# Square on Straight Line which produces Medial Whole with Rational Area applied to Rational Straight Line

## Theorem

In the words of Euclid:

*The square on the straight line which produces with a rational area a medial whole, if applied to a rational straight line, produces as breadth a fifth apotome.*

(*The Elements*: Book $\text{X}$: Proposition $101$)

## Proof

Let $AB$ be a straight line which produces with a rational area a medial whole.

Let $CD$ be a rational straight line.

Let the rectangle $CE$ be applied to $CD$ equal to $AB^2$ producing $CF$ as breadth.

It is to be demonstrated that $CF$ is a fifth apotome.

Let $BG$ be the annex to $AB$.

Therefore, by definition, $AG$ and $GB$ are straight lines which are incommensurable in square which make $AG^2 + GB^2$ medial but $2 \cdot AG \cdot GB$ rational.

Let the rectangle $CH$ be applied to $CD$ equal to the square on $AG$, producing $CK$ as breadth.

Let the rectangle $KL$ be applied to $CD$ equal to the square on $BG$, producing $KM$ as breadth.

Then the whole $CL$ is equal to the squares on $AG$ and $GB$.

But $AG^2 + GB^2$ medial.

Therefore $CL$ is medial.

We have that $CL$ is applied to the rational straight line $CD$ producing $CM$ as breadth.

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

- $CM$ is rational and incommensurable in length with $CD$.

We have that:

- $CL = AG^2 + GB^2$

and:

- $AB^2 = CE$

Therefore by Proposition $7$ of Book $\text{II} $: Square of Difference:

- $2 \cdot AG \cdot GB = FL$

Let $FM$ be bisected at the point $N$.

Let $NO$ be drawn through $N$ parallel to $CD$.

Therefore each of the rectangles $FO$ and $LN$ is equal to the rectangle contained by $AB$ and $GB$.

We have that $2 \cdot AG \cdot GB$ is rational.

Therefore $FL$ is rational.

Also $FL$ is applied to the rational straight line $FE$, producing $FM$ as breadth.

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

- $FM$ is rational and commensurable in length with $CD$.

We have that $CL$ is medial and $FL$ is rational.

Therefore $CL$ is incommensurable with $FL$.

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

- $CL : FL = CM : FM$

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

- $CM$ is incommensurable in length with $FM$.

But both $CM$ and $FM$ are rational.

Therefore $CM$ and $FM$ are rational straight lines which are commensurable in square only.

Therefore, by definition, $CF$ is an apotome.

It remains to be shown that $CF$ is a fifth apotome.

It can be proved similarly that:

- $CK \cdot KM = NM^2 = \dfrac {FM^2} 4$

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We have that $AG^2$ is incommensurable with $GB^2$.

We also have:

- $CH = AG^2$

and:

- $KL = BG^2$

Therefore $CH$ is incommensurable with $KL$.

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

- $CH : KL = CK : KM$

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

- $CK$ is incommensurable with $KM$.

We have that:

- $CM$ and $MF$ are unequal straight lines

and:

- the rectangle $CK \cdot KM$ has been applied to $CM$ equal to $\dfrac {FM^2} 4$ and deficient by a square figure

while:

- $CK$ is incommensurable with $KM$.

Therefore from Proposition $18$ of Book $\text{X} $: Condition for Incommensurability of Roots of Quadratic Equation:

- $CM^2$ is greater than $MF^2$ by the square on a straight line which is incommensurable in length with $CM$.

Also the annex $FM$ is commensurable in length with the rational straight line $CD$.

Therefore, by definition, $CF$ is a fifth apotome.

$\blacksquare$

## Historical Note

This proof is Proposition $101$ of Book $\text{X}$ of Euclid's *The Elements*.

It is the converse of Proposition $95$: Side of Area Contained by Rational Straight Line and Fifth Apotome.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 3*(2nd ed.) ... (previous) ... (next): Book $\text{X}$. Propositions