# 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.

### 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 $AC, CB$.

## 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.

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 $CD$.

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 $EF^2$.

$AG : EF = EF : EG$
$AH : EL = EL : KG$

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 $GE$.

$AE$ is commensurable in length with both $AG$ and $GE$.

But by hypothesis $AE$ is also commensurable in length with $AB$.

$AB$ 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.

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 $EF$
$AG$ is incommensurable in length with $EF$.

From:

Proposition $1$ of Book $\text{VI}$: Areas of Triangles and Parallelograms Proportional to Base

and:

Proposition $11$ of Book $\text{X}$: Commensurability of Elements of Proportional Magnitudes:

it follows that:

$AH$ is incommensurable with $EL$.

Therefore $SN$ is also incommensurable with $MR$.

$SN : MR = PN : NR$
$PN$ 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 $AC$.

$\blacksquare$