Commensurability of Squares

Proof
Let $A$ and $B$ be commensurable in length.

Then $A$ has to $B$ the ratio which a number has to a number.

Let $\dfrac A B = \dfrac C D$.

From Similar Polygons are composed of Similar Triangles: Porism:
 * $\dfrac {A^2} {B^2} = \left({\dfrac C D}\right)^2 = \dfrac {C^2} {D^2}$

Let:
 * $\dfrac {A^2} {B^2} = \dfrac {C^2} {D^2}$

Then it follows that:
 * $\dfrac A B = \dfrac C D$

and from Magnitudes with Rational Ratio are Commensurable it follows that $A$ and $B$ are commensurable in length.

Let $A$ and $B$ be incommensurable in length.

Suppose:
 * $\dfrac {A^2} {B^2} = \dfrac {C^2} {D^2}$

where $C$ and $D$ are integers.

Then $A$ and $B$ would be commensurable in length, which they are not.

So the square on $A$ and the square on $B$ do not have to one another the ratio which a square number has to a square number.

Let the square on $A$ not have to the square on $B$ not have the ratio which a square number has to a square number.

Suppose $A$ was commensurable in length with $B$.

Then the square on $A$ would have to the square on $B$ the ratio which a square number has to a square number, which it does not.

So $A$ is not commensurable in length with $B$.