Number does not divide Number iff Cube does not divide Cube

Theorem

Let $a, b \in \Z$ be integers.

Then:

$a \nmid b \iff a^3 \nmid b^3$

where $a \nmid b$ denotes that $a$ is not a divisor of $b$.

In the words of Euclid:

If a cube number do not measure a cube number, neither will the side measure the side; and, if the side do not measure the side, neither will the cube measure the cube.

(The Elements: Book $\text{VIII}$: Proposition $17$)

Proof

Let $a \nmid b$.

Aiming for a contradiction, suppose:

$a^3 \divides b^3$

where $\divides$ denotes divisibility.

Then by Number divides Number iff Cube divides Cube:

$a \divides b$

From Proof by Contradiction it follows that $a^2 \divides b^2$ is false.

Thus $a^3 \nmid b^3$.

$\Box$

Let $a^3 \nmid b^3$.

Aiming for a contradiction, suppose

$a \divides b$

Then by Number divides Number iff Cube divides Cube:

$a^3 \divides b^3$

From Proof by Contradiction it follows that $a \divides b$ is false.

Thus $a \nmid b$.

$\blacksquare$

Historical Note

This proof is Proposition $17$ of Book $\text{VIII}$ of Euclid's The Elements.

Sources

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