Commentary on Proposition VII.34

Back to Proposition VII.34.

This is the problem of finding the least common multiple of two numbers, as $a, b$.

  1. If $a, b$ be prime to one another, the L.C.M. is $ab$.

For, if not, let it be $d$, some number less than $ab$. Then

$$d = ma = nb\text{, where } m, n \text{ are integers.}$$

Therefore

$$a : b = n : m, \quad \text{[VII. 19]}$$

and hence, $a, b$ being prime to one another,

$$b \text{ measures } m. \quad \text{[VII. 20, 21]}$$

But

$$b : m = ab : am \quad \text{[VII. 17]}$$

$$= ab : d.$$

Therefore $ab$ measures $d$: which is impossible.

  1. If $a, b$ be not prime to one another, find the numbers which are the least of those having the ratio of $a$ to $b$, say $m, n$; [VII. 33] then

$$a : b = m : n,$$

and

$$an = bm \text{ } (= c\text{, say}); \quad \text{[VII. 19]}$$

$c$ is then the L.C.M.

For, if not, let it be $d$ ($< c$), so that

$$ap = bq = d\text{, where } p, q \text{ are integers.}$$

Then

$$a : b = q : p, \quad \text{[VII. 19]}$$

whence

$$m : n = q : p,$$

so that

$$n \text{ measures } p. \quad \text{[VII. 20, 21]}$$

And

$$n : p = an : ap = c : d,$$

so that

$$c \text{ measures } d\text{:}$$

which is impossible.

Therefore etc.

By VII. 33,

$$ \left. \begin{aligned} m &= \frac{a}{g} \\ n &= \frac{b}{g} \end{aligned} \right\}, \text{ where } g \text{ is the G.C.M. of } a, b. $$

Hence the L.C.M. is $\frac{ab}{g}$.