Multiple of Divisor Divides Multiple

Theorem

Let $\left({D, +, \times}\right)$ be an integral domain.

Let $a, b, c \in D$.

Let $a$ be a divisor of $b$, that is, $a \backslash b$.

Then $a \times c$ is a divisor of $b \times c$.

Corollary

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

Let $a$ be a divisor of $b$, that is, $a \backslash b$.

Then $a c \backslash b c$.

Proof

By definition, if $a \backslash b$ then $\exists d \in D: a \times d = b$.

Then $\left({a \times d}\right) \times c = b \times c$, that is:

$\left({a \times c}\right) \times d = b \times c$

which follows because $\times$ is commutative and associative in an integral domain.

Hence the result.

$\blacksquare$

Proof of Corollary

Follows directly from the fact that Integers form Integral Domain.

$\blacksquare$

Sources

• C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 3.10$: Theorem $16 \ \text{(ii)}$
• C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 6.26$: Theorem $49 \ \text{(ii)}$