Construction of Third Binomial Straight Line
Theorem
In the words of Euclid:
- To find the third binomial straight line.
(The Elements: Book $\text{X}$: Proposition $50$)
Proof
Let $AC$ and $CB$ be straight lines constructed such that $AB = AC + CB$ is itself a straight line.
- $AB : BC = m^2 : n^2$
where $m$ and $n$ are numbers such that $m^2 - n^2$ is not square.
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:
- $AB : BC = m^2 : n^2$
where $m$ and $n$ are numbers.
Therefore:
- $FG : K = m^2 : n^2$
Therefore from Proposition $9$ of Book $\text{X} $: Commensurability of Squares:
- $FG$ is commensurable in length with $K$.
Therefore $FG^2$ is greater than $GH^2$ by the square on a straight line which is commensurable 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 third binomial straight line.
$\blacksquare$
Historical Note
This proof is Proposition $50$ 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