# Divisor Divides Multiple

## Theorem

Let $a, b$ be integers.

Let:

$a \divides b$

where $\divides$ denotes divisibility.

Then:

$\forall c \in \Z: a \divides b c$

## Proof 1

Let $a \divides b$.

From Integer Divides Zero:

$a \divides 0$

Thus $a$ is a common divisor of $b$ and $0$.

$\forall p, q \in \Z: a \divides \paren {p \cdot b + q \cdot 0}$

Putting $p = c$ and $q = 1$ (for example):

$a \divides \paren {c b + 0}$

Hence the result.

$\blacksquare$

## Proof 2

 $\displaystyle a$ $\divides$ $\displaystyle b$ $\displaystyle \leadsto \ \$ $\, \displaystyle \exists x \in \Z: \,$ $\displaystyle b$ $=$ $\displaystyle x a$ Definition of Divisor of Integer $\displaystyle \leadsto \ \$ $\displaystyle b c$ $=$ $\displaystyle x c a$ $\displaystyle \leadsto \ \$ $\, \displaystyle \exists z \in \Z: \,$ $\displaystyle b c$ $=$ $\displaystyle z a$ where $z = x c$ $\displaystyle \leadsto \ \$ $\displaystyle a$ $\divides$ $\displaystyle b c$ Definition of Divisor of Integer

$\blacksquare$