LCM from Prime Decomposition/General Result

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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:

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


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

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.


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

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:

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

Basis for the Induction

$\map P 2$ is the case:

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

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:

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

from which it is to be shown that:

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

Induction Step

This is the induction step:

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

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


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