# Square on Rational Straight Line applied to Apotome

## Theorem

In the words of Euclid:

*The square on a rational straight line, if applied to an apotome, produces as breadth the binomial straight line the terms of which are commensurable with the terms of the apotome and in the same ratio; and further the binomial so arising has the same order as the apotome.*

(*The Elements*: Book $\text{X}$: Proposition $113$)

## Proof

Let $A$ be a rational straight line.

Let $BD$ be an apotome.

Let $BD \cdot KH$ be the rectangle on $BD$ equal to $A^2$.

It is to be demonstrated that:

- $KH$ is a binomial straight line whose terms are commensurable with the terms of $BD$ and in the same ratio

and:

Let $DC$ be the annex of $BD$.

By the definition of apotome, $BC$ and $CD$ are rational straight lines which are commensurable in square only.

Let $BC \cdot G$ be the rectangle on $BC$ equal to $A^2$.

But $A^2$ is rational.

Therefore $BC \cdot G$ is also rational.

We have that $BC \cdot G$ has been applied to a rational straight line $BC$.

Thus from Proposition $20$ of Book $\text{X} $: Quotient of Rationally Expressible Numbers is Rational:

- $G$ is rational and commensurable in length with $BC$.

We have that:

- $BC \cdot G = BD \cdot KH$

Therefore by Proposition $16$ of Book $\text{VI} $: Rectangles Contained by Proportional Straight Lines:

- $CD \cdot BD = KH \cdot G$

But:

- $BC > BD$

and so from:

and:

it follows that:

- $KH > G$

Let $KE = G$.

Then $KE$ is commensurable in length with $BC$.

We have that:

- $CB : BD = HK : KE$

So from Porism to Proposition $19$ of Book $\text{V} $: Proportional Magnitudes have Proportional Remainders:

- $CD : BD = HF : FE$

Let it be contrived that $KH : HE = HF : FE$.

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

- $KF : FH = KH : HE$

That is:

- $KF : FH = BC : CD$

But $BC$ and $CD$ are commensurable in square only.

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

- $KF$ and $FH$ are commensurable in square only.

We have that:

- $KH : HE = KF : FH$

and:

- $KH : HE = HF : FE$

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

- $KF : HE = HF : FE$

So from Book $\text{V}$ Definition $9$: Duplicate Ratio:

- $KF : FE = KF^2 : FH^2$

But $KF$ and $FH$ are commensurable in square.

So $KF^2$ is commensurable with $FH^2$.

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

- $KF$ is commensurable in length with $FE$.

So from Proposition $15$ of Book $\text{X} $: Commensurability of Sum of Commensurable Magnitudes:

- $KF$ is commensurable in length with $KE$.

But $KE$ is rational and commensurable in length with $BC$.

Therefore from Proposition $12$ of Book $\text{X} $: Commensurability is Transitive Relation:

- $KF$ is rational and commensurable in length with $BC$.

We have that:

- $BC : CD = KF : FH$

Thus from Proposition $16$ of Book $\text{V} $: Proportional Magnitudes are Proportional Alternately:

- $BC : KF = DC : FH$

But $BC$ is commensurable in length with $KF$.

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

- $FH$ is commensurable in length with $CD$.

But $BC$ and $CD$ are rational straight lines which are commensurable in square only.

Therefore $KF$ and $FH$ are rational straight lines which are commensurable in square only.

Therefore, by definition, $KH$ is binomial.

$\Box$

We have that:

- $BC^2 = CD^2 + \lambda^2$

where either:

- $\lambda$ is commensurable in length with $BC$

or:

- $\lambda$ is incommensurable in length with $BC$.

First suppose $\lambda$ is commensurable in length with $BC$.

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

- $KF^2 = FH^2 + \mu^2$

where $\mu$ is commensurable in length with $KF$.

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

Then by:

and:

it follows that:

- $KF$ is commensurable in length with $\alpha$.

Let $CD$ be commensurable in length with $\alpha$.

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

- $FH$ is commensurable in length with $\alpha$.

Let neither $BC$ nor $CD$ be commensurable in length with $\alpha$.

Then neither $KF$ nor $FH$ is commensurable in length with $\alpha$.

Next suppose $\lambda$ is incommensurable in length with $BC$.

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

- $KF^2 = FH^2 + \mu^2$

where $\mu$ is incommensurable in length with $KF$.

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

Then by:

and:

it follows that:

- $KF$ is commensurable in length with $\alpha$.

Let $CD$ be commensurable in length with $\alpha$.

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

- $FH$ is commensurable in length with $\alpha$.

Let neither $BC$ nor $CD$ be commensurable in length with $\alpha$.

Then neither $KF$ nor $FH$ is commensurable in length with $\alpha$.

It follows that:

- the terms of $KH$ are commensurable with the terms of $BD$ and in the same ratio

and:

$\blacksquare$

## Historical Note

This proof is Proposition $113$ 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