Theorem 1.20 (Rudin)
(a) If \(x \in \mathbb{R}\), \(y \in \mathbb{R}\) and \(x > 0\), then there is a positive integer \(n\) such that
$$
nx > y.
$$
Proof
Let \(A\) be the set of all \(nx\) where \(n\) runs through the positive integers. So
$$
A = \{nx \mid n \in 1,2,3,\ldots\}
$$
Suppose that \((a)\) is false. That is, there is a \(y \in \mathbb{R}\) such that \(y > nx\) for all \(a \in A\). Then, this implies that \(y\) is an upper bound of \(A\). Since \(A\) is bounded above, then \(A\) has a least upper bound in \(\mathbb{R}\). So let \(\alpha = \sup A\). We are given that \(x > 0\) so observe that
$$
\begin{align*}
x &> 0 \\
-x &< 0 \quad \text{(By Proposition 1.18)} \\
\alpha - x &< \alpha \quad \text{(By Definition 1.17)}
\end{align*}
$$
This implies that \(\alpha - x\) is not an upper bound of \(A\). This means that there exists some integer \(m\) such that \(\alpha - x < mx\). Re-arranging the terms
$$
\begin{align*}
\alpha &< mx + x \\
\alpha &< x(m + 1)
\end{align*}
$$
But \(m + 1\) is an integer. So \(x(m+1) \in A\). This is a contradiction since \(\alpha\) is an upper bound of \(A\). \(\ \blacksquare\)
References