# 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.*

(*The Elements*: Book $\text{VIII}$: Proposition $4$)

## Proof

Let the given ratios in least numbers be:

- $a : b$, $c : d$, $e : f$

From Proposition $34$ of Book $\text{VII} $: Existence of Lowest Common Multiple, we can find:

- $g = \lcm set {b, c}$

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:

- $h = r a$
- $g = r b$

thus from Book $\text{VII}$ Definition $20$: Proportional:

- $a : h = b : g$

So from Proposition $13$ of Book $\text{VII} $: Proportional Numbers are Proportional Alternately:

- $a : b = h : g$

Similarly:

- $c : d = g : k$

and further:

- $e : f = k : l$

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

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.

By Proposition $20$ of Book $\text{VII} $: Ratios of Fractions in Lowest Terms:

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

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

Therefore by Proposition $35$ of Book $\text{VII} $: LCM Divides Common Multiple:

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

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:

- $m = v e$
- $p = v f$

Thus from Book $\text{VII}$ Definition $20$: Proportional:

- $h : n = g : o$

From Proposition $13$ of Book $\text{VII} $: Proportional Numbers are Proportional Alternately:

- $h : g = n : o$

Similarly, let: But:

- $h : g = a : b$

and so:

- $a : b = n : o$

For the same reason:

- $c : d = o : m$

Since:

- $m = v e$
- $p = v f$

from Book $\text{VII}$ Definition $20$: Proportional:

- $e : m = f : p$

and so from Proposition $13$ of Book $\text{VII} $: Proportional Numbers are Proportional Alternately:

- $e : f = m : p$

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

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.

By Proposition $20$ of Book $\text{VII} $: Ratios of Fractions in Lowest Terms:

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

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

Therefore by Proposition $35$ of Book $\text{VII} $: LCM Divides Common Multiple:

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

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.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 2*(2nd ed.) ... (previous) ... (next): Book $\text{VIII}$. Propositions