Proof: By Euclid
(related to Proposition: Prop. 9.10: Elements of Geometric Progression from One where First Element is not Power of Number)
- 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, $A$, not be square.
- I say that no other (number) will be square either, apart from the third from the unit [and (all) those (numbers after that) which leave an interval of one (number)].
- For, if possible, let $C$ be square.
- And $B$ is also square [Prop. 9.8].
- Thus, $B$ and $C$ have to one another (the) ratio which (some) square number (has) to (some other) square number.
- And as $B$ is to $C$, (so) $A$ (is) to $B$.
- Thus, $A$ and $B$ have to one another (the) ratio which (some) square number has to (some other) square number.
- Hence, $A$ and $B$ are similar plane (numbers) [Prop. 8.26].
- And $B$ is square.
- Thus, $A$ is also square.
- The very opposite thing was assumed.
- $C$ is thus not square.
- So, similarly, we can show that no other (number is) square either, apart from the third from the unit, and (all) those (numbers after that) which leave an interval of one (number).
- And so let $A$ not be cube.
- I say that no other (number) will be cube either, apart from the fourth from the unit, and (all) those (numbers after that) which leave an interval of two (numbers).
- For, if possible, let $D$ be cube.
- And $C$ is also cube [Prop. 9.8].
- For it is the fourth (number) from the unit.
- And as $C$ is to $D$, (so) $B$ (is) to $C$.
- And $B$ thus has to $C$ the ratio which (some) cube (number has) to (some other) cube (number) .
- And $C$ is cube.
- Thus, $B$ is also cube [Prop. 7.13], [Prop. 8.25].
- And since as the unit is to $A$, (so) $A$ (is) to $B$, and the unit measures $A$ according to the units in it, $A$ thus also measures $B$ according to the units in ($A$).
- Thus, $A$ has made the cube (number) $B$ (by) multiplying itself.
- And if a number makes a cube (number by) multiplying itself then it itself will be cube [Prop. 9.6].
- Thus, $A$ (is) also cube.
- The very opposite thing was assumed.
- Thus, $D$ is not cube.
- So, similarly, we can show that no other (number) is cube either, apart from the fourth from the unit, and (all) those (numbers after that) which leave an interval of two (numbers).
- (Which is) the very thing it was required to show.
∎
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References
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"
