Proof: Suppose not. Then \(\mathbb{N}\) is a subset of \(\mathbb{R}\) bounded above by \(x\). Now let \(s = \sup \mathbb{N}\). According to the Axiom of Completeness, \(s\) is finite. Now \(s-1\) is not an upper bound of \(\mathbb{N}\) since \(s\) is the least upper bound. So there is a natural number \(n_0\) such that \(s-1 < n_0\).

Thus \(s < n_0 + 1\ (\in\mathbb{N})\) and \(s\) is not an upper bound. This is a contradiction. The result follows.

Note: To see that this is equivalent to the textbook version, we suppose that \(a, b > 0\). Then \(b/a > 0\) and by AP, there is a natural number \(n\) such that \(n > b/a\). That is, \(na > b\), as desired.

Proof: Let \(0 < a < b\) be real numbers. Then \(b-a > 0\) and by AP, there is a natural number \(q\) such that \(q(b-a) > 1\). Rearranging we have \(qa < 1 + qa < qb\).

Unfortunately, \(1+qa\) is not necessarily an integer. Let \(W = \{n\in\mathbb{N}: n\geq qb\}\). According to AP, \(W\) is not empty and so it must contain a least element, \(p_0\). We have \(p_0 - 1 < qb < p_0\) since \(p_0\) is the smallest element in \(W\). Putting the two inequalities together yields

\begin{align*} 1 + qa &< qb < p_0 \\ qa < qb - 1 &< p_0 - 1 < qb \\ a <\ &\frac{p_0 - 1}{q} < b \end{align*}

Since \(p_0 - 1\) is a natural number, we are done. Notice that we assumed that \(a\) and \(b\) were both positive.

Exercise: Complete the proof by eliminating this assumption. (Hint: If \(a<0\) then there is a natural number \(n\) such that \(n > -a\).)