Akukho \(n \in \mathbb{Z}^+\) ukuba \(0 < n < 1\) .
Ubungqina: Masicinge ukuba eli bango libubuxoki. Emva koko kukho \(n \in \mathbb{Z}^+\) ukuba \(0 < n < 1\) . Makhe siqwalasele iseti \(S := \{ m \in \mathbb{Z}^+ : 0 < m < 1 \}\) . Ekubeni \( n \in S\) , \(S\) ayinanto. Ngokomgaqo-nkqubo wolungelelwaniso kakuhle (lonke iseti engaphantsi engenanto ye \(\mathbb{Z}^+\) ineyona elementi incinane), \(S\) kufuneka ibe neyona elementi incinci, eyile \(b := min(S)\) . Emva koko \(b \in S\) , oko kukuthi \(b \in \mathbb{Z}^+\) kunye \(0 < b < 1\) . Inombolo ephozithivu ephinda-phinda nge-positive integer inika i-positive integer, ngoko \(b^2 \in \mathbb{Z}^+\) . Ukuba uphinda-phinda \(b\) nge \(0<b<1\) , ufumana \(0<b^2<b\) , ukuze \(0<b^2<1\) - ngoko \(b^2 \in S\) . Kodwa ekubeni \(b = min(S)\) , sinayo \(b \leq b^2\) , ephikisanayo \(b^2 < b\) . Ngoko ke, emva kobungqina ngokuchasa, uluvo luyinyaniso.