Ora ana \(n \in \mathbb{Z}^+\) sing \(0 < n < 1\) .
Bukti: Ayo kita nganggep manawa pratelan iki salah. Banjur ana \(n \in \mathbb{Z}^+\) sing \(0 < n < 1\) . Ayo nimbang himpunan \(S := \{ m \in \mathbb{Z}^+ : 0 < m < 1 \}\) . Wiwit \( n \in S\) , \(S\) ora kosong. Miturut prinsip urutan sing apik (saben subset sing ora kosong saka \(\mathbb{Z}^+\) nduweni unsur sing paling cilik), \(S\) kudu nduweni unsur sing paling cilik, yaiku \(b := min(S)\) . Banjur \(b \in S\) , yaiku \(b \in \mathbb{Z}^+\) lan \(0 < b < 1\) . Integer positif dikalikan karo integer positif menehi integer positif, dadi \(b^2 \in \mathbb{Z}^+\) . Yen sampeyan tikel \(b\) karo \(0<b<1\) , sampeyan entuk \(0<b^2<b\) , supaya \(0<b^2<1\) - mulane \(b^2 \in S\) . Nanging wiwit \(b = min(S)\) , kita duwe \(b \leq b^2\) , sing mbantah \(b^2 < b\) . Dadi, sawise bukti kanthi kontradiksi, pratelan kasebut bener.