# Straight Line Commensurable with Major Straight Line is Major

## Theorem

In the words of Euclid:

A straight line commensurable with a major straight line is itself also major.

## Proof

Let $AB$ be a major.

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

It is to be shown that $CD$ is also a major.

Let $AB$ be divided into its terms by $E$.

Let $AE$ be the greater term.

By definition, $AE$ and $EB$ are straight lines such that:

$AE$ and $EB$ are incommensurable in square
$AE^2 + EB^2$ is rational
$AE \cdot EB$ is medial.

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

$AB : CD = AE : CF$
$EB : FD = AB : CD$
$AE : CF = EB : FD$

But $AB$ is commensurable in length with $CD$.

$AE$ is commensurable in length with $CF$

and:

$EB$ is commensurable in length with $FD$.

We have that:

$AE : CF = EB : FD$
$AE : EB = CF : FD$
$AB : BE = CD : DF$
$AB^2 : BE^2 = CD^2 : DF^2$

Using a similar line of reasoning to the above:

$AB^2 : AE^2 = CD^2 : CF^2$

and:

$AB^2 : AE^2 + EB^2 = CD^2 : CF^2 + FD^2$
$AB^2 : CD^2 = AE^2 + EB^2 : CF^2 + FD^2$

But $AB^2$ is commensurable with $CD^2$.

Therefore $AE^2 + EB^2$ is commensurable with $CF^2 + FD^2$.

We have that $AE^2 + EB^2$ is rational.

Therefore $CF^2 + FD^2$ is rational.

Similarly, $2 \cdot AE \cdot EB$ is commensurable with $2 \cdot CF \cdot FD$.

$2 \cdot AE \cdot EB$ is medial.

Therefore $2 \cdot CF \cdot FD$ is medial.

Thus it has been demonstrated that $CF$ and $FD$ are straight lines such that:

$CF$ and $FD$ are incommensurable in square
$CF^2 + FD2$ is rational
$CF \cdot FD$ is medial.

Thus by definition $CD$ is a major.

$\blacksquare$

## Historical Note

This proof is Proposition $68$ of Book $\text{X}$ of Euclid's The Elements.