# 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.

## 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:

the order of $KH$ is the same as the order of $BD$.

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 $BC$.

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

We have that:

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

But:

$BC > BD$

and so from:

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

and:

Proposition $14$ of Book $\text{V}$: Relative Sizes of Components of Ratios

it follows that:

$KH > G$

Let $KE = G$.

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

We have that:

$CB : BD = HK : KE$
$CD : BD = HF : FE$

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

$KF : FH = KH : HE$

That is:

$KF : FH = BC : CD$

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

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

We have that:

$KH : HE = KF : FH$

and:

$KH : HE = HF : FE$
$KF : HE = HF : FE$
$KF : FE = KF^2 : FH^2$

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

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

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

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

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

We have that:

$BC : CD = KF : FH$
$BC : KF = DC : FH$

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

$FH$ 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 $BC$.

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

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:

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

and:

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

it follows that:

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

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

$FH$ 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 $BC$.

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

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:

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

and:

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

it follows that:

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

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

$FH$ 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:

the order of $KH$ is the same as the order of $BD$.

$\blacksquare$

## Historical Note

This proof is Proposition $113$ of Book $\text{X}$ of Euclid's The Elements.