Proving that $a \mathbb{Z} \cap b\mathbb{Z} = \operatorname{lcm}(a,b)\mathbb Z$

Question

Given $m \in \mathbb{Z}$, let $m\mathbb{Z}$ denote the set of integer multiples of $m$, i.e. $m\mathbb{Z} := \{mk\mid k \in \mathbb{Z}\}$. Now let $a,b \in \mathbb{Z}$ with $a,b$ not both $0$. Prove that $a\mathbb{Z} \cap b\mathbb{Z} = \operatorname{lcm}(a,b)\mathbb{Z}$.

I am trying to write a proof for this, but I am unsure of what method to use. Also I am confused by $mk\mid k$, because wouldn't $m=1$ for this to be true.

Answer 1

Hint:

Step 1: Can you prove that $\operatorname{lcm}(a,b)\in a\mathbb{Z}\cap b\mathbb{Z}$?

Step 2: Can you prove that if $a$ and $b$ both divide $c$, then $c\in a\mathbb{Z}\cap b\mathbb{Z}$?

Step 3: How do steps 1 and 2, together, imply your result?

Answer 2

First of all, to clear up your confusion:

$$m\mathbb Z=\{mk\mid k\in\mathbb Z\}$$

does not mean that $mk$ divides $k$. The vertical line can be read as "so that" or "where". It means that the set $m\mathbb Z$ is the set of numbers in the form of $mk$, where $k$ is any element of $\mathbb Z$.

That said, to prove that $a\mathbb Z\cap b\mathbb Z=\operatorname{lcm}(a,b)\mathbb Z$, you need to prove that

1. if $x\in a\mathbb Z\cap b\mathbb Z$, then $x\in \operatorname{lcm}(a,b)\mathbb Z$.
2. if $x\in\operatorname{lcm}(a,b)\mathbb Z$, then $x\in a\mathbb Z\cap b\mathbb Z$.