# Construction of Fifth Binomial Straight Line

## Theorem

In the words of Euclid:

To find the fifth binomial straight line.

## Proof 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.

$CA : AB = EF^2 : FG^2$

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

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

But from Proposition $9$ of Book $\text{X}$: Commensurability of Squares: $EF$ 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:

$CA : AB = EF^2 : FG^2$

while:

$AB > CA$

Therefore:

$GF^2 > FE^2$

Let:

$EF^2 + H^2 = GF^2$

for some $H$.

$AB : BC = GF^2 : H^2$

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

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

$FG$ is incommensurable in length with $H$.

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

We have that:

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

and:

the lesser term $EF$ is commensurable in length with $D$.

Therefore $EG$ is a fifth binomial straight line.

$\blacksquare$

## Historical Note

This proof is Proposition $52$ of Book $\text{X}$ of Euclid's The Elements.