Root of Area contained by Rational Straight Line and First Binomial

== Theorem ==
=== 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 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 , $AE$ 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 $AE$.
Using :
:Let a parallelogram be applied to $AE$ equal to $EF^2$ and deficient by a square.
By , 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 $CD$.
Using :
: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.
From :
:$SQ$ is a square.
We have that the rectangle contained by $AG$ and $GE$ is equal to $EF^2$.
Then by :
:$AG : EF = EF : EG$
and so by :
:$AH : EL = EL : KG$
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 $NQ$.
But from :
:$MR$ 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 :
:$AE$ is commensurable in length with both $AG$ and $GE$.
But by hypothesis $AE$ is also commensurable in length with $AB$.
Therefore by :
:$AB$ 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 :
: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 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 :
:$SN : MR = PN : NR$
and so from :
:$PN$ 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 $AC$.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory