Construction of Fourth Binomial Straight Line

Proof

 * Euclid-X-51.png

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

Let neither $AB : AC$ nor $AB : BC$ be the ratio which a square number has to a square number.

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:
 * $AB : AC = EF^2 : FG^2$

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

From :
 * $EF$ and $FG$ are commensurable in square.

Therefore $FG$ is also a rational straight line.

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.

Since:
 * $AB : AC = EF^2 : FG^2$

while:
 * $BA > AC$

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

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

for some $H$.

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

But $AB : BC$ is not the ratio that a square number has to a square number.

Therefore $EF^2 : H^2$ is not the ratio that a square number has to a square number.

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

Therefore $EF^2 > GF^2$ by the square on a straight line which is incommensurable in length with $EF$.

We have that:
 * $EF$ and $FG$ are rational straight lines which are commensurable in square only

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

Therefore $EG$ is a fourth binomial straight line.