Ideal of $ \Bbb Z$ 
Assume that J is an ideal of $ \Bbb Z$. Prove that there exists $n \in J $ such that J=(n)

Not sure how to prove it. Any help is appropriated. 
 A: It is enough to prove that every additive subgroup $J$ of $\mathbb Z$ is of the form $J=n\mathbb Z$ for some integer $n$. (Why is the additive subgroup $n\mathbb Z$ the same as the ideal $(n)$? And why is it OK to just make this argument of additive subgroups?)
Now let's prove the above claim. Let $J<\mathbb Z$. Let $a$ be the least positive element of $J$ (why does such an $a$ exist?). To show that every element of $J$ is a multiple of $a$, we suppose for contradiction that there exists some $b\in J$ is not a multiple of $a$. Since $b$ is not a multiple of $a$, we can divide $b$ by $a$, and see that $b=aq+r$ with $r<a$. Since $r$ is a nonzero element of $J$ (Why is it in $J$? Why is it not $0$?), we have just contradicted our choice of $a$, so there cannot be such a $b$.
A: HINT: let $J\neq 0$ be any ideal and argue that there has to be a smallest positive element $b$ of $J$. Use the division algorithm to argue that any other element $a\in J$ can be written as $a=q\cdot b+ r$ where $0\leq r<b$. What can now be said about $r$?
