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.

\(\Box\)