# Construction of Fourth Binomial Straight Line

## Theorem

In the words of Euclid:

*To find the fourth binomial straight line.*

(*The Elements*: Book $\text{X}$: Proposition $51$)

## Proof

Let $AC$ and $CB$ be straight lines constructed such that $AB = AC + CB$ is itself a straight line.

Let neither $AB : AC$ nor $AB : BC$ be the ratio which a square number has to a square number.

Let $D$ be a rational straight line.

Let $EF$ be constructed commensurable in length with $D$.

Then $EF$ is also a rational straight line.

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

- $AB : AC = EF^2 : FG^2$

where $FG$ is a straight line constructed such that $EG = EF + FG$ is itself a straight line.

From Proposition $6$ of Book $\text{X} $: Magnitudes with Rational Ratio are Commensurable:

- $EF$ and $FG$ are commensurable in square.

Therefore $FG$ is also a rational straight line.

But from Proposition $9$ of Book $\text{X} $: Commensurability of Squares: $EF$ and $FG$ are incommensurable in length.

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

Therefore by definition $EG$ is a binomial.

Since:

- $AB : AC = EF^2 : FG^2$

while:

- $BA > AC$

Therefore:

- $EF^2 > FG^2$

Let:

- $FG^2 + H^2 = EF^2$

for some $H$.

From Porism to Proposition $19$ of Book $\text{V} $: Proportional Magnitudes have Proportional Remainders:

- $AB : BC = EF^2 : H^2$

But $AB : BC$ is not the ratio that a square number has to a square number.

Therefore $EF^2 : H^2$ is not the ratio that a square number has to a square number.

Therefore by Proposition $9$ of Book $\text{X} $: Commensurability of Squares:

- $EF$ is incommensurable in length with $H$.

Therefore $EF^2 > GF^2$ by the square on a straight line which is incommensurable in length with $EF$.

We have that:

- $EF$ and $FG$ are rational straight lines which are commensurable in square only

and:

- $EF$ is commensurable in length with $D$.

Therefore $EG$ is a fourth binomial straight line.

$\blacksquare$

## Historical Note

This proof is Proposition $51$ 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