Multiple of Divisor Divides Multiple

Theorem

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

Let:

$a \divides b$

where $\divides$ denotes divisibility.


Then:

$a c \divides b c$


Proof 1

We have that Integers form Integral Domain.

The result then follows from Multiple of Divisor in Integral Domain Divides Multiple.

$\blacksquare$


Proof 2

By definition, if $a \divides b$ then $\exists d \in \Z: a d = b$.

Then:

$\paren {a d} c = b c$

that is:

$\paren {a c} d = b c$

which follows because Integer Multiplication is Commutative and Integer Multiplication is Associative.

Hence the result.

$\blacksquare$


Sources

  • 1979: G.H. Hardy and E.M. Wright: An Introduction to the Theory of Numbers (5th ed.) ... (previous) ... (next): $\text I$: The Series of Primes: $1.1$ Divisibility of integers
  • 1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $2 \ \text{(c)}$
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