Construction of Sequence of Numbers with Given Ratios

Proof
Let the given ratios in least numbers be:
 * $a : b$, $c : d$, $e : f$

From, 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 :
 * $a : h = b : g$

So from :
 * $a : b = h : g$

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.

By :
 * $b \divides o$
 * $c \divides o$

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

Therefore by :
 * $\lcm \set {b, c} \divides o$

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.

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 :
 * $h : n = g : o$

From :
 * $h : g = n : o$

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 :
 * $e : m = f : p$

and so from :
 * $e : f = m : p$

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.

By :
 * $b \divides o'$
 * $c \divides o'$

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

Therefore by :
 * $\lcm \set {b, c} \divides o'$

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.