Akukho \(n \in \mathbb{Z}^+\) kangangokuthi \(0 < n < 1\) .
Ubufakazi: Ake sicabange ukuthi lesi simangalo singamanga. Bese kuba khona \(n \in \mathbb{Z}^+\) okuthi \(0 < n < 1\) . Ake sicabangele isethi \(S := \{ m \in \mathbb{Z}^+ : 0 < m < 1 \}\) . Njengoba \( n \in S\) , \(S\) ayinalutho. Ngokwesimiso sokuhleleka kahle (yonke isethi engaphansi engenalutho ye \(\mathbb{Z}^+\) inesici esincane kunazo zonke), \(S\) kufanele ibe nento encane kunazo zonke, okungukuthi \(b := min(S)\) . Bese \(b \in S\) , okungukuthi \(b \in \mathbb{Z}^+\) kanye \(0 < b < 1\) . Inombolo ephozithivu ephindwe ngenombolo ephozithivu inikeza inombolo ephelele, ngakho \(b^2 \in \mathbb{Z}^+\) . Uma uphindaphinda \(b\) ngokuthi \(0<b<1\) , uthola \(0<b^2<b\) , ukuze \(0<b^2<1\) - ngakho-ke \(b^2 \in S\) . Kodwa njengoba \(b = min(S)\) , sine \(b \leq b^2\) , ephikisana \(b^2 < b\) . Ngakho, ngemva kobufakazi ngokuphikisana, ukugomela kuyiqiniso.