# GCD and LCM Distribute Over Each Other

## Theorem

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

Then:

- $\lcm \set {a, \gcd \set {b, c} } = \gcd \set {\lcm \set {a, b}, \lcm \set {a, c} }$
- $\gcd \set {a, \lcm \set {b, c} } = \lcm \set {\gcd \set {a, b}, \gcd \set {a, c} }$

That is, greatest common divisor and lowest common multiple are distributive over each other.

## Proof

### LCM Distributive over GCD

Let $p_s$ be any of the prime divisors of $a, b$ or $c$, and let $s_a, s_b$ and $s_c$ be its exponent in each of those numbers.

Let $x = \lcm \set {a, \gcd \set {b, c} }$.

Then from GCD and LCM from Prime Decomposition, the exponent of $p_s$ in $x$ is $\max \set {s_a, \min \set {s_b, s_c} }$.

From Max and Min Operations are Distributive over Each Other, $\max$ distributes over $\min$.

Therefore:

- $\max \set {s_a, \min \set {s_b, s_c} } = \min \set {\max \set {s_a, s_b}, \max \set {s_a, s_c} }$

Hence it follows that LCM is distributive over GCD.

$\Box$

### GCD Distributive over LCM

Let $p_s$ be any of the prime divisors of $a, b$ or $c$, and let $s_a, s_b$ and $s_c$ be its exponent in each of those numbers.

Let $x = \gcd \set {a, \lcm \set {b, c} }$.

Then from GCD and LCM from Prime Decomposition, the exponent of $p_s$ in $x$ is $\min \set {s_a, \max \set {s_b, s_c} }$.

From Max and Min Operations are Distributive over Each Other, $\min$ distributes over $\max$.

Therefore:

- $\min \set {s_a, \max \set {s_b, s_c} } = \max \set {\min \set {s_a, s_b}, \min \set {s_a, s_c} }$

and the result follows.

Hence it follows that GCD is distributive over LCM.

$\blacksquare$

## Sources

- 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $1$: Properties of the Natural Numbers: $\S 23 \epsilon$