Square of Cube Number is Cube/Proof 1

Theorem

Let $a \in \N$ be a natural number.

Let $a$ be a cube number.

Then $a^2$ is also a cube number.

In the words of Euclid:

If a cube number by multiplying itself make some number the product will be cube.

(The Elements: Book $\text{IX}$: Proposition $3$)

Proof

By the definition of cube number:

$\exists k \in \N: k^3 = a$

Thus:

\(\ds a^2\)

\(=\)

\(\ds \paren {k^3}^2\)

\(\ds \)

\(=\)

\(\ds k^6\)

\(\ds \)

\(=\)

\(\ds \paren {k^2}^3\)

Thus:

$\exists r = k^2 \in \N: a = r^3$

Hence the result by definition of cube number.

$\blacksquare$