Product of GCD and LCM/Proof 3
Jump to navigation
Jump to search
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
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:
- $a b = d l = \gcd \set {a, b} \cdot \lcm \set {a, b}$
$\blacksquare$