# 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.*

(*The Elements*: Book $\text{X}$: Proposition $66$)

## 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:

- $AB : CD = AE : CF$

Therefore by Proposition $19$ of Book $\text{V} $: Proportional Magnitudes have Proportional Remainders:

- $EB : FD = AB : CD$

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

Therefore from Proposition $11$ of Book $\text{X} $: Commensurability of Elements of Proportional Magnitudes:

- $AE$ 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.

From Proposition $11$ of Book $\text{V} $: Equality of Ratios is Transitive:

- $AE : CF = EB : FD$

Therefore by Proposition $16$ of Book $\text{V} $: Proportional Magnitudes are Proportional Alternately:

- $AE : EB = CF : FD$

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

Therefore by Proposition $11$ of Book $\text{X} $: Commensurability of Elements of Proportional Magnitudes:

- $CF$ 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 $AE$.

Then by Proposition $14$ of Book $\text{X} $: Commensurability of Squares on Proportional Straight Lines:

- $CF^2$ 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.

Then by Proposition $12$ of Book $\text{X} $: Commensurability is Transitive Relation:

- $CF$ 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.

Then by Proposition $12$ of Book $\text{X} $: Commensurability is Transitive Relation:

- $FD$ 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 $AE$.

Then by Proposition $14$ of Book $\text{X} $: Commensurability of Squares on Proportional Straight Lines:

- $CF^2$ 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.

Then by Proposition $12$ of Book $\text{X} $: Commensurability is Transitive Relation:

- $CF$ 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.

Then by Proposition $12$ of Book $\text{X} $: Commensurability is Transitive Relation:

- $FD$ 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*.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 3*(2nd ed.) ... (previous) ... (next): Book $\text{X}$. Propositions