Construction of Third Binomial Straight Line

Proof

 * Euclid-X-50.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 number which is not square.

Let $D$ have to neither $AB$ nor $AC$ the ratio that a square number has to another square number.

Let $E$ be a rational straight line.

Using, let:
 * $D : AB = E^2 : FG^2$

where $FG$ is a straight line.

From :
 * $E^2$ is commensurable with $FG^2$.

We have that $E$ is rational.

Therefore $FG$ is also rational.

Since:
 * $D$ does not have to $AB$ the ratio that a square number has to another square number

then:
 * $E^2$ does not have to $FG^2$ the ratio that a square number has to another square number.

From :
 * $E$ is incommensurable in length with $FG$.

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

where $GH$ is a straight line constructed such that $FH = FG + GH$ is itself a straight line.

From :
 * $FG^2$ is commensurable with $GH^2$.

We have that $FG$ is rational.

Therefore $GH$ is also rational.

Since:
 * $BA$ does not have to $AC$ the ratio that a square number has to another square number

then:
 * $FG^2$ does not have to $GH^2$ the ratio that a square number has to another square number.

From :
 * $FG$ is incommensurable in length with $GH$.

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

Therefore by definition $FH$ is a binomial.

We have that:
 * $D : AB = E^2 : FG^2$

and
 * $BA : AC = FG^2 : GH^2$

From :
 * $D : AC = E^2 : GH^2$

But $D$ does not have to $AC$ the ratio that a square number has to another square number.

Therefore, neither does $E$ have to $GH$ the ratio that a square number has to another square number.

Therefore from :
 * $E$ is incommensurable in length with $GH$.

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

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

Let:
 * $FG^2 = GH^2 + K^2$

for some $K$.

From :
 * $AB : BC = FG : K$

But:
 * $AB : BC = m^2 : n^2$

where $m$ and $n$ are numbers.

Therefore:
 * $FG : K = m^2 : n^2$

Therefore from :
 * $FG$ is commensurable in length with $K$.

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

But $FG$ and $GH$ are rational straight lines which are commensurable in square only.

Also, neither $FG$ nor $GH$ is commensurable in length with $E$.

Therefore $FH$ is a third binomial straight line.