# Proof: By Euclid

• Let any multitude whatsoever of numbers, $A$, $B$, $C$, $D$, $E$, $F$, be in continued proportion, (starting) from a unit.
• And let the (number) after the unit, $A$, be square.
• I say that all the remaining (numbers) will also be square. • In fact, it has (already) been shown that the third (number) from the unit, $B$, is square, and all those (numbers after that) which leave an interval of one (number) [Prop. 9.8].

• So I say that all the remaining (numbers) are also square.
• For since $A$, $B$, $C$ are in continued proportion, and $A$ (is) square, $C$ is [thus] also square [Prop. 8.22].
• Again, since $B$, $C$, $D$ are [also] in continued proportion, and $B$ is square, $D$ is [thus] also square [Prop. 8.22].
• So, similarly, we can show that all the remaining (numbers) are also square.
• And so let $A$ be cube.
• I say that all the remaining (numbers) are also cube.
• In fact, it has (already) been shown that the fourth (number) from the unit, $C$, is cube, and all those (numbers after that) which leave an interval of two (numbers) [Prop. 9.8].
• So I say that all the remaining (numbers) are also cube.
• For since as the unit is to $A$, so $A$ (is) to $B$, the unit thus measures $A$ the same number of times as $A$ (measures) $B$.
• And the unit measures $A$ according to the units in it.
• Thus, $A$ also measures $B$ according to the units in ($A$).
• $A$ has thus made $B$ (by) multiplying itself.
• And $A$ is cube.
• And if a cube number makes some (number by) multiplying itself then the created (number) is cube [Prop. 9.3].
• Thus, $B$ is also cube.
• And since the four numbers $A$, $B$, $C$, $D$ are in continued proportion, and $A$ is cube, $D$ is thus also cube [Prop. 8.23].
• So, for the same (reasons), $E$ is also cube, and, similarly, all the remaining (numbers) are cube.
• (Which is) the very thing it was required to show.

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