# 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.

This article, or a section of it, needs explaining.In particular: "it divides it" -- not quite sure "what" divides "what" here.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Explain}}` from the code. |

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