# Construction of Sequence of Numbers with Given Ratios

## Theorem

In the words of Euclid:

Given as many ratios as we please in least numbers, to find numbers in continued proportion which are the least in the given ratios.

## Proof

Let the given ratios in least numbers be:

$a : b$, $c : d$, $e : f$
$g = \lcm set {b, c}$

Let:

$g = r b$
$g = s c$

Then let:

$h = r a$
$k = s d$

Now either $e \divides k$ or $e \nmid k$.

First, suppose $e \divides k$.

Let $k = t e$.

Then let $l = t f$.

We have:

$h = r a$
$g = r b$
$a : h = b : g$
$a : b = h : g$

Similarly:

$c : d = g : k$

and further:

$e : f = k : l$

Thus $h, g, k, l$ are continuously proportional:

in the ratio of $a : b$
in the ratio of $c : d$
in the ratio of $e : f$.

It is necessary to prove that these are the least natural numbers with this property.

Suppose $h, g, k, l$ are not the least natural numbers in the ratios of $a : b$, $c : d$, $e : f$.

Let these numbers be $n, o, m, p$.

Then:

$a : b = n : o$

while $a$ and $b$ are least.

$b \divides o$
$c \divides o$

and so $o$ is a common multiple of $b$ and $c$.

$\lcm \set {b, c} \divides o$

But $g = \lcm \set {b, c}$ from above.

Therefore $g \divides o$, the greater the less, which is impossible.

Therefore there are no numbers $n, o, m, p$ less than $h, g, k, l$ which have the properties in question.

$\Box$

Now suppose $e \nmid k$.

Let $m = \lcm \set {e, k}$.

Thus let:

$m = u k$
$n = u h$
$o = u g$

and:

$m = v e$
$p = v f$
$h : n = g : o$
$h : g = n : o$

Similarly, let: But:

$h : g = a : b$

and so:

$a : b = n : o$

For the same reason:

$c : d = o : m$

Since:

$m = v e$
$p = v f$
$e : m = f : p$
$e : f = m : p$

Therefore $n, o, m, p$ are continuously proportional:

in the ratio of $a : b$
in the ratio of $c : d$
in the ratio of $e : f$.

It is necessary to prove that these are the least natural numbers with this property.

Suppose $n, o, m, p$ are not the least natural numbers in the ratios of $a : b$, $c : d$, $e : f$.

Let these numbers be $n', o', m', p'$.

Then:

$a : b = n' : o'$

while $a$ and $b$ are least.

$b \divides o'$
$c \divides o'$

and so $o'$ is a common multiple of $b$ and $c$.

$\lcm \set {b, c} \divides o'$

But $g = \lcm \set {b, c}$ from above.

Therefore $g \divides o'$, the greater the less, which is impossible.

Therefore there are no numbers $n', o', m', p'$ less than $n, o, m, p$ which have the properties in question.

$\blacksquare$

## Historical Note

This proof is Proposition $4$ of Book $\text{VIII}$ of Euclid's The Elements.
It needs to be pointed out that Euclid uses the term continued proportion here to mean something different from geometric sequence; the ratios $a : b$, $c : d$ and $e : f$ are not necessarily equal.