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

== Theorem ==
== 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 , let it be contrived that:
:$AB : CD = AE : CF$
Therefore by :
:$EB : FD = AB : CD$
But $AB$ is commensurable in length with $CD$.
Therefore from :
:$AE$ 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 :
:$AE : CF = EB : FD$
Therefore by :
:$AE : EB = CF : FD$
But by hypothesis $AE$ and $EB$ are commensurable in square only.
Therefore by :
:$CF$ 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 :
:$CF^2$ 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 :
:$CF$ 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 :
:$FD$ 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. 
Then from :
: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 :
:$CF^2$ 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 :
:$CF$ 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 :
:$FD$ 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. 
Then from :
: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]].
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory