Square on Rational Straight Line applied to Binomial Straight Line
Theorem
In the words of Euclid:
- The square on a rational straight line applied to the binomial straight line produces as breadth an apotome the terms of which are commensurable with the terms of the binomial and moreover in the same ratio; and further the apotome so arising will have the same order as the binomial straight line.
(The Elements: Book $\text{X}$: Proposition $112$)
Proof
Let $A$ be a rational straight line.
Let $BC$ be a binomial.
Let $DC$ be the greater term of $BC$.
Let $BC \cdot EF$ be the rectangle on $BC$ equal to $A^2$.
It is to be demonstrated that:
- $EF$ is an apotome whose terms are commensurable with $CD$ and $DB$ and in the same ratio
and:
Let $BD \cdot G$ be the rectangle on $BD$ equal to $A^2$.
Then:
- $BC \cdot EF = BD \cdot G$
So from Proposition $16$ of Book $\text{VI} $: Rectangles Contained by Proportional Straight Lines:
- $CB : BD = G : EF$
But:
- $CB > BD$
and so from:
and:
it follows that:
- $G > EF$
Let $EH = G$.
Then:
- $CB : BD = HE : EF$
So from Proposition $17$ of Book $\text{V} $: Magnitudes Proportional Compounded are Proportional Separated:
- $CD : BD = HF : FE$
Let it be contrived that $HF : FE = FK : KE$.
Therefore from Proposition $12$ of Book $\text{V} $: Sum of Components of Equal Ratios:
- $HK : KF = FK : KE$
Therefore from Proposition $11$ of Book $\text{V} $: Equality of Ratios is Transitive:
- $FK : KE = CD : DB$
and:
- $HK : KF = CD : DB$
But from the definition of binomial, $CD^2$ is commensurable with $DB^2$.
Therefore from:
and:
it follows that:
- $HK^2$ is commensurable with $KF^2$.
We have that $HK$, $KF$ and $KE$ are proportional.
So:
- $HK^2 : KF^2 = HK : KE$
Thus $HK$ is commensurable in length with $KE$.
Therefore from Proposition $15$ of Book $\text{X} $: Commensurability of Sum of Commensurable Magnitudes:
- $HE$ is commensurable in length with $EK$.
We have that $A^2 = EH \cdot BD$.
But $A^2$ is rational.
Therefore $EH \cdot BD$ is also rational.
We have that $EH \cdot BD$ is applied to the rational straight line $BD$.
Therefore from Proposition $20$ of Book $\text{X} $: Quotient of Rationally Expressible Numbers is Rational:
- $EH$ is rational and commensurable in length with $BD$.
We have that $EK$ is rational and commensurable in length with $EH$.
So $EK$ is rational and commensurable in length with $BD$.
We have that:
- $CD : DB = FK : KE$
and:
- $CD$ and $DB$ are straight lines which are commensurable in square only.
Therefore by Proposition $11$ of Book $\text{X} $: Commensurability of Elements of Proportional Magnitudes:
- $FK$ and $KE$ are straight lines which are commensurable in square only.
But $KE$ is rational.
Therefore $FK$ is also rational.
Therefore $FK$ and $KE$ are rational straight lines which are commensurable in square only.
Therefore $EF$ is an apotome.
$\Box$
We have that:
- $CD^2 = DB^2 + \lambda^2$
where either:
- $\lambda$ is commensurable in length with $CD$
or:
- $\lambda$ is incommensurable in length with $CD$.
First suppose $\lambda$ is commensurable in length with $CD$.
Then by Proposition $14$ of Book $\text{X} $: Commensurability of Squares on Proportional Straight Lines:
- $FK^2 = KE^2 + \mu^2$
where $\mu$ is commensurable in length with $FK$.
Let $CD$ be commensurable in length with a rational straight line $\alpha$ which has been set out.
Then by:
and:
it follows that:
- $FK$ is commensurable in length with $\alpha$.
Let $BD$ be commensurable in length with $\alpha$.
Then by Proposition $12$ of Book $\text{X} $: Commensurability is Transitive Relation:
- $KE$ is commensurable in length with $\alpha$.
Let neither $CD$ nor $BD$ be commensurable in length with $\alpha$.
Then neither $FK$ nor $KE$ is commensurable in length with $\alpha$.
Next suppose $\lambda$ is incommensurable in length with $HF$.
Then by Proposition $14$ of Book $\text{X} $: Commensurability of Squares on Proportional Straight Lines:
- $FK^2 = KE^2 + \mu^2$
where $\mu$ is incommensurable in length with $FK$.
Let $CD$ be commensurable in length with a rational straight line $\alpha$ which has been set out.
Then by:
and:
it follows that:
- $FK$ is commensurable in length with $\alpha$.
Let $BD$ be commensurable in length with $\alpha$.
Then by Proposition $12$ of Book $\text{X} $: Commensurability is Transitive Relation:
- $KE$ is commensurable in length with $\alpha$.
Let neither $CD$ nor $BD$ be commensurable in length with $\alpha$.
Then neither $FK$ nor $KE$ is commensurable in length with $\alpha$.
It follows that:
- the terms of $EF$ are commensurable with the terms of $BD$ and in the same ratio
and:
$\blacksquare$
Historical Note
This proof is Proposition $112$ 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