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