Construction of First Apotome

Proof

 * Euclid-X-85.png

Let $A$ be a rational straight line.

Let $BG$ be commensurable in length with $A$.

Therefore, by definition, $BG$ is also rational.

Let $DE$ and $EF$ be square numbers such that $FD = DE - EF$ is not square.

Then $ED : FD$ is not the ratio that a square number has to another square number.

Using, let it be contrived that:
 * $EF : DF = BG^2 : CG^2$

Therefore by :
 * $BG^2$ is commensurable with $CG^2$.

But $BG^2$ is rational.

Therefore $GC^2$ is rational.

Therefore $GC$ is rational.

We have that $ED : FD$ is not the ratio that a square number has to another square number.

Therefore $BG^2 : GC^2$ is not the ratio that a square number has to another square number.

Therefore from :
 * $BG$ is incommensurable in length with $GC$.

Both $BG$ and $GC$ are rational.

Therefore $BG$ and $GC$ rational straight lines which are commensurable in square only.

Therefore $BC$ is an apotome.

It remains to be shown that $BC$ is a first apotome.

Let $H$ be a straight line such that $H^2 = BG^2 - GC^2$.

We have that:
 * $ED : FD = BG^2 : GC^2$

So by :
 * $DE : EF = GB^2 : H^2$

But $DE$ and $EF$ are both square numbers.

So $DE$ has to $EF$ the ratio that a square number has to another square number.

Therefore $GB^2$ has to $H^2$ the ratio that a square number has to another square number.

Therefore from :
 * $BG$ is commensurable in length with $H$.

We have that:
 * $BG^2 = GC^2 + H^2$

Therefore $BG^2$ is greater than $GC^2$ by the square on a straight line which is commensurable in length with $BG$.

Also, the whole $BG$ is commensurable in length with the rational straight line $A$.

Therefore, by definition, $BC$ is a first apotome.