtreqless qb$, $pa \gtreqless qd$.[1]

And $pa, pc$ are any equimultiples of $a, c$, and $qb, qd$ any equimultiples of $b, d$.

Therefore

$$a : b = c : d. \qquad \text{[V. Def. 5.]}$$