There is no \(n \in \mathbb{Z}^+\), so that \(0 < n < 1\).
Proof: Let us assume that this claim is false. Then there is a \(n \in \mathbb{Z}^+\) such that \(0 < n < 1\) . Let's consider the set \(S := \{ m \in \mathbb{Z}^+ : 0 < m < 1 \}\) . Since \( n \in S\) , \(S\) is not empty. According to the well-ordering principle (every nonempty subset of \(\mathbb{Z}^+\) has a smallest element) \(S\) must have a smallest element, namely \(b := min(S)\) . Then \(b \in S\) , that is \(b \in \mathbb{Z}^+\) and \(0 < b < 1\) . A positive integer multiplied by a positive integer gives a positive integer, so \(b^2 \in \mathbb{Z}^+\) . If you multiply \(b\) by \(0<b<1\) , you get \(0<b^2<b\) , so that \(0<b^2<1\) - therefore \(b^2 \in S\) . But since \(b = min(S)\) , we have \(b \leq b^2\) , which contradicts \(b^2 < b\) . Thus, after proof by contradiction, the assertion is true.