# Sum of Rational Area and Medial Area gives rise to four Irrational Straight Lines

## Theorem

In the words of Euclid:

*If a rational and a medial area be added together, four irrational straight lines arise, namely a binomial or a first bimedial or a major or a side of a rational plus a medial area.*

(*The Elements*: Book $\text{X}$: Proposition $71$)

## Proof

Let $AB$ be a rational area.

Let $CD$ be a medial area.

It is to be demonstrated that the "side" of their combined area $AD$ is either:

- $(1): \quad$ a binomial
- $(2): \quad$ a first bimedial
- $(3): \quad$ a major

or:

- $(4): \quad$ a side of a rational plus a medial area.

We have that $AB$ is either greater than or less than $CD$.

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 $AB$ is rational and equals $EG$.

Therefore $EG$ is also rational.

By Proposition $20$ of Book $\text{X} $: Quotient of Rationally Expressible Numbers is Rational:

- $EH$ is a rational straight line which is commensurable in length with $EF$.

We have that $CD$ is medial and equals $HI$.

Therefore $HI$ is also medial.

By Proposition $22$ of Book $\text{X} $: Square on Medial Straight Line:

- $HK$ is a rational straight line which is incommensurable in length with $EF$.

We have that:

- $CD$ is medial

and: $AB$ is rational.

Therefore $AB$ is incommensurable with $CD$.

Thus $EG$ is incommensurable with $HI$.

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 $HK$.

Both $EH$ and $HK$ are rational straight lines.

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 $EH$.

We have that the greater straight line $HE$ is commensurable in length with the rational straight line $EF$.

Then by definition $EK$ is a first 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 binomial.

Therefore the "side" of $EI$ is binomial.

That is, the "side" of $AD$ is binomial.

$\Box$

Next, let $\lambda$ be incommensurable in length with $EH$.

As before, the greater straight line $HE$ is commensurable in length with the rational straight line $EF$.

Then by definition $EK$ is a fourth binomial.

But $EF$ is rational.

- the "side" of the square equal to the rectangle contained by a rational straight line and a fourth binomial is major.

Therefore the "side" of $EI$ is major.

That is, the "side" of $AD$ is major.

$\Box$

Let $AB < CD$.

Then $EG < HI$ and so $EH < HK$.

Thus $HK^2$ is greater than $EH^2$ by either:

- the square on a straight line which is commensurable in length with $HK$

or:

- the square on a straight line which is incommensurable in length with $HK$.

Let $HK^2 = EK^2 + \lambda^2$.

First, let $\lambda$ be commensurable in length with $HK$.

We have that $EH$ is commensurable in length with $EF$.

Therefore by definition $EK$ is a second binomial.

But $EF$ is rational.

- the "side" of the square equal to the rectangle contained by a rational straight line and a second binomial is first bimedial.

Therefore the "side" of $EI$ is first bimedial.

That is, the "side" of $AD$ is first bimedial.

$\Box$

Next, let $\lambda$ be incommensurable in length with $HK$.

We have that $EH$ is commensurable in length with $EF$.

Therefore by definition $EK$ is a fifth binomial.

But $EF$ is rational.

- the "side" of the square equal to the rectangle contained by a rational straight line and a fifth binomial is first bimedial.

Therefore the "side" of $EI$ is the side of a rational plus a medial area.

That is, the "side" of $AD$ is the side of a rational plus a medial area.

$\blacksquare$

## Historical Note

This proof is Proposition $71$ 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