Root of Area contained by Rational Straight Line and Second Binomial
Theorem
In the words of Euclid:
- If an area be contained by a rational straight line and the second binomial, the "side" of the area is the irrational straight line which is called a first bimedial.
(The Elements: Book $\text{X}$: Proposition $55$)
Proof
Let the rectangular area $ABCD$ be contained by the rational straight line $AB$ and the second 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 $2$: Second Binomial, the lesser term $ED$ 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 $AE$.
By Proposition $17$ of Book $\text{X} $: Condition for Commensurability of Roots of Quadratic Equation:
- $AG$ 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 first bimedial.
We have that:
- $AE$ is incommensurable in length with $ED$
and:
- $ED$ is commensurable in length with $AB$
it follows from Proposition $13$ of Book $\text{X} $: Commensurable Magnitudes are Incommensurable with Same Magnitude:
- $AE$ is incommensurable in length with $AB$.
As:
- $AG$ is commensurable in length with $EG$
from Proposition $15$ of Book $\text{X} $: Commensurability of Sum of Commensurable Magnitudes:
- $AE$ is commensurable in length with each of $AG$ and $GE$.
But $AE$ is incommensurable in length with $AB$.
- $AG$ and $GE$ are incommensurable in length with $AB$.
Therefore:
- $BA$ and $AG$ are rational straight lines which are commensurable in square only
and:
- $BA$ and $GE$ are rational straight lines which are commensurable in square only.
Therefore by definition rectangles $AH$ and $GK$ are both medial.
Therefore squares $SN$ and $NQ$ are both medial.
Therefore $MN$ and $NO$ are both medial straight lines.
We have that $AG$ is commensurable in length with $GE$.
From:
and:
it follows that:
- $AH$ is commensurable with $GK$.
That is:
- $SN$ is commensurable with $NQ$.
That is:
- $MN^2$ is commensurable with $NO^2$.
We have that:
- $AE$ is incommensurable in length with $ED$
and:
- $AE$ is commensurable in length with $AG$
and:
- $ED$ is commensurable in length with $EF$.
Therefore from Proposition $13$ of Book $\text{X} $: Commensurable Magnitudes are Incommensurable with Same Magnitude:
- $AG$ is incommensurable in length with $EF$
so that:
- $AH$ is incommensurable in length with $EL$.
That is:
- $SN$ is incommensurable with $MR$.
From:
and:
it follows that:
- $PN$ is incommensurable in length with $NR$.
That is:
- $MN$ is incommensurable in length with $NO$.
But $MN$ and $NO$ have been shown to be medial and commensurable in square.
Therefore, by definition, $MN$ and $NO$ are medial and commensurable in square only.
Finally it remains to be shown that $MN$ and $NO$ contain a rational rectangle.
We have by hypothesis:
- $DE$ is commensurable in length with each of $AB$ and $EF$.
Therefore by Proposition $11$ of Book $\text{X} $: Commensurability of Elements of Proportional Magnitudes:
- $EF$ is commensurable in length with each of $EK$.
Both $EF$ and $EK$ are rational straight lines.
Therefore $EL = MR$ is rational.
But $MR$ is the rectangle contained by $MN$ and $MO$.
Therefore $MN$ and $NO$ are:
By definition, $MO$ is a first bimedial straight line.
$\blacksquare$
Historical Note
This proof is Proposition $55$ of Book $\text{X}$ of Euclid's The Elements.
It is the converse of Proposition $61$: Square on First Bimedial 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