Izinombolo ezinengqondo \(\mathbb{Q}\) zibonakala ekuqaleni ziyinto ephelele eqhubekayo: phakathi kwanoma yiziphi izingxenyana ezimbili, kuhlala kukhona enye. Kodwa lo mbono uyakhohlisa: kunezinqoqo zezinombolo ezinengqondo ezilinganiselwe ngempela, kodwa eziphakeme noma ezingenalutho zazo ezingekho ku \(\mathbb{Q}\) . Isizathu sisekubeni khona kwezinombolo ezingenangqondo ezifana ne \(\sqrt{2}\) , ezithi, ngandlela thile, zidale izimbobo ezingabonakali emgqeni wezinombolo ezinengqondo.
Ukuze sibonise ukuthi \(\mathbb{Q}\) ayiphelele, sizocacisa ama-subset amabili angewona angenalutho e \(\mathbb{Q}\) : eyodwa eboshiwe ngenhla kodwa engenayo i-supremum ku \(\mathbb{Q}\) , kanye nenye eboshiwe ngezansi kodwa engenayo i-infimum ku- \(\mathbb{Q}\) .
Cabanga ngesethi \(X = \{x \in \mathbb{Q} \mid x \geq 0,\; x^2 < 2\}\) . Isethi \(X\) ayilona elingenalutho, njengoba \(1 \in X\) . Uma \(x \geq 2\) , khona-ke \(x^2 \geq 4 > 2\) , okuphikisana \(x^2 < 2\) . Ngakho-ke, \(x < 2\) ibamba yonke \(x \in X\) . Ngakho-ke \(X \subset [0, 2]\) , kanye \(2\) kungomunye wemingcele ephezulu ye \(X\) . Ngakho-ke, \(X\) iboshwe ngenhla.
Ake \(x \in X\) . Sibonisa ukuthi kukhona \(n \in \mathbb{N}\) kangangokuthi \(x + \frac{1}{n} \in X\) . Kutholakale ukuthi
\[\left(x + \frac{1}{n}\right)^2 = x^2 + \frac{2x}{n} + \frac{1}{n^2} \leq x^2 + \frac{2x}{n} + \frac{1}{n} = x^2 + \frac{1}{n}(2x + 1).\]
Ngakho-ke \(x^2 + \frac{1}{n}(2x + 1) < 2 \iff \frac{1}{n} < \frac{2 - x^2}{2x + 1}\) . Njengoba \(2 - x^2 > 0\) kanye \(2x + 1 > 0\) , i-axiom ka-Archimedes iqinisekisa ukuba khona kwe \(n \in \mathbb{N}\) . Ngakho-ke, \(\left(x + \frac{1}{n}\right)^2 < 2\) , futhi ngaleyo ndlela \(x + \frac{1}{n} \in X\) .
Manje cabanga ngesethi \(Y = \{y \in \mathbb{Q} \mid y > 0,\; y^2 > 2\}\) . Ngokusobala, \(Y \neq \varnothing\) . Ngaphezu kwalokho, \(Y\) iboshwe ngezansi futhi ayinamingcele ngenhla, okungukuthi \(Y \subset (0, +\infty)\) . Ake \(y \in Y\) . Sibonisa ukuthi kukhona \(m \in \mathbb{N}\) kangangokuthi \(y - \frac{1}{m} \in Y\) . Kuyaphawuleka ukuthi
\[\left(y - \frac{1}{m}\right)^2 = y^2 - \frac{2y}{m} + \frac{1}{m^2} > y^2 - \frac{2y}{m}.\]
Ngakho-ke \(y^2 - \frac{2y}{m} > 2 \iff \frac{1}{m} < \frac{y^2 - 2}{2y}\) . Njengoba \(y^2 - 2 > 0\) kanye \(2y > 0\) , i-axiom ka-Archimedes iphinde iqinisekise ukuba khona kwe- \(m \in \mathbb{N}\) . Ngakho-ke, \(\left(y - \frac{1}{m}\right)^2 > 2\) , futhi ngaleyo ndlela \(y - \frac{1}{m} \in Y\) .
Ake sithi \(w = \sup X\) . Akunakuba yilokho \(w^2 < 2\) , ngoba-ke \(w \in X\) , futhi kuzoba khona \(n \in \mathbb{N}\) no \(w + \frac{1}{n} \in X\) kanye no \(w < w + \frac{1}{n}\) , okuphikisana neqiniso lokuthi \(w\) umkhawulo ophezulu we- \(X\) . Akunakuba yiqiniso futhi ukuthi \(w^2 > 2\) , ngoba-ke \(w \in Y\) , futhi kuzoba khona \(m \in \mathbb{N}\) no \(w - \frac{1}{m} \in Y\) kanye no \(w - \frac{1}{m} < w\) , okuphikisana neqiniso lokuthi \(w\) umkhawulo ophezulu omncane we- \(X\) .
Manje ake sicabange ukuthi \(v = \inf Y\) . Akunakuba yiqiniso ukuthi \(v^2 > 2\) , ngoba-ke \(v \in Y\) , kanye \(v\) bekungeke kube umkhawulo ophansi we \(Y\) . Ngokufanayo, akunakuba yiqiniso ukuthi \(v^2 < 2\) , ngoba-ke \(v \in X\) , futhi ngaleyo ndlela \(v\) bekungeke kube umkhawulo ophansi kakhulu we \(Y\) .
Ngakho-ke okuwukuphela kwamathuba asele yi \(w^2 = 2\) kanye ne \(v^2 = 2\) . Kodwa-ke, ayikho inombolo enengqondo elingana nesikwele sayo ne \(2\) . Ngakho-ke \(w = \sup X\) ayikwazi ukuba khona ku \(\mathbb{Q}\) . Ngesizathu esifanayo, \(v = \inf Y\) ayikwazi ukuba khona ku \(\mathbb{Q}\) .
Ngakho-ke, \(\mathbb{Q}\) akuyona inkambu ehlelwe kahle. \(\square\)