Multiple of Divisor Divides Multiple
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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 $2.2$: $2 \ \text{(c)}$