Infimum of a set > If $a$ is a positive number, show that $\inf \\{a/n: n \in \mathbb{Z}^{+} \\} = 0$. So let $A = \\{a/n: n \in \mathbb{Z}^{+} \\}$. Then $A$ is bounded below by $0$. Hence $\alpha = \inf(A)$ exists. So $0 \leq \alpha$. Now $\alpha \leq a/2n$ which implies that $2 \alpha$ is a lower bound for $A$. Thus $2 \alpha \leq \alpha$. This can only be true if $\alpha \leq 0$. By trichotomy, $\alpha = 0$. Does this seem correct?

Your argument seems fine to me, however the wording is a bit strange. (See Arturo's comment).

Here is an alternative: (It is not better or worse, just different)

We know $0\leq \alpha=\inf \\{ a/n:\ n\in\mathbb{Z} \\}$. Suppose $\alpha>0$. Then choose $N\in\mathbb{Z}$ so large that $\frac{a}{N}<\alpha$. That is, choose $N>\frac{a}{\alpha}$. Then $\frac{a}{N}$ is in the set, but is smaller then $\alpha$. Hence $\alpha>0$ is impossible, so we conclude $\alpha =0 $.