# Construction of Sixth Binomial Straight Line

## Theorem

In the words of Euclid:

*To find the sixth binomial straight line.*

(*The Elements*: Book $\text{X}$: Proposition $53$)

## 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 number which is not square.

Let $D$ have to neither $AB$ nor $AC$ the ratio that a square number has to another square number.

Let $E$ be a rational straight line.

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

- $D : AB = E^2 : FG^2$

where $FG$ is a straight line.

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

- $E^2$ is commensurable with $FG^2$.

We have that $E$ is rational.

Therefore $FG$ is also rational.

Since:

- $D$ does not have to $AB$ the ratio that a square number has to another square number

then:

- $E^2$ does not have to $FG^2$ the ratio that a square number has to another square number.

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

- $E$ is incommensurable in length with $FG$.

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

- $BA : AC = FG^2 : GH^2$

where $GH$ is a straight line constructed such that $FH = FG + GH$ is itself a straight line.

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

- $FG^2$ is commensurable with $GH^2$.

We have that $FG$ is rational.

Therefore $GH$ is also rational.

Since:

- $BA$ does not have to $AC$ the ratio that a square number has to another square number

then:

- $FG^2$ does not have to $GH^2$ the ratio that a square number has to another square number.

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

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

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

Therefore by definition $FH$ is a binomial.

We have that:

- $D : AB = E^2 : FG^2$

and

- $BA : AC = FG^2 : GH^2$

From Proposition $22$ of Book $\text{V} $: Equality of Ratios Ex Aequali:

- $D : AC = E^2 : GH^2$

But $D$ does not have to $AC$ the ratio that a square number has to another square number.

Therefore, neither does $E$ have to $GH$ the ratio that a square number has to another square number.

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

- $E$ is incommensurable in length with $GH$.

We have that:

- $BA : AC = FG^2 : GH^2$

Therefore:

- $FG^2 > GH^2$

Let:

- $FG^2 = GH^2 + K^2$

for some $K$.

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

- $AB : BC = FG : K$

But $AB$ does not have to $AC$ the ratio that a square number has to another square number.

Therefore, neither does $FG^2$ have to $k^2$ the ratio that a square number has to another square number.

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

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

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

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

Also, neither $FG$ nor $GH$ is commensurable in length with $E$.

Therefore $FH$ is a sixth binomial straight line.

$\blacksquare$

## Historical Note

This proof is Proposition $53$ 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