# Product of GCD and LCM

## 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 $a, b \in \Z_{>0}$.

 $\ds d$ $=$ $\ds \gcd \set {a, b}$ $\ds \leadsto \ \$ $\ds d$ $\divides$ $\ds a b$ $\ds \leadsto \ \$ $\ds \exists n \in \Z_{>0}: \,$ $\ds a b$ $=$ $\ds d n$

 $\ds d \divides a$ $\land$ $\ds d \divides b$ $\ds \leadsto \ \$ $\ds \exists u, v \in \Z: \,$ $\ds a = d u$ $\land$ $\ds b = d v$ $\ds \leadsto \ \$ $\ds d u b = d n$ $\land$ $\ds a d v = d n$ $\ds \leadsto \ \$ $\ds n = b u$ $\land$ $\ds n = a v$ $\ds \leadsto \ \$ $\ds a \divides n$ $\land$ $\ds b \divides n$

Now we have:

$a \divides m \land b \divides m \implies m = a r = b s$

Also, by Bézout's Identity we have:

$d = a x + b y$

So:

 $\ds m d$ $=$ $\ds a x m + b y m$ $\ds$ $=$ $\ds b s a x + a r b y$ $\ds$ $=$ $\ds a b \paren {s x + r y}$ $\ds$ $=$ $\ds d n \paren {s x + r y}$

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.

It follows that:

$a = g m \divides g m n$

and:

$b = g n \divides g m n$

So $g m n$ is a common multiple of $a$ and $b$.

Hence there exists an integer $g k \le g m n$ that is divisible by both $a$ and $b$.

Then:

 $\ds a$ $=$ $\ds g m$ $\ds \leadsto \ \$ $\ds g m$ $\divides$ $\ds g k$ $\ds \leadsto \ \$ $\ds m$ $\divides$ $\ds k$

 $\ds b$ $=$ $\ds g n$ $\ds \leadsto \ \$ $\ds g n$ $\divides$ $\ds g k$ $\ds \leadsto \ \$ $\ds n$ $\divides$ $\ds k$

As $g k \le g m n$, it follows that:

$k \le m n$

But $m, n$ are coprime.

So:

 $\ds \map \lcm {m, n}$ $=$ $\ds m n$ $\ds \leadsto \ \$ $\ds k$ $\not <$ $\ds m n$ $\ds \leadsto \ \$ $\ds k$ $=$ $\ds m n$ $\ds \leadsto \ \$ $\ds g k$ $=$ $\ds g m n$ $\ds$ $=$ $\ds \map \lcm {a, b}$ $\ds \leadsto \ \$ $\ds \lcm \set {a, b} \times \gcd \set {a, b}$ $=$ $\ds g m n \times g$ $\ds$ $=$ $\ds g m \times g n$ $\ds$ $=$ $\ds \size {a b}$

$\blacksquare$

## 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:

 $\ds d l$ $=$ $\, \ds \paren {d j_1} b \,$ $\, \ds = \,$ $\ds a \paren {d j_2}$ $\ds \leadsto \ \$ $\ds l$ $=$ $\, \ds j_1 b \,$ $\, \ds = \,$ $\ds a j_2$

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 $m = a k_1 = b k_2$.

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

So:

 $\ds m d$ $=$ $\ds m a x + m b y$ $\ds$ $=$ $\ds \paren {b k_2} a x + \paren {a k_1} b y$ $\ds$ $=$ $\ds a b \paren {b k_2 + a k_1}$ $\ds$ $=$ $\ds d l \paren {b k_2 + a k_1}$

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

From Fundamental Theorem of Arithmetic, let:

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

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:

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

$\blacksquare$

## Proof 5

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

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

Let $m = \dfrac {a b} d$.

Then:

$m = a s = r b$

which makes $m$ a common multiple of $a$ and $b$.

Let $c \in \Z_{>0}$ be a common multiple of $a$ and $b$.

Let us say that:

$c = a u = b v$

From Bézout's Identity:

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

Then:

 $\ds \dfrac c m$ $=$ $\ds \dfrac {c d} {a b}$ $\ds$ $=$ $\ds \dfrac {c \paren {a x + b y} } {a b}$ $\ds$ $=$ $\ds \dfrac c b x + \dfrac c a y$ $\ds$ $=$ $\ds v x + n y$

That is:

$m \divides c$

where $\divides$ denotes divisibility.

$m \le c$

Hence by definition of the LCM:

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

In conclusion:

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

and the result follows.

$\blacksquare$