Divisor Divides Multiple

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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 $0$.


From Common Divisor Divides Integer Combination:

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

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

$\blacksquare$


Sources