Yes, you're right that the completeness axiom doesn't hold in \(\mathbb{Q}\)!
Here's a counterexample:
Consider the sequence \(\{a_n\}_{n=1}^\infty\) where \(a_n = (1+\frac{1}{n})^n\). It can be shown that this sequence is increasing,\(^1\) and bounded. By the monotone convergence theorem, \( \sup_{n \in \mathbb{N}}a_n = \lim_{n \to \infty}a_n\) and it so happens that this limit is \(e = 2.718... \notin \mathbb{Q} \). Therefore, \(\{a_n\}_{n=1}^\infty\) has no supremum in \(\mathbb{Q}\), and \(\{-a_n\}_{n=1}^\infty\) has no infimum in \(\mathbb{Q}\).
\(^1\) Think of it as the accumulation of a unit under an effective interest rate \(i = 1\) compounded \(n\) times per year. The more frequent the compounding, the more it accumulates.