Let $I$ be an ideal of $\Bbb{Z}$. Prove that $I=\langle m \rangle$ for some $m \in \Bbb{ Z}$ 
Let $I$ be an ideal of $\Bbb{Z}$. Prove that $I=\langle m \rangle$ for
  some $m \in \Bbb{ Z}$.

Definition: Let $m \in \Bbb{Z}$ we define the principal ideal generated by $m $ as $\langle m \rangle=\{mx:x \in \Bbb{Z}\}$.
Thoughts: I know $\langle m \rangle=m\Bbb{Z}$ so all I need to show is that every ideal of $\Bbb{Z}$ can be written as $m\Bbb{Z}$.
I want to give some more input but I really can't think of where to begin.
I have dealt with a specific case of $2\Bbb{Z} \unlhd \Bbb{Z}$ and shown this to be true so for $I=2\Bbb{Z}$ we have $I=\langle 2 \rangle$. 
I understand it for these specific cases but I'm not sure on how to generalise it.

Suppose $I$ is not the zero ideal so there exists a positive element. By the well-ordering principle there exists a smallest positive element $b\in I$. We claim that $I=\langle b \rangle$.
One inclusion is obvious. For the other proceed by contradiction. Suppose there exists $a\in I$ such that $a\notin\langle b \rangle$. Use euclidean division with remainder to get a contradicition.
