Root of Area contained by Rational Straight Line and Fourth Binomial
Theorem
In the words of Euclid:
- If an area be contained by a rational straight line and the fourth binomial, the "side" of the area is the irrational straight line called major.
(The Elements: Book $\text{X}$: Proposition $57$)
Proof
Let the rectangular area $ABCD$ be contained by the rational straight line $AB$ and the fourth binomial $AD$.
Let $E$ divide $AD$ into its terms such that $AE > ED$.
From Proposition $42$ of Book $\text{X} $: Binomial Straight Line is Divisible into Terms Uniquely this is possible at only one place.
By definition of binomial, $AE$ and $ED$ are rational straight lines which are commensurable in square only.
Thus by definition of fourth binomial:
- $AE^2$ is greater than $ED^2$ by the square on a straight line which is incommensurable in length with $AE$.
By Book $\text{X (II)}$ Definition $4$: Fourth Binomial, the greater term $AE$ is commensurable in length with $AB$.
Let $ED$ be bisected at $F$.
- Let a parallelogram be applied to $AE$ equal to $EF^2$ and deficient by a square.
Let the rectangle contained by $AG$ and $GE$ equal to $EF^2$ be applied to $AE$.
- $AG$ is incommensurable in length with $EG$.
Let $GH$, $EK$ and $FL$ be drawn from $G$, $E$ and $F$ parallel to $AB$ and $CD$.
Using Proposition $14$ of Book $\text{II} $: Construction of Square equal to Given Polygon:
- let the square $SN$ be constructed equal to the parallelogram $AH$
and:
- let the square $NQ$ be constructed equal to the parallelogram $GK$.
Let $SN$ and $NQ$ be placed so that $MN$ is in a straight line with $NO$.
Therefore $RN$ is also in a straight line with $NP$.
Let the parallelogram $SQ$ be completed.
- $SQ$ is a square.
We have that $AD$ is a fourth binomial straight line.
Therefore $AE$ and $ED$ are rational straight lines which are commensurable in square only.
We have that the rectangle contained by $AG$ and $GE$ is equal to $EF^2$.
Then by Proposition $17$ of Book $\text{VI} $: Rectangles Contained by Three Proportional Straight Lines:
- $AG : EF = EF : EG$
and so by Proposition $1$ of Book $\text{VI} $: Areas of Triangles and Parallelograms Proportional to Base:
- $AH : EL = EL : KG$
Therefore $EL$ is a mean proportional between $AH$ and $GK$.
But:
- $AH = SN$
and:
- $GK = NQ$
therefore $EL$ is a mean proportional between $SN$ and $NQ$.
- $MR$ is a mean proportional between $SN$ and $NQ$.
Therefore:
- $EL = MR$
and so:
- $EL = PO$
But:
- $AH + GK = SN + NQ$
Therefore:
- $AC = SQ$
That is:
- $AC$ is an area contained by a rational straight line $AB$ and the first binomial $AD$, while the side of the area $SQ$ is $MO$.
We have that:
- $AG$ is incommensurable in length with $EG$
and so by:
and:
it follows that:
- $AH$ is incommensurable with $GK$
that is:
- $SN$ is incommensurable with $NQ$.
Therefore $MN$ and $NO$ are incommensurable in square.
We have that:
- $AE$ is commensurable in length with $AB$
and therefore from Proposition $19$ of Book $\text{X} $: Product of Rationally Expressible Numbers is Rational:
- $AK$ is rational.
But $AK = MN^2 + NO^2$.
Therefore $MN^2 + NO^2$ is rational.
Since:
- $DE$ is incommensurable in length with $AB = EK$
while:
- $DE$ is commensurable in length with $EF$
- $EF$ is incommensurable in length with $EK$.
Therefore $EF$ and $EK$ are rational straight lines which are commensurable in square only.
Therefore $LE = MR$ is medial.
But $MR$ is a rectangle contained by $MN$ and $NO$.
Therefore the rectangle contained by $MN$ and $NO$ is medial.
We also have that:
- $MN^2 + NO^2$ is rational
- $MN$ and $NO$ are incommensurable in square.
Thus by definition $MN + NO$ is major.
$\blacksquare$
Historical Note
This proof is Proposition $57$ of Book $\text{X}$ of Euclid's The Elements.
It is the converse of Proposition $63$: Square on Major Straight Line applied to Rational Straight Line.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): Book $\text{X}$. Propositions