Integer Coprime to Factors is Coprime to Whole/Proof 1

Theorem

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

Let:

$a \perp b$

$a \perp c$

where $\perp$ denotes coprimality.

Then:

$a \perp b c$

In the words of Euclid:

If two numbers be prime to any number, their product also will be prime to the same.

(The Elements: Book $\text{VII}$: Proposition $24$)

Proof

This proof follows the structure of Euclid's proof, if not its detail.

Let $a, b, c \in \Z$ such that $c$ is coprime to each of $a$ and $b$.

Let $d = a b$.

Aiming for a contradiction, suppose $c$ and $d$ are not coprime.

Then:

$\exists e \in \Z_{>1}: e \divides c, e \divides d$

We have that $c \perp a$ and $e \divides c$

From Proposition $23$ of Book $\text{VII} $: Divisor of One of Coprime Numbers is Coprime to Other:

$e \perp a$

As $e \divides d$ it follows by definition of divisibility that:

$\exists f \in \Z: e f = d$

We also have that:

$a b = d$

So:

$a b = e f$

But from Proposition $19$ of Book $\text{VII} $: Relation of Ratios to Products:

$e : a = b : f$

or in more contemporary language:

$\dfrac a e = \dfrac b f$

But $a \perp e$.

From Proposition $21$ of Book $\text{VII} $: Coprime Numbers form Fraction in Lowest Terms, $\dfrac a e$ is in canonical form.

From Proposition $20$ of Book $\text{VII} $: Ratios of Fractions in Lowest Terms:

$e \divides b$

But we already have that:

$e \divides c$

So $e$ is a divisor of both $b$ and $c$.

But this is a contradiction of the assertion that $c$ and $b$ are coprime.

Hence the result.

$\blacksquare$

Historical Note

This proof is Proposition $24$ of Book $\text{VII}$ 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 $\text{VII}$. Propositions