Root of Area contained by Rational Straight Line and First Binomial
Theorem
In the words of Euclid:
- If an area be contained by a rational straight line and the first binomial, the "side" of the area is the irrational straight line which is called binomial.
(The Elements: Book $\text{X}$: Proposition $54$)
Lemma
In the words of Euclid:
- Let there be two squares $AB, BC$, and let them be placed so that $DB$ is in a straight line with $BE$; therefore $FB$ is also in a straight line with $BG$.
Let the parallelogram $AC$ be completed; I say that $AC$ is a square, that $DG$ is a mean proportional between $AB, BC$, and further, that $DC$ is a mean proportional between $AC, CB$.
(The Elements: Book $\text{X}$: Proposition $54$ : Lemma)
Proof
Let the rectangular area $AC$ be contained by the rational straight line $AB$ and the first binomial $AD$.
Let $E$ divide $AD$ into its terms.
From Proposition $42$ of Book $\text{X} $: Binomial Straight Line is Divisible into Terms Uniquely this is possible at only one place.
Let $AE$ be its greater term.
By definition of binomial, $AE$ and $ED$ are rational straight lines which are commensurable in square only.
Thus $AE^2$ is greater than $ED^2$ by the square on a straight line which is commensurable with $AE$.
By Book $\text{X (II)}$ Definition $1$: First Binomial, $AE$ is commensurable in length with $AB$.
Let $ED$ be bisected at $F$.
We have that $AE^2$ is greater than $ED^2$ by the square on a straight line which is commensurable with $AE$.
- Let a parallelogram be applied to $AE$ equal to $EF^2$ and deficient by a square.
By Proposition $17$ of Book $\text{X} $: Condition for Commensurability of Roots of Quadratic Equation, it divides it into commensurable parts.
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Let the rectangle contained by $AG$ and $GE$ equal to $EF^2$ be applied to $AE$.
Then $AG$ is commensurable 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 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 $MO$.
It remains to be demonstrated that $MO$ is binomial.
We have that $AG$ is commensurable in length with $GE$.
Therefore by Proposition $15$ of Book $\text{X} $: Commensurability of Sum of Commensurable Magnitudes:
- $AE$ is commensurable in length with both $AG$ and $GE$.
But by hypothesis $AE$ is also commensurable in length with $AB$.
Therefore by Proposition $12$ of Book $\text{X} $: Commensurability is Transitive Relation:
- $AB$ is commensurable in length with both $AG$ and $GE$.
We have that $AB$ is a rational straight line.
Therefore $AG$ and $GE$ are rational straight lines.
Therefore each of the rectangles $AH$ and $GK$ are rational.
Therefore by Proposition $19$ of Book $\text{X} $: Product of Rationally Expressible Numbers is Rational:
- each of the rectangles $AH$ and $GK$ are rational.
and:
- $AH$ is commensurable with $GK$.
But:
- $AH = SN$
and:
- $GK = NQ$
Therefore $SN$ and $NQ$, the squares on $MN$ and $NO$, are rational and commensurable.
Since:
- $AE$ is incommensurable in length with $ED$
while:
- $AE$ is commensurable in length with $AG$
and:
- $DE$ is commensurable in length with $EF$
it follows from Proposition $13$ of Book $\text{X} $: Commensurable Magnitudes are Incommensurable with Same Magnitude that:
- $AG$ is incommensurable in length with $EF$.
From:
and:
it follows that:
- $AH$ is incommensurable with $EL$.
Therefore $SN$ is also incommensurable with $MR$.
But from Proposition $1$ of Book $\text{VI} $: Areas of Triangles and Parallelograms Proportional to Base:
- $SN : MR = PN : NR$
and so from Proposition $11$ of Book $\text{X} $: Commensurability of Elements of Proportional Magnitudes:
- $PN$ is incommensurable with $NR$.
But:
- $PN = MN$ and $NR = NO$
Therefore $MN$ is incommensurable with $NO$.
Also:
- $MN^2$ is commensurable with $NO^2$
and:
- $MN^2$ and $NO^2$ are both rational.
Therefore $MN$ and $NO$ are rational straight lines which are commensurable in square only.
Therefore by definition $MO$ is binomial and the "side" of the area $AC$.
$\blacksquare$
Historical Note
This proof is Proposition $54$ of Book $\text{X}$ of Euclid's The Elements.
It is the converse of Proposition $60$: Square on Binomial 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