Product of GCD and LCM

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Theorem

$\lcm \set {a, b} \times \gcd \set {a, b} = \size {a b}$

where:

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


Proof 1

It is sufficient to prove that $\lcm \set {a, b} \times \gcd \set {a, b} = a b$, where $a, b \in \Z_{>0}$.

\(\displaystyle d\) \(=\) \(\displaystyle \gcd \set {a, b}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle d\) \(\divides\) \(\displaystyle a b\)
\(\displaystyle \leadsto \ \ \) \(\, \displaystyle \exists n \in \Z_{>0}: \, \) \(\displaystyle a b\) \(=\) \(\displaystyle d n\)


\(\displaystyle d \divides a\) \(\land\) \(\displaystyle d \divides b\)
\(\displaystyle \leadsto \ \ \) \(\, \displaystyle \exists u, v \in \Z: \, \) \(\displaystyle a = d u\) \(\land\) \(\displaystyle b = d v\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle d u b = d n\) \(\land\) \(\displaystyle a d v = d n\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle n = b u\) \(\land\) \(\displaystyle n = a v\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle a \divides n\) \(\land\) \(\displaystyle b \divides n\)


Now we have $a \divides m \land b \divides m \implies m = a r = b s$.

Also, by Bézout's Lemma we have $d = a x + b y$.

So:

\(\displaystyle m d\) \(=\) \(\displaystyle a x m + b y m\)
\(\displaystyle \) \(=\) \(\displaystyle b s a x + a r b y\)
\(\displaystyle \) \(=\) \(\displaystyle a b \paren {s x + r y}\)
\(\displaystyle \) \(=\) \(\displaystyle d n \paren {s x + r y}\)


So:

$m = n \paren {s x + r y}$

Thus:

$n \divides m \implies n \le \size m$

while:

$a b = d n = \gcd \set {a, b} \times \lcm \set {a, b}$

as required.

$\blacksquare$


Proof 2

Let $a = g m$ and $b = g n$, where $g = \gcd \set {a, b}$ and $m$ and $n$ are coprime.

The existence of $m$ and $n$ are proved by Integers Divided by GCD are Coprime.

Since $a = g m \divides g m n$ and $b = g n \divides g m n$, $g m n$ is the LCM of $a$ and $b$.


Then it follows that:

$\lcm \set {a, b} \times \gcd \set {a, b} = g m n \times g = g m \times g n = \size {a b}$

$\blacksquare$


Proof 3

Let $d := \gcd \set {a, b}$.

Then by definition of the GCD, there exist $j_1, j_2 \in \Z$ such that $a = d j_1$ and $b = d j_2$.

Because $d$ divides both $a$ and $b$, it must divide their product:

$\exists l \in \Z$ such that $a b = d l$

Then we have:

\(\displaystyle d l\) \(=\) \(\, \displaystyle \paren {d j_1} b \, \) \(\, \displaystyle =\, \) \(\displaystyle a \paren {d j_2}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle l\) \(=\) \(\, \displaystyle j_1 b \, \) \(\, \displaystyle =\, \) \(\displaystyle a j_2\)

showing that $a \divides l$ and $b \divides l$.

That is, $l$ is a common multiple of $a$ and $b$.


Now it must be shown that $l$ is the least such number.

Let $m$ be any common multiple of $a$ and $b$.

Then there exist $k_1, k_2 \in \Z$ such that $m = a k_1 = b k_2$.

By Bézout's Lemma:

$\exists x, y \in \Z: d = a x + b y$

So:

\(\displaystyle m d\) \(=\) \(\displaystyle m a x + m b y\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {b k_2} a x + \paren {a k_1} b y\)
\(\displaystyle \) \(=\) \(\displaystyle a b \paren {b k_2 + a k_1}\)
\(\displaystyle \) \(=\) \(\displaystyle d l \paren {b k_2 + a k_1}\)

Thus:

$m = l \paren {b k_2 + a k_1}$

that is, $l \divides m$.

Hence by definition of the LCM:

$\lcm \set {a, b} = l$

In conclusion:

$a b = d l = \gcd \set {a, b} \cdot \lcm \set {a, b}$

$\blacksquare$


Proof 4

Let:

\(\displaystyle m\) \(=\) \(\displaystyle {p_1}^{k_1} {p_2}^{k_2} \dotsm {p_r}^{k_r}\)
\(\displaystyle n\) \(=\) \(\displaystyle {p_1}^{l_1} {p_2}^{l_2} \dotsm {p_r}r^{l_r}\)


From LCM from Prime Decomposition:

$\lcm \set {m, n} = p_1^{\max \set {k_1, l_1} } p_2^{\max \set {k_2, l_2} } \dotsm p_r^{\max \set {k_r, l_r} }$


From GCD from Prime Decomposition:

$\gcd \set {m, n} = p_1^{\min \set {k_1, l_1} } p_2^{\min \set {k_2, l_2} } \dotsm p_r^{\min \set {k_r, l_r} }$


From Sum of Maximum and Minimum, for all $i \in \set {1, 2, \ldots, r}$:

$\min \set {k_i, l_i} + \max \set {k_i, l_i} = k_i + l_i$


Hence:

\(\displaystyle \gcd \set {m, n} \times \lcm \set {m, n}\) \(=\) \(\displaystyle p_1^{k_1 + l_1} p_2^{k_2 + l_2} \dotsm p_r^{k_r + l_r}\)
\(\displaystyle \) \(=\) \(\displaystyle p_1^{k_1} p_1^{l_1} p_2^{k_2} p_2^{l_2} \dotsm p_r^{k_r} p_r^{l_r}\)
\(\displaystyle \) \(=\) \(\displaystyle p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r} \times p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}\)
\(\displaystyle \) \(=\) \(\displaystyle m n\)

$\blacksquare$


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