Let $a_1$, $a_2$$\mathbb Z$ and let $N$ = lcm($a_1$, $a_2$).


$a_1 \mathbb Z$$a_2 \mathbb Z$ = $N \mathbb Z$

where $c \mathbb Z$ = {$c · n : n$$\mathbb Z$}.

My attempt:

let $x$$a_1 \mathbb Z$$a_2 \mathbb Z$, thus $x$$a_1 \mathbb Z$ and $x$$a_2 \mathbb Z$, so we can write $x$ = $a_1.n$ and $x$ = $a_2.m$, where $n,m$$\mathbb Z$.

Since $N$ = lcm($a_1$, $a_2$), then $a_1$/$N$ and $a_2$/$N$, thus there exists $l,s$$\mathbb Z$ such that $l.a_1 = N$ and $s.a_2 = N$.

Let $y$$N \mathbb Z$, thus we can write $y$ = $N.b$ for some $b$$\mathbb Z$

But $x$ = $a_1.n$ = $\frac Nl . n$, and $x$ = $a_2.m$ = $\frac Ns . m$.

But $x=x$, thus $\frac Nl . n$ = $\frac Ns . m$ and hence $\frac nl$ = $\frac ms$, and since $b$$\mathbb Z$, it can be written as $\frac nl$.

Therefore $y = N.b = N. \frac nl$ = $x$.

Hence for all $x$$a_1 \mathbb Z$$a_2 \mathbb Z$, $x$$N \mathbb Z$ and so, $a_1 \mathbb Z$$a_2 \mathbb Z$ = $N \mathbb Z$


marked as duplicate by Javi, ThorWittich, Yanior Weg, Matthew Daly, nmasanta Sep 16 at 2:17

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  • $\begingroup$ But is my answer correct?? $\endgroup$ – cbc bc Sep 14 at 15:17
  • $\begingroup$ $\operatorname{lcm}(a,b)$ is the unique positive integer $k$ such that $a \Bbb Z \cap b \Bbb Z = k \Bbb Z $. This is the abstract way to define the lcm. $\endgroup$ – Henno Brandsma Sep 15 at 7:21