Construction of First Binomial Straight Line

Proof

 * Euclid-X-48.png

Let $AC$ and $CB$ be straight lines constructed such that $AB = AC + CB$ is itself a straight line.

Using, let:
 * $AB : BC = m^2 : n^2$

where $m$ and $n$ are numbers such that $m^2 - n^2$ is not square.

Let $D$ be a rational straight line.

Let $EF$ be constructed commensurable in length with $D$.

Then $EF$ is also a rational straight line.

Using, let:
 * $BA : AC = EF^2 : FG^2$

where $FG$ is a straight line constructed such that $EG = EF + FG$ is itself a straight line.

But:
 * $AB : AC = m^2 : \left({m^2 - n^2}\right)$

where $\left({m^2 - n^2}\right)$ is not square.

So from :
 * $EF^2 : FG^2 = m^2 : \left({m^2 - n^2}\right)$

$EF$ and $FG$ are commensurable in square.

But from : $EF$ and $FG$ are incommensurable in length.

Therefore $EF$ and $FG$ are rational straight lines which are commensurable in square only.

Therefore by definition $EG$ is a binomial.

We have that:
 * $BA : AC = EF^2 : FG^2$

while:
 * $BA > AC$

Therefore:
 * $EF^2 > FG^2$

Let:
 * $FG^2 + H^2 = EF^2$

for some $H$.

From :
 * $AB : BC = EF^2 : H$

But $AB : BC$ has the ratio that a square number has to a square number.

Therefore by :
 * $EF$ is commensurable in length with $H$.

Therefore $EF^2$ is greater than $FG^2$ by the square on a straight line commensurable with $EF$.

We have that:
 * $EF$ and $FG$ are rational straight lines

and:
 * $EF$ is commensurable in length with $D$.

Therefore $EF$ is a first binomial straight line.