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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?

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Your argument about $2\alpha$ is a bit confusing as it is written. Rather, you should say that "since $\alpha\leq a/n$ for all $n$, then $\alpha\leq a/2n$ for all $n$, hence $2\alpha\leq a/n$ for all $n$. Thus..." (Note the specification that the inequalities hold *for all $n$*. The way it's written, you only have $\alpha\leq a/2n$ for some $n$, which means $2\alpha\leq a/n$ for some $n$, so there is a gap between that assertion and saying it is a lower bound for $A$. Otherwise, it seems fine. –  Arturo Magidin Jun 21 '11 at 18:36
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Also, rather than trichotomy, the correct term is antisymmetry. –  Apostolos Jun 21 '11 at 18:40
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1 Answer 1

up vote 2 down vote accepted

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 $.

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