# Straight Line Commensurable with Apotome

## Theorem

In the words of Euclid:

*A straight line commensurable in length with an apotome is an apotome and the same in order.*

(*The Elements*: Book $\text{X}$: Proposition $103$)

## Proof

Let $AB$ be an apotome.

Let $CD$ be commensurable in length with $AB$.

It is to be demonstrated that:

- $CD$ is an apotome

and:

Let $BE$ be the annex of $CD$.

This can be constructed uniquely by Construction of Apotome is Unique.

By definition, $AE$ and $EB$ are rational straight lines which are commensurable in square only.

From Proposition $12$ of Book $\text{VI} $: Construction of Fourth Proportional Straight Line, let it be contrived that:

- $BE : DF = AB : CD$

From Proposition $12$ of Book $\text{V} $: Sum of Components of Equal Ratios:

- $AE : CF = AB : CD$

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

- $AE : CF = BE : DF$

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:

- $BE$ is commensurable in length with $DF$.

We have that $AE$ and $EB$ are rational straight lines which are commensurable in square only.

Therefore from Proposition $13$ of Book $\text{X} $: Commensurable Magnitudes are Incommensurable with Same Magnitude:

- $CF$ and $FD$ are rational straight lines which are commensurable in square only.

Thus $CD$ is an apotome.

$\Box$

It remains to be shown that the order of $CD$ is the same as the order of $AB$.

We have that:

- $AE : CF = BE : DF$

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

- $AE : EB = CF : FD$

We have that:

- $AE^2 = EB^2 + \lambda^2$

where either:

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

or:

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

Suppose $\lambda$ is commensurable in length with $AE$.

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

- $CF^2 > FD^2 + \mu^2$

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

Let $AE$ be commensurable in length with a rational straight line $\alpha$ set out.

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

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

Similarly, let $BE$ be commensurable in length with $\alpha$.

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

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

Suppose neither $AE$ nor $BE$ is commensurable in length with $\alpha$.

- neither $CF$ nor $FD$ will be commensurable in length with $\alpha$.

Let $\lambda$ be incommensurable in length with $AE$.

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

- $CF^2 > FD^2 + \mu^2$

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

Let $AE$ be commensurable in length with a rational straight line $\alpha$ set out.

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

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

Similarly, let $BE$ be commensurable in length with $\alpha$.

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

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

Suppose neither $AE$ nor $BE$ is commensurable in length with $\alpha$.

- neither $CF$ nor $FD$ will be commensurable in length with $\alpha$.

It follows that $CD$ is an apotome of the same as the order as $AB$.

$\blacksquare$

## Historical Note

This proof is Proposition $103$ 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