# Construction of First Binomial Straight Line

## Theorem

In the words of Euclid:

To find the first binomial straight line.

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

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

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.

$BA : AC = EF^2 : FG^2$

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.

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

$EF$ 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.

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 $H$.

$AB : BC = EF^2 : H$

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

$EF$ 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.

$\blacksquare$

## Historical Note

This proof is Proposition $48$ of Book $\text{X}$ of Euclid's The Elements.

## Sources

in which a mistake appears.