# Square on Rational Straight Line applied to Binomial Straight Line

## Theorem

In the words of Euclid:

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

## Proof

Let $A$ be a rational straight line.

Let $BC$ be a binomial.

Let $DC$ be the greater term of $BC$.

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

It is to be demonstrated that:

$EF$ is an apotome whose terms are commensurable with $CD$ and $DB$ and in the same ratio

and:

the order of $EF$ is the same as the order of $BC$.

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

Then:

$BC \cdot EF = BD \cdot G$
$CB : BD = G : EF$

But:

$CB > BD$

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:

$G > EF$

Let $EH = G$.

Then:

$CB : BD = HE : EF$
$CD : BD = HF : FE$

Let it be contrived that $HF : FE = FK : KE$.

$HK : KF = FK : KE$
$FK : KE = CD : DB$

and:

$HK : KF = CD : DB$

But from the definition of binomial, $CD^2$ is commensurable with $DB^2$.

Therefore from:

Proposition $22$ of Book $\text{VI}$: Similar Figures on Proportional Straight Lines

and:

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

it follows that:

$HK^2$ is commensurable with $KF^2$.

We have that $HK$, $KF$ and $KE$ are proportional.

So:

$HK^2 : KF^2 = HK : KE$

Thus $HK$ is commensurable in length with $KE$.

$HE$ is commensurable in length with $EK$.

We have that $A^2 = EH \cdot BD$.

But $A^2$ is rational.

Therefore $EH \cdot BD$ is also rational.

We have that $EH \cdot BD$ is applied to the rational straight line $BD$.

$EH$ is rational and commensurable in length with $BD$.

We have that $EK$ is rational and commensurable in length with $EH$.

So $EK$ is rational and commensurable in length with $BD$.

We have that:

$CD : DB = FK : KE$

and:

$CD$ and $DB$ are straight lines which are commensurable in square only.
$FK$ and $KE$ are straight lines which are commensurable in square only.

But $KE$ is rational.

Therefore $FK$ is also rational.

Therefore $FK$ and $KE$ are rational straight lines which are commensurable in square only.

Therefore $EF$ is an apotome.

$\Box$

We have that:

$CD^2 = DB^2 + \lambda^2$

where either:

$\lambda$ is commensurable in length with $CD$

or:

$\lambda$ is incommensurable in length with $CD$.

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

$FK^2 = KE^2 + \mu^2$

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

Let $CD$ 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:

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

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

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

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

Then neither $FK$ nor $KE$ is commensurable in length with $\alpha$.

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

$FK^2 = KE^2 + \mu^2$

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

Let $CD$ 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:

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

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

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

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

Then neither $FK$ nor $KE$ is commensurable in length with $\alpha$.

It follows that:

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

and:

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

$\blacksquare$

## Historical Note

This proof is Proposition $112$ of Book $\text{X}$ of Euclid's The Elements.