Product of Rationally Expressible Numbers is Rational

Theorem
Let $a, b \in \set {x \in \R_{>0} : x^2 \in \Q}$ be the lengths of rational line segments.

Furthermore, let $\dfrac a b \in \Q$.

Then, $a b \in \Q$.

Proof 1

 * Euclid-X-19.png

Let the rectangle $AC$ be contained by the rational straight lines $AB$ and $BC$.

Let $AB$ and $BC$ be commensurable in length.

Let the square $AD$ be described on $AB$.

From, $AD$ is rational.

Since:
 * $AB$ is commensurable in length with $BC$

and:
 * $AB = BD$

it follows that
 * $BD$ is commensurable in length with $BC$.

From Areas of Triangles and Parallelograms Proportional to Base:
 * $BD : BC = DA : AC$

Therefore $DA$ is commensurable in length with $AC$.

But $DA$ is rational.

It follows from that $AC$ is also rational.

Proof 2
We have that $b^2 \in \Q$ by definition.

We also have that $\dfrac a b \in \Q$ by hypothesis.

Therefore, by Rational Multiplication is Closed, their product, $a b$, is also rational.