# Lowest Common Multiple of Integers with Common Divisor

## Theorem

Let $b, d \in \Z_{>0}$ be (strictly) positive integers

Then:

$\lcm \set {a b, a d} = a \lcm \set {b, d}$

where:

$a \in \Z_{>0}$
$\lcm \set {b, d}$ denotes the lowest common multiple of $m$ and $n$.

## Proof

We have that:

 $\ds b$ $\divides$ $\ds \lcm \set {b, d}$ Definition of Lowest Common Multiple of Integers $\, \ds \land \,$ $\ds d$ $\divides$ $\ds \lcm \set {b, d}$ $\ds \leadsto \ \$ $\ds r b$ $=$ $\ds \lcm \set {b, d}$ for some $r \in \Z$ $\, \ds \land \,$ $\ds s d$ $=$ $\ds \lcm \set {b, d}$ for some $s \in \Z$ $\ds \leadsto \ \$ $\ds r \paren {a b}$ $=$ $\ds a \lcm \set {b, d}$ $\, \ds \land \,$ $\ds s \paren {a d}$ $=$ $\ds a \lcm \set {b, d}$ $\ds \leadsto \ \$ $\ds a b$ $\divides$ $\ds a \lcm \set {b, d}$ Definition of Divisor of Integer $\, \ds \land \,$ $\ds a d$ $\divides$ $\ds a \lcm \set {b, d}$

Suppose $n \in \Z$ such that $a b \divides n$ and $a d \divides n$.

It will be shown that $a \lcm \set {b, d} \divides n$.

So:

 $\ds a b$ $\divides$ $\ds n$ by hypothesis $\, \ds \land \,$ $\ds a d$ $\divides$ $\ds n$ $\ds \leadsto \ \$ $\ds a r b$ $=$ $\ds n$ for some $r \in \Z$ $\, \ds \land \,$ $\ds a s d$ $=$ $\ds n$ for some $s \in \Z$ $\ds \leadsto \ \$ $\ds r b$ $=$ $\ds \dfrac n a$ $\, \ds \land \,$ $\ds s d$ $=$ $\ds \dfrac n a$ $\ds \leadsto \ \$ $\ds b$ $\divides$ $\ds \dfrac n a$ Definition of Divisor of Integer $\, \ds \land \,$ $\ds d$ $\divides$ $\ds \dfrac n a$ $\ds \leadsto \ \$ $\ds \lcm \set {b, d}$ $\divides$ $\ds \dfrac n a$ LCM Divides Common Multiple $\, \ds \land \,$ $\ds a \lcm \set {b, d}$ $\divides$ $\ds n$

Thus we have:

$a b \divides a \lcm \set {b, d} \land a d \divides a \lcm \set {b, d}$

and:

$a b \divides n \land a d \divides n \implies a \lcm \set {b, d} \divides n$

It follows from LCM iff Divides All Common Multiples that:

$\lcm \set {a b, a d} = a \lcm \set {b, d}$

$\blacksquare$
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