Albatta, bu yerda ziddiyat orqali isbot: 2^{1/n} 2 dan katta yoki teng bo'lgan n butun son mavjud, deb taxmin qiling, ya'ni 2^{1/n} = p/q. Agar p va q umumiy bo'linmas sonlar bo'lsa, tenglamaning ikkala tomonini ham n-chi darajaga ko'taramiz: 2 = (p/q) ^n = p^n/q^n Ikkala tomonni q^n bilan ko'paytirsak, biz quyidagilarni olamiz: 2q^n = p^n Bu p^n juft ekanligini anglatadi, ya'ni p juft bo'lishi kerak (chunki butun sonlarni ko'paytirishning yagona usuli kamida bitta juft bo'linmaga ega bo'lishi kerak). p = 2k uchun ba'zi bir butun sonlar k. Keyin, biz ushbu ifodani yuqoridagi teng
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