# Proof

(related to Proposition: Least Common Multiple)

• Assume $$a,b\in\mathbb Z$$ are integers and $$M_{a,b}$$ is a set defined by $M:=\left\{m\in\mathbb N : a\mid m\wedge b\mid m\right\}.$
• $M_{a,b}$ is not empty, since it contains e.g. the numbers $$\pm ab, \pm 2ab, \pm 3ab$$ etc.
• Moreover, the intersection $M_{a,b}\cap \mathbb N$ with the set $\mathbb N$ of natural numbers is not empty, since it contains, e.g. the numbers $$ab, 2ab, 3ab$$ etc.
• From the well-ordering principle of natural numbers, it follows that $M_{a,b}\cap \mathbb N$ contains a unique minimum $$m_0$$.
• We set $\operatorname{lcm}(a,b):=m_0.$
• It remains to be shown that $$\operatorname{lcm}(a,b)$$ is a divisor of any $$m\in M_{a,b}$$, in particular $\operatorname{lcm}(a,b)\neq 0.$
• According to division with quotient and remainder, the equation $m=q\cdot \operatorname{lcm}(a,b)+r$ is uniquely solved by two other natural numbers $$q,r$$ with $$0\le r < \operatorname{lcm}(a,b).\quad\quad( * ).$$
• Because $r = m-q\cdot \operatorname{lcm}(a,b)=m\cdot 1 + \operatorname{lcm}(a,b)(-q)$ and because $a\mid m,$ $a\mid \operatorname{lcm}(a,b),$ $b\mid m,$ and $b\mid \operatorname{lcm}(a,b),$ it follows from the divisibility laws that $a\mid r,$ and $b\mid r.$
• If we assume that $$r > 0$$, then $$r\in M_{a,b}$$.
• However, we defined $$\operatorname{lcm}(a,b)$$ as the minimum element of $$M_{a,b}$$, thus $$\operatorname{lcm}(a,b)\le r$$, in contradiction to $$( * )$$ requiring $$\operatorname{lcm}(a,b) > r$$.
• Thus, our assumption $$r > 0$$ must be wrong and we have $$r=0$$. This means that $m=q\cdot \operatorname{lcm}(a,b).$
• This demonstrates that $\operatorname{lcm}(a,b)\mid m$ for all $m\in M_{a,b}.$

Github: ### References

#### Bibliography

1. Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927