Ma jiro wax la mid ah \(n \in \mathbb{Z}^+\) \(0 < n < 1\) .
Caddeyn: Aan u qaadanno in sheegashadani ay been tahay. Markaas waxaa jira \(n \in \mathbb{Z}^+\) oo ah \(0 < n < 1\) . Aynu tix-gelinno habka \(S := \{ m \in \mathbb{Z}^+ : 0 < m < 1 \}\) . Mar haddii \( n \in S\) , \(S\) waa wax aan madhnayn. Marka loo eego mabda'a si wanaagsan u dalbiyay (qayb kasta oo aan faaruqin oo ah \(\mathbb{Z}^+\) waxa uu leeyahay curiyaha ugu yar), \(b := min(S)\) \(S\) \(b := min(S)\) Dabadeed \(b \in S\) , waa \(b \in \mathbb{Z}^+\) iyo \(0 < b < 1\) . Isku dhufashada togan ee lagu dhufto tiro togan waxa ay keentaa tiro togan, markaa \(b^2 \in \mathbb{Z}^+\) . Haddaad ku dhufato \(b\) \(0<b<1\) , waxaad helaysaa \(0<b^2<b\) , si \(0<b^2<1\) - sidaas darteed \(b^2 \in S\) . Laakiin tan iyo \(b = min(S)\) , waxaan leenahay \(b \leq b^2\) , kaas oo khilaafsan \(b^2 < b\) . Haddaba, ka dib caddayn is burinaya, sheegashadu waa run.