Sum of two Incommensurable Medial Areas give rise to two Irrational Straight Lines
Theorem
In the words of Euclid:
- If two medial areas incommensurable with one another be added together, the remaining two irrational straight lines arise, namely either a second bimedial or a side of the sum of two medial areas.
(The Elements: Book $\text{X}$: Proposition $72$)
Proof
Let $AB$ and $CD$ be medial areas.
It is to be demonstrated that the "side" of their combined area $AD$ is either:
- $(1): \quad$ a second bimedial
or:
- $(2): \quad$ a side of the sum of two medial areas.
WLOG let $AB > CD$.
Let a rational straight line $EF$ be set out.
Let the rectangle $EG = AB$ be applied to $EF$, producing $EH$ as breadth.
Let the rectangle $HI = DC$ be applied to $EF$, producing $HK$ as breadth.
We have that $AB$ and $CD$ are medial areas and equals $EG$ and $HI$ respectively.
Therefore $EG$ and $HI$ are medial areas.
By Proposition $22$ of Book $\text{X} $: Square on Medial Straight Line:
- $EH$ and $HK$ are rational straight lines which are both incommensurable in length with $EF$.
We have that:
- $AB$ is incommensurable with $CD$
and:
- $AB = EG, CD = HI$
Therefore:
- $EG$ is incommensurable with $HI$.
But from Proposition $1$ of Book $\text{VI} $: Areas of Triangles and Parallelograms Proportional to Base:
- $EG : HI = EH : HK$
Therefore from Proposition $11$ of Book $\text{X} $: Commensurability of Elements of Proportional Magnitudes:
- $EH$ is incommensurable in Length with $HK$.
Therefore $EH$ and $HK$ are rational straight lines which are commensurable in square only.
Therefore, by definition, $EK$ is a binomial straight line which is divided at $H$.
We have that:
- $AB > CD$
and:
- $AB = EG, CD = HI$
Therefore:
- $EH > HK$
Thus $EH^2$ is greater than $HK^2$ by either:
- the square on a straight line which is commensurable in length with $EH$
or:
- the square on a straight line which is incommensurable in length with $EH$.
Let $EH^2 = HK^2 + \lambda^2$.
First, let $\lambda$ be commensurable in length with $EH$.
Neither of $EH$ or $HK$ is commensurable in length with the rational straight line $EF$.
Therefore $EK$ is a third binomial.
But $EF$ is rational.
- the "side" of the square equal to the rectangle contained by a rational straight line and a first binomial is second bimedial.
Therefore the "side" of $EI$ is second bimedial.
That is, the "side" of $AD$ is second bimedial.
$\Box$
Next, let $\lambda$ be incommensurable in length with $EH$.
Neither of $EH$ or $HK$ is commensurable in length with the rational straight line $EF$.
Therefore $EK$ is a sixth binomial.
But $EF$ is rational.
- the "side" of the square equal to the rectangle contained by a rational straight line and a first binomial is the side of the sum of two medial areas.
Therefore the "side" of $EI$ is the side of the sum of two medial areas.
That is, the "side" of $AD$ is the side of the sum of two medial areas.
$\blacksquare$
Historical Note
This proof is Proposition $72$ of Book $\text{X}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): Book $\text{X}$. Propositions