Amanani anengqondo \(\mathbb{Q}\) abonakala ekuqaleni njenge-continuous whole: phakathi kwamaqela amabini, kusoloko kukho elinye. Kodwa le ngcamango iyakhohlisa: kukho iiseti zamanani anengqondo alinganiselweyo, kodwa angaphezulu kwawo okanye i-infimum engekhoyo kwi \(\mathbb{Q}\) . Isizathu sikukubakho kwamanani angenangqondo afana ne \(\sqrt{2}\) , athi, ngandlela ithile, adale imingxunya engabonakaliyo kumgca wamanani anengqondo.
Ukubonisa ukuba \(\mathbb{Q}\) ayiphelelanga, siza kuchaza ii-subsets ezimbini ezingezizo ezingenanto ze- \(\mathbb{Q}\) : enye ebotshelelwe ngasentla kodwa ingenaso i-supremum kwi- \(\mathbb{Q}\) , kunye nenye ebotshelelwe ngezantsi kodwa ingenaso i-infimum kwi- \(\mathbb{Q}\) .
Cinga ngeseti \(X = \{x \in \mathbb{Q} \mid x \geq 0,\; x^2 < 2\}\) . Iseti \(X\) ayilonto ingenanto, kuba \(1 \in X\) . Ukuba \(x \geq 2\) , ngoko \(x^2 \geq 4 > 2\) , ephikisana \(x^2 < 2\) . Ke ngoko, \(x < 2\) ibamba yonke \(x \in X\) . Ngoko ke \(X \subset [0, 2]\) , kunye \(2\) ngomnye wemida ephezulu ye \(X\) . Ke ngoko, \(X\) ibotshelelwe ngasentla.
Masibonise ukuba kukho \(x \in X\) . Sibonisa ukuba kukho \(n \in \mathbb{N}\) kangangokuba \(x + \frac{1}{n} \in X\) . Kufunyaniswe ukuba
\[\left(x + \frac{1}{n}\right)^2 = x^2 + \frac{2x}{n} + \frac{1}{n^2} \leq x^2 + \frac{2x}{n} + \frac{1}{n} = x^2 + \frac{1}{n}(2x + 1).\]
Ngoko ke \(x^2 + \frac{1}{n}(2x + 1) < 2 \iff \frac{1}{n} < \frac{2 - x^2}{2x + 1}\) . Ekubeni \(2 - x^2 > 0\) kunye \(2x + 1 > 0\) , i-axiom ka-Archimedes iqinisekisa ukubakho kwe- \(n \in \mathbb{N}\) . Ngoko ke, \(\left(x + \frac{1}{n}\right)^2 < 2\) , kwaye ngaloo ndlela \(x + \frac{1}{n} \in X\) .
Ngoku qwalasela iseti \(Y = \{y \in \mathbb{Q} \mid y > 0,\; y^2 > 2\}\) . Ngokucacileyo, \(Y \neq \varnothing\) . Ngaphezu koko, \(Y\) ibotshelelwe ngezantsi kwaye ayinabotshelelwe ngasentla, oko kukuthi \(Y \subset (0, +\infty)\) . Let \(y \in Y\) . Sibonisa ukuba kukho \(m \in \mathbb{N}\) kangangokuba \(y - \frac{1}{m} \in Y\) . Kuyabonwa ukuba
\[\left(y - \frac{1}{m}\right)^2 = y^2 - \frac{2y}{m} + \frac{1}{m^2} > y^2 - \frac{2y}{m}.\]
Ngoko ke \(y^2 - \frac{2y}{m} > 2 \iff \frac{1}{m} < \frac{y^2 - 2}{2y}\) . Ekubeni \(y^2 - 2 > 0\) kunye \(2y > 0\) , i-axiom ka-Archimedes iqinisekisa kwakhona ukubakho kwe- \(m \in \mathbb{N}\) . Ngoko ke, \(\left(y - \frac{1}{m}\right)^2 > 2\) , kwaye ngaloo ndlela \(y - \frac{1}{m} \in Y\) .
Masithi \(w = \sup X\) . Akunakuba yiloo nto \(w^2 < 2\) , kuba emva koko \(w \in X\) , kwaye kuya kubakho \(n \in \mathbb{N}\) kunye \(w + \frac{1}{n} \in X\) kunye \(w < w + \frac{1}{n}\) , ephikisana nenyaniso yokuba \(w\) ngumda ophezulu we \(X\) . Akunakuba yinyani ukuba \(w^2 > 2\) , kuba emva koko \(w \in Y\) , kwaye kuya kubakho \(m \in \mathbb{N}\) kunye \(w - \frac{1}{m} \in Y\) kunye \(w - \frac{1}{m} < w\) , ephikisana nenyaniso yokuba \(w\) ngumda omncinci ophezulu we \(X\) .
Masicinge ukuba \(v = \inf Y\) . Akunakuba yinyani ukuba \(v^2 > 2\) , kuba emva koko \(v \in Y\) , kunye \(v\) aziyi kuba ngumda osezantsi we \(Y\) . Ngokufanayo, akunakuba yinyani ukuba \(v^2 < 2\) , kuba emva koko \(v \in X\) , kwaye ngaloo ndlela \(v\) aziyi kuba ngumda ophantsi kakhulu we \(Y\) .
Ezona zinto ziseleyo ke ngoko zezi \(w^2 = 2\) kunye \(v^2 = 2\) . Nangona kunjalo, akukho nani liqiqileyo elilingana nesikwere salo \(2\) . Ke ngoko \(w = \sup X\) ayinakubakho kwi \(\mathbb{Q}\) . Ngesizathu esifanayo, \(v = \inf Y\) ayinakubakho kwi \(\mathbb{Q}\) .
Ngoko ke, \(\mathbb{Q}\) ayilocandelo elicwangciswe ngokupheleleyo. \(\square\)