# Construction of Third Apotome

## Theorem

In the words of Euclid:

To find the third apotome.

## Proof

Let $A$ be a rational straight line.

Let $E$, $BC$ and $CD$ be numbers set out which do not have pairwise between them the ratio that a square number has to another square number.

But let $CB : BD$ be the ratio that a square number has to a square number.

Using Porism to Proposition $6$ of Book $\text{X}$: Magnitudes with Rational Ratio are Commensurable, let it be contrived that:

$E : BC = A^2 : FG^2$

and:

$BC : CD = FG^2 : GH^2$

We have that:

$E : BC = A^2 : FG^2$.
$A^2$ is commensurable with $FG^2$.

But $A^2$ is rational.

Therefore $FG^2$ is rational.

Therefore $FG$ is rational.

We have that $E : BC$ is not the ratio that a square number has to another square number.

Therefore $A^2 : FG^2$ is not the ratio that a square number has to another square number.

$A$ is incommensurable in length with $FG$.

We have that:

$BC : CD = FG^2 : GH^2$
$FG^2$ is commensurable with $GH^2$.

But $FG^2$ is rational.

Therefore $GH^2$ is rational.

Therefore $GH$ is rational.

We have that $BC : CD$ is not the ratio that a square number has to another square number.

Therefore $FG^2 : GH^2$ is not the ratio that a square number has to another square number.

$FG$ is incommensurable in length with $GH$.

Both $FG$ and $GH$ are rational.

Therefore $FG$ and $GH$ rational straight lines which are commensurable in square only.

Therefore $FH$ is an apotome.

It remains to be shown that $FH$ is a third apotome.

We have that:

$E : BC = A^2 : FG^2$.

and:

$BC : CD = FG^2 : GH^2$
$E : CD = A^2 : HG^2$

But $E : CD$ is not the ratio that a square number has to another square number.

Therefore neither is $A^2 : GH^2$ the ratio that a square number has to another square number.

$A$ is incommensurable in length with $GH$.

Therefore neither $FG$ and $GH$ is incommensurable in length with the rational straight line $A$.

Now let $K^2 = FG^2 - GH^2$.

We have that:

$BC : CD = FG^2 GH^2$
$BC : BD = FG^2 : K^2$

But $BC$ has to $BD$ the ratio that a square number has to another square number.

Therefore $FG^2$ has to $K^2$ the ratio that a square number has to another square number.

$FG$ is incommensurable in length with $K$.

So:

$FG^2$ is greater than $GH^2$ by the square on a straight line which is commensurable in length with the rational straight line $A$.

Therefore, by definition, $FH$ is a third apotome.

$\blacksquare$

## Historical Note

This proof is Proposition $87$ of Book $\text{X}$ of Euclid's The Elements.