LCM from Prime Decomposition/Proof 1
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Theorem
Let $a, b \in \Z$.
From Expression for Integers as Powers of Same Primes, let:
\(\ds a\) | \(=\) | \(\ds p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}\) | ||||||||||||
\(\ds b\) | \(=\) | \(\ds p_1^{l_1} p_2^{l_2} \ldots p_r^{l_r}\) | ||||||||||||
\(\ds \forall i \in \set {1, 2, \dotsc, r}: \, \) | \(\ds p_i\) | \(\divides\) | \(\ds a\) | |||||||||||
\(\, \ds \lor \, \) | \(\ds p_i\) | \(\divides\) | \(\ds b\) |
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
\(\ds \lcm \set {a, b}\) | \(=\) | \(\ds \frac {a b} {\gcd \set {a, b} }\) | Product of GCD and LCM | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r} } \paren {p_1^{l_1} p_2^{l_2} \ldots p_r^{l_r} } } {p_1^{\min \set {k_1, l_1} } p_2^{\min \set {k_2, l_2} } \cdots p_r^{\min \set {k_r, l_r} } }\) | GCD from Prime Decomposition | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {p_1^{k_1 + l_1} p_2^{k_2 + l_2} \cdots p_r^{k_r + l_r} } } {p_1^{\min \set {k_1, l_1} } p_2^{\min \set {k_2, l_2} } \cdots p_r^{\min \set {k_r, l_r} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds p_1^{k_1 + l_1 - \min \set {k_1, l_1} } p_2^{k_2 + l_2 - \min \set {k_2, l_2} } \cdots p_r^{k_r + l_r - \min \set {k_r, l_r} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds p_1^{\max \set {k_1, l_1} } p_2^{\max \set {k_2, l_2} } \cdots p_r^{\max \set {k_r, l_r} }\) | Sum Less Minimum is Maximum |
$\blacksquare$