Between two Similar Solid Numbers exist two Mean Proportionals

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Theorem

In the words of Euclid:

Between two similar solid numbers there fall two mean proportional numbers, and the solid number has to the solid number the ratio triplicate of that which the corresponding side has to the corresponding side.

(The Elements: Book $\text{VIII}$: Proposition $19$)


Proof

Let $m$ and $n$ be similar solid numbers.

Then for some $a, b, c, d, e, f \in \Z$ such that $a \le b \le c$ and $d \le e \le f$:

$m = a b c$
$n = d e f$

such that:

$\dfrac a d = \dfrac b e = \dfrac c f$

Let:

$r := a e = b d = b f = c e$

So:

\(\ds \dfrac {a b} r\) \(=\) \(\ds \dfrac {a b} {a e}\)
\(\ds \) \(=\) \(\ds \dfrac b e\)

and:

\(\ds \dfrac r {d e}\) \(=\) \(\ds \dfrac {b d} {d e}\)
\(\ds \) \(=\) \(\ds \dfrac b e\)

Thus by definition, $\tuple {a b, r, d e}$ is a geometric sequence.

By definition, $r$ is a mean proportional between $a b$ and $d e$.


Now consider the products $r c$ and $r f$.

We have:

\(\ds \frac {a b c} {r c}\) \(=\) \(\ds \frac {a b c} {c b d}\)
\(\ds \) \(=\) \(\ds \frac {a b} {b d}\)
\(\ds \) \(=\) \(\ds \frac a d\)

and:

\(\ds \frac {r c} {r f}\) \(=\) \(\ds \frac {c b d} {f b d}\)
\(\ds \) \(=\) \(\ds \frac c f\)

and:

\(\ds \frac {f b d} {d e f}\) \(=\) \(\ds \frac {b d} {d e}\)
\(\ds \) \(=\) \(\ds \frac b e\)

That is:

$\tuple {a b c, b c d, b d f, d e f}$ is a geometric sequence.

Also from the above:

$\dfrac a d = \dfrac b e = \dfrac c f$

showing that $m$ is in triplicate ratio to $n$ as their sides.

$\blacksquare$


Historical Note

This proof is Proposition $19$ of Book $\text{VIII}$ of Euclid's The Elements.


Sources

(in which a mistake apppears)
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