# Proof: By Euclid

• For let the two numbers $C$ and $D$ fall between the two numbers $A$ and $B$ in mean proportion.
• I say that $A$ and $B$ are similar solid (numbers) . • For let the three least numbers $E$, $F$, $G$ having the same ratio as $A$, $C$, $D$ have been taken [Prop. 8.2].

• Thus, the outermost of them, $E$ and $G$, are prime to one another [Prop. 8.3].
• And since one number, $F$, has fallen (between) $E$ and $G$ in mean proportion, $E$ and $G$ are thus similar plane numbers [Prop. 8.20].
• Therefore, let $H$, $K$ be the sides of $E$, and $L$, $M$ (the sides) of $G$.
• Thus, it is clear from the (proposition) before this (one) that $E$, $F$, $G$ are in continued proportion in the ratio of $H$ to $L$, and of $K$ to $M$.
• And since $E$, $F$, $G$ are the least (numbers) having the same ratio as $A$, $C$, $D$, and the multitude of $E$, $F$, $G$ is equal to the multitude of $A$, $C$, $D$, thus, via equality, as $E$ is to $G$, so $A$ (is) to $D$ [Prop. 7.14].
• And $E$ and $G$ (are) prime (to one another), and prime (numbers) are also the least (of those numbers having the same ratio as them) [Prop. 7.21], and the least (numbers) measure those (numbers) having the same ratio as them an equal number of times, the greater (measuring) the greater, and the lesser the lesser - that is to say, the leading (measuring) the leading, and the following the following [Prop. 7.20].
• Thus, $E$ measures $A$ the same number of times as $G$ (measures) $D$.
• So as many times as $E$ measures $A$, so many units let there be in $N$.
• Thus, $N$ has made $A$ (by) multiplying $E$ [Def. 7.15] .
• And $E$ is the (number created) from (multiplying) $H$ and $K$.
• Thus, $N$ has made $A$ (by) multiplying the (number created) from (multiplying) $H$ and $K$.
• Thus, $A$ is solid, and its sides are $H$, $K$, $N$.
• Again, since $E$, $F$, $G$ are the least (numbers) having the same ratio as $C$, $D$, $B$, thus $E$ measures $C$ the same number of times as $G$ (measures) $B$ [Prop. 7.20].
• So as many times as $E$ measures $C$, so many units let there be in $O$.
• Thus, $G$ measures $B$ according to the units in $O$.
• Thus, $O$ has made $B$ (by) multiplying $G$.
• And $G$ is the (number created) from (multiplying) $L$ and $M$.
• Thus, $O$ has made $B$ (by) multiplying the (number created) from (multiplying) $L$ and $M$.
• Thus, $B$ is solid, and its sides are $L$, $M$, $O$.
• Thus, $A$ and $B$ are (both) solid.
• So I say that (they are) also similar.
• For since $N$, $O$ have made $A$, $C$ (by) multiplying $E$, thus as $N$ is to $O$, so $A$ (is) to $C$ - that is to say, $E$ to $F$ [Prop. 7.18].
• But, as $E$ (is) to $F$, so $H$ (is) to $L$, and $K$ to $M$.
• And thus as $H$ (is) to $L$, so $K$ (is) to $M$, and $N$ to $O$.
• And $H$, $K$, $N$ are the sides of $A$, and $L$, $M$, $O$ the sides of $B$.
• Thus, $A$ and $B$ are similar solid numbers [Def. 7.21] .
• (Which is) the very thing it was required to show.

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