# Show $a_1 \mathbb Z$ ∩ $a_2 \mathbb Z$ = $N \mathbb Z$ [duplicate]

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

Show

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

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, nmasantaSep 16 at 2:17

• $\operatorname{lcm}(a,b)$ 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. – Henno Brandsma Sep 15 at 7:21