Proving $a\mathbb{Z}\cap b\mathbb{Z}=[a,b]\mathbb{Z}$ The question is 

Prove that $a\mathbb{Z}\cap b\mathbb{Z}=[a,b]\mathbb{Z}$.
  Hint:First prove $b\mid a \Leftrightarrow a\mathbb Z\le b\mathbb Z$ and then prove $a\mathbb Z+b\mathbb Z=(a,b)\mathbb Z$

I managed to prove $b\mid a\Leftrightarrow a\mathbb Z\le b\mathbb Z$. About the next claim: From the first statement ($b\mid a...$) it follows that since $(a,b)\mid a$ and $(a,b)\mid b, a\mathbb Z+b\mathbb Z\le(a,b)\mathbb Z$. My problem is proving the opposite side (a.k.a $a\mathbb Z+b\mathbb Z\ge(a,b)\mathbb Z$). How can I do so and how can I proceed from $(a,b)$ to $[a,b]$?

$\newcommand{\lcm}{\operatorname{lcm}}$

$$a\mid b\iff a\Bbb Z\supset b\Bbb Z$$

P First, suppose $b=ak$. Pick $x\in b\Bbb Z$. Then $x=by=a(ky)\in a\Bbb Z$. Conversely, if $a\Bbb Z\geq b\Bbb Z$ we have $b\in a\Bbb Z$ so $b=ak$, $a\mid b$.

$$a\Bbb Z\cap b\Bbb Z=\ell\Bbb Z\;,\; \ell=\lcm(a,b)$$

P Since $a\mid \ell$, $a\Bbb Z \supset \ell \Bbb Z$. Similarily,  $b\Bbb Z \supset \ell \Bbb Z$, so $a\Bbb Z\cap b\Bbb Z\supset \ell \Bbb Z$. Now pick $x\in a\Bbb Z\cap b\Bbb Z$. Then $x=ak=bj$ for some $k,j$. Thus $a,b\mid x$. So $x$ is a common multiple, whence $ \ell\mid x$, that is $\ell =xm$ for some $m$ and $x\in\ell\Bbb Z$

$$a\Bbb Z+b\Bbb Z=d\Bbb Z\;,\;d=\gcd(a,b)$$

P Pick $y\in a\Bbb Z+b\Bbb Z$. Then $y=am+bn$. But then $d\mid y$, since $d\mid a,b$, and thus $y=kd\in d\Bbb Z$. Now pick $y\in d\Bbb Z$. Then $y=dk$ for some $k$. Bezout tells us we can write $d=an+bm$ so $y=adn+bdm\in a\Bbb Z+b\Bbb Z$.
