The notion of an ideal of $\mathbb{Z}$ is a non-empty set of integers that is closed under addition, and closed under multiplication by an arbitrary integer. Let $I$ be a non-empty set of integers that is closed under addition (i.e., $a + b \in I$ for all $a, b \in I$ ). Show that $I$ is an ideal if and only if $-a \in I$ for all $a \in I$.

== Attempt ==

$\implies$

If $I$ is an ideal, clearly $-a \in I$, as $a \cdot (-1) \in I$, picking -1 as an arbitrary integer.

$\impliedby$

Given $I$ is closed under addition, for all $a \in I, a + a = 2a \in I, 2a + a = 3a \in I$, etc. Clearly, if $-a \in I, a\mathbb{Z} \subseteq I$. In other words, if $-a \in I, I$ becomes closed under multiplication by an arbitrary integer.

$\Box$