# Straight Line Commensurable with Binomial Straight Line is Binomial and of Same Order

## Theorem

In the words of Euclid:

A straight line commensurable in length with a binomial straight line is itself also binomial and the same in order.

## Proof

Let $AB$ be a binomial.

Let $CD$ be commensurable in length with $AB$.

It is to be shown that $CD$ is a binomial, and that the order of $CD$ is the same as the order of $AB$.

Let $AB$ be divided into its terms by $E$.

Let $AE$ be the greater term.

By definition, $AE$ and $EB$ are rational straight lines which are commensurable in square only.

Using Proposition $12$ of Book $\text{VI}$: Construction of Fourth Proportional Straight Line, let it be contrived that:

$AB : CD = AE : CF$
$EB : FD = AB : CD$

But $AB$ is commensurable in length with $CD$.

$AE$ is commensurable in length with $CF$

and:

$EB$ is commensurable in length with $FD$.

But by hypothesis $AE$ and $EB$ are rational.

Therefore $CF$ and $FD$ are rational.

$AE : CF = EB : FD$
$AE : EB = CF : FD$

But by hypothesis $AE$ and $EB$ are commensurable in square only.

$CF$ and $FD$ are commensurable in square only.

But $CF$ and $FD$ are rational.

Therefore, by definition, $CD$ is a binomial.

It remains to be demonstrated that $CD$ is of the same order as $AB$.

We have that $AE > EB$.

Then $AE^2$ is greater than $EB^2$ by either:

the square on a straight line which is commensurable with $AE$

or:

the square on a straight line which is incommensurable with $AE$.

Suppose $AE^2 > EB^2$ by the square on a straight line which is commensurable with $AE$.

$CF^2$ will greater than $FD^2$ by the square on a straight line which is commensurable with $CF$.

Let $AE$ be commensurable in length with a rational straight line which has been set out.

$CF$ will likewise be commensurable in length with that same rational straight line.

Thus both $AB$ and $CD$ are by definition first binomial, and therefore of the same [Definition:Order of Binomial|order]].

Let $EB$ be commensurable in length with a rational straight line which has been set out.

$FD$ will likewise be commensurable in length with that same rational straight line.

Thus both $AB$ and $CD$ are by definition second binomial, and therefore of the same [Definition:Order of Binomial|order]].

Suppose that neither $AE$ nor $EB$ is commensurable in length with a rational straight line which has been set out.

Neither $CF$ nor $FD$ will be commensurable in length with that rational straight line.

It follows that both $AB$ and $CD$ are by definition third binomial, and therefore of the same [Definition:Order of Binomial|order]].

Suppose $AE^2 > EB^2$ by the square on a straight line which is incommensurable in length with $AE$.

$CF^2$ will greater than $FD^2$ by the square on a straight line which is incommensurable in length with $CF$.

Let $AE$ be commensurable in length with a rational straight line which has been set out.

$CF$ will likewise be commensurable in length with that same rational straight line.

It follows that both $AB$ and $CD$ are by definition fourth binomial, and therefore of the same [Definition:Order of Binomial|order]].

Let $EB$ be commensurable in length with a rational straight line which has been set out.

$FD$ will likewise be commensurable in length with that same rational straight line.

Thus both $AB$ and $CD$ are by definition fifth binomial, and therefore of the same [Definition:Order of Binomial|order]].

Suppose that neither $AE$ nor $EB$ is commensurable in length with a rational straight line which has been set out.

Neither $CF$ nor $FD$ will be commensurable in length with that rational straight line.

It follows that both $AB$ and $CD$ are by definition sixth binomial, and therefore of the same [Definition:Order of Binomial|order]].

$\blacksquare$

## Historical Note

This proof is Proposition $66$ of Book $\text{X}$ of Euclid's The Elements.