Product of Divisors is Divisor of Product

Theorem

Let $a, b, c, d \in \Z$ be integers such that $a, c \ne 0$.

Let $a \divides b$ and $c \divides d$, where $\divides$ denotes divisibility.

Then:

$a c \divides b d$

By definition of divisibility:

$\ds \exists k_1 \in \Z: \, $

$\ds b$

$=$

$\ds a k_1$

$\ds \exists k_2 \in \Z: \, $

$\ds d$

$=$

$\ds c k_2$

Then:

$\ds b d$

$=$

$\ds \paren {a k_1} \paren {c k_2}$

$\ds $

$=$

$\ds k_1 k_2 \paren {a c}$

and the result follows by definition of divisibility.

$\blacksquare$

1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Theorem $2 \text{-} 2 \ (3)$