Proof: By Euclid

For let $A$ and $B$ be (straight lines which are) commensurable in length.
I say that the square on $A$ has to the square on $B$ the ratio which (some) square number (has) to (some) square number.

fig009e

For since $A$ is commensurable in length with $B$ , $A$ thus has to $B$ the ratio which (some) number (has) to (some) number [Prop. 10.5].
Let it have (that) which $C$ (has) to $D$ .
Therefore, since as $A$ is to $B$ , so $C$ (is) to $D$ .
But the (ratio) of the square on $A$ to the square on $B$ is the square of the ratio of $A$ to $B$ .
For similar figures are in the squared ratio of (their) corresponding sides [Prop. 6.20 corr.] 5.
And the (ratio) of the square on $C$ to the square on $D$ is the square of the ratio of the [number] $C$ to the [number] $D$ .
For there exists one number in mean proportion3 to two square numbers, and (one) square (number) has to the (other) square [number] a squared ratio with respect to (that) the side (of the former has) to the side (of the latter) [Prop. 8.11].
And, thus, as the square on $A$ is to the square on $B$ , so the square [number] on the (number) $C$ (is) to the square [number] on the [number] $D$ .¹
And so let the square on $A$ be to the (square) on $B$ as the square (number) on $C$ (is) to the [square] (number) on $D$ .
I say that $A$ is commensurable in length with $B$ .
For since as the square on $A$ is to the [square] on $B$ , so the square (number) on $C$ (is) to the [square] (number) on $D$ .
But, the ratio of the square on $A$ to the (square) on $B$ is the square of the (ratio) of $A$ to $B$ [Prop. 6.20 corr.] 5.
And the (ratio) of the square [number] on the [number] $C$ to the square [number] on the [number] $D$ is the square of the ratio of the [number] $C$ to the [number] $D$ [Prop. 8.11].
Thus, as $A$ is to $B$ , so the [number] $C$ also (is) to the [number] $D$ .
$A$ , thus, has to $B$ the ratio which the number $C$ has to the number $D$ .
Thus, $A$ is commensurable in length with $B$ " [Prop. 10.6] ":https://www.bookofproofs.org/branches/magnitudes-with-rational-ratio-are-commensurable/.²
And so let $A$ be incommensurable in length with $B$ .
I say that the square on $A$ does not have to the [square] on $B$ the ratio which (some) square number (has) to (some) square number.
For if the square on $A$ has to the [square] on $B$ the ratio which (some) square number (has) to (some) square number then $A$ will be commensurable (in length) with $B$ .
But it is not.
Thus, the square on $A$ does not have to the [square] on the $B$ the ratio which (some) square number (has) to (some) square number.
So, again, let the square on $A$ not have to the [square] on $B$ the ratio which (some) square number (has) to (some) square number.
I say that $A$ is incommensurable in length with $B$ .
For if $A$ is commensurable (in length) with $B$ then the (square) on $A$ will have to the (square) on $B$ the ratio which (some) square number (has) to (some) square number.
But it does not have (such a ratio).
Thus, $A$ is not commensurable in length with $B$ .
Thus, (squares) on (straight lines which are) commensurable in length, and so on ....

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References

Adapted from (subject to copyright, with kind permission)

Fitzpatrick, Richard: Euclid's "Elements of Geometry"

Footnotes

There is an unstated assumption here that if $\alpha/\beta=\gamma/\delta$ then $\alpha^2/\beta^2=\gamma^2/\delta$ (translator's note). ↩
There is an unstated assumption here that if $\alpha^2/\beta^2=\gamma^2/\delta^2$ then $\alpha/\beta=\gamma/\delta$ (translator's note). ↩

◀ ▲ ▶Branches / Geometry / Elements-euclid / Book-10-incommensurable-magnitudes / Proof: By Euclid

Proof: By Euclid

References

Adapted from (subject to copyright, with kind permission)

Footnotes