LCM from Prime Decomposition

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Theorem

Let $a, b \in \Z$.

From Expression for Integers as Powers of Same Primes, let:

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


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 $b$.


General Result

Let $n \in \N$ be a natural number such that $n \ge 2$.

Let $\N_n$ be defined as:

$\N_n := \set {1, 2, \dotsc, n}$

Let $A_n = \set {a_1, a_2, \dotsc, a_n} \subseteq \Z$ be a set of $n$ integers.


From Expression for Integers as Powers of Same Primes, let:

$\displaystyle \forall i \in \N_n: a_i = \prod_{p_j \mathop \in T} {p_j}^{e_{i j} }$

where:

$T = \set {p_j: j \in \N_r}$

such that:

$\forall j \in \N_{r - 1}: p_j < p_{j - 1}$
$\forall j \in \N_r: \exists i \in \N_n: p_j \divides a_i$

where $\divides$ denotes divisibility.


Then:

$\displaystyle \map \lcm {A_n} = \prod_{j \mathop \in \N_r} {p_j}^{\max \set {e_{i j}: \, i \in \N_n} }$

where $\map \lcm {A_n}$ denotes the greatest common divisor of $a_1, a_2, \dotsc, a_n$.

The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:

$\displaystyle \map \lcm {A_n} = \prod_{j \mathop \in \N_r} {p_j}^{\max \set {e_{i j}: \, i \in \N_n} }$


Basis for the Induction

$\map P 2$ is the case:

$\displaystyle \gcd \set {a_1, a_2} = \prod_{j \mathop \in \N_r} {p_j}^{\max \set {e_{1 j}, e_{2 j} } }$

This is GCD from Prime Decomposition.


Thus $\map P 2$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$\displaystyle \map \lcm {A_k} = \prod_{j \mathop \in \N_r} {p_j}^{\max \set {e_{i j}: \, i \in \N_k} }$


from which it is to be shown that:

$\displaystyle \map \lcm {A_{k + 1} } = \prod_{j \mathop \in \N_r} {p_j}^{\max \set {e_{i j}: \, i \in \N_{k + 1} } }$


Induction Step

This is the induction step:

\(\displaystyle \map \lcm {A_{k + 1} }\) \(=\) \(\displaystyle \map \lcm {A_k \cup a_{k + 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \)



So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \Z_{\ge 2}: \displaystyle \map \lcm {A_n} = \prod_{j \mathop \in \N_r} {p_j}^{\max \set {e_{i j}: \, i \in \N_n} }$


Proof 1

\(\displaystyle \lcm \set {a, b}\) \(=\) \(\displaystyle \frac {a b} {\gcd \set {a, b} }\) Product of GCD and LCM
\(\displaystyle \) \(=\) \(\displaystyle \frac {\paren {p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r} } \paren {p_1^{l_1} p_2^{l_2} \ldots p_r^{l_r} } } {p_1^{\min \set {k_1, l_1} } p_2^{\min \set {k_2, l_2} } \cdots p_r^{\min \set {k_r, l_r} } }\) GCD from Prime Decomposition
\(\displaystyle \) \(=\) \(\displaystyle \frac {\paren {p_1^{k_1 + l_1} p_2^{k_2 + l_2} \cdots p_r^{k_r + l_r} } } {p_1^{\min \set {k_1, l_1} } p_2^{\min \set {k_2, l_2} } \cdots p_r^{\min \set {k_r, l_r} } }\)
\(\displaystyle \) \(=\) \(\displaystyle p_1^{k_1 + l_1 - \min \set {k_1, l_1} } p_2^{k_2 + l_2 - \min \set {k_2, l_2} } \cdots p_r^{k_r + l_r - \min \set {k_r, l_r} }\)
\(\displaystyle \) \(=\) \(\displaystyle p_1^{\max \set {k_1, l_1} } p_2^{\max \set {k_2, l_2} } \cdots p_r^{\max \set {k_r, l_r} }\) Sum Less Minimum is Maximum

$\blacksquare$


Proof 2

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:

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

$\blacksquare$


Examples

$125$ and $150$

The lowest common multiple of $125$ and $150$ is:

$\lcm \set {125, 150} = 750$


$132$ and $154$

The lowest common multiple of $132$ and $154$ is:

$\lcm \set {132, 154} = 924$


$39$ and $143$

The lowest common multiple of $39$ and $143$ is:

$\lcm \set {39, 143} = 429$


$253$ and $506$

The lowest common multiple of $253$ and $506$ is:

$\lcm \set {253, 506} = 506$


$111$ and $1221$

The lowest common multiple of $111$ and $1221$ is:

$\lcm \set {111, 1221} = 1221$


$39$, $102$ and $75$

The lowest common multiple of $39$, $102$ and $75$ is:

$\lcm \set {39, 102, 75} = 33 \, 150$


$p^2 q$ and $p q r$

The lowest common multiple of $p^2 q$ and $p q r$, where $p$, $q$ and $r$ are all primes, is:

$\lcm \set {p^2 q, p q r} = p^2 q r$


Also see


Sources