# LCM from Prime Decomposition/Proof 2

## Theorem

Let $a, b \in \Z$.

 $\ds a$ $=$ $\ds p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$ $\ds b$ $=$ $\ds p_1^{l_1} p_2^{l_2} \ldots p_r^{l_r}$ $\ds \forall i \in \set {1, 2, \dotsc, r}: \,$ $\ds p_i$ $\divides$ $\ds a$ $\, \ds \lor \,$ $\ds p_i$ $\divides$ $\ds b$

That is, the primes given in these prime decompositions may be divisors of either of the numbers $a$ or $b$.

Then:

$\lcm \set {a, b} = p_1^{\max \set {k_1, l_1} } p_2^{\max \set {k_2, l_2} } \ldots p_r^{\max \set {k_r, l_r} }$

where $\lcm \set {a, b}$ denotes the lowest common multiple of $a$ and $b$.

## Proof

Note that if one of the primes $p_i$ does not appear in the prime decompositions of either one of $a$ or $b$, then its corresponding index $k_i$ or $l_i$ will be zero.

Let $a \divides m$.

Then:

$m$ is of the form $p_1^{h_1} p_2^{h_2} \ldots p_r^{h_r}, \forall i: 1 \le i \le r, 0 \le k_i \le h_i$
$a \divides l \iff \forall i: 1 \le i \le r, 0 \le l_i \le h_i$

So:

$a \divides m \land b \divides m \iff \forall i: 1 \le i \le r, 0 \le \max \set {k_i, l_i} \le h_i$

For $m$ to be at its smallest, we want the smallest possible exponent for each of these primes.

So for each $i \in \closedint 1 r$, $h_i$ needs to equal $\max \set {k_i, l_i}$.

Hence the result:

$\lcm \set {a, b} = p_1^{\max \set {k_1, l_1} } p_2^{\max \set {k_2, l_2} } \ldots p_r^{\max \set {k_r, l_r} }$

$\blacksquare$