Tirooyinka caqliga leh \(\mathbb{Q}\) waxay marka hore u muuqdaan inay yihiin kuwo isku xiran: inta u dhaxaysa laba jajab kasta, waxaa had iyo jeer jira mid kale. Laakiin aragtidani waa mid khiyaano leh: waxaa jira tirooyin caqli-gal ah oo runtii xaddidan, laakiin kuwa ugu sarreeya ama kuwa aan dhammaystirnayn si fudud ugama jiraan \(\mathbb{Q}\) . Sababtu waxay ku jirtaa jiritaanka tirooyin aan caqli-gal ahayn sida \(\sqrt{2}\) , kuwaas oo, macno ahaan, abuura godad aan la arki karin oo ku jira xariiqda tirada caqliga leh.
Si aan u muujinno in \(\mathbb{Q}\) uusan dhammaystirnayn, waxaan qeexi doonnaa laba qaybood oo aan madhnayn oo ah \(\mathbb{Q}\) : mid ka mid ah oo kor ku xusan laakiin aan lahayn wax sare oo ku jira \(\mathbb{Q}\) , iyo mid kale oo hoos ku xaddidan laakiin aan lahayn wax daciif ah oo ku jira \(\mathbb{Q}\) .
Ka fiirso setka \(X = \{x \in \mathbb{Q} \mid x \geq 0,\; x^2 < 2\}\) . setka \(X\) ma madhan yahay, maadaama \(1 \in X\) . Haddii \(x \geq 2\) , markaas \(x^2 \geq 4 > 2\) , kaas oo ka soo horjeeda \(x^2 < 2\) . Sidaa darteed, \(x < 2\) wuxuu u taagan yahay \(x \in X\) kasta. Sidaas darteed \(X \subset [0, 2]\) , iyo \(2\) waa mid ka mid ah xuduudaha sare ee \(X\) . Sidaa darteed, \(X\) waa xuduud sare.
Aan \(x \in X\) . Waxaan tusinaynaa inuu jiro \(n \in \mathbb{N}\) oo ah in \(x + \frac{1}{n} \in X\) . Waxaa la ogaaday in
\[\left(x + \frac{1}{n}\right)^2 = x^2 + \frac{2x}{n} + \frac{1}{n^2} \leq x^2 + \frac{2x}{n} + \frac{1}{n} = x^2 + \frac{1}{n}(2x + 1).\]
Sidaas darteed \(x^2 + \frac{1}{n}(2x + 1) < 2 \iff \frac{1}{n} < \frac{2 - x^2}{2x + 1}\) . Maadaama \(2 - x^2 > 0\) iyo \(2x + 1 > 0\) , axiom-ka Archimedes wuxuu dammaanad qaadayaa jiritaanka \(n \in \mathbb{N}\) . Sidaa darteed, \(\left(x + \frac{1}{n}\right)^2 < 2\) , sidaas darteedna \(x + \frac{1}{n} \in X\) .
Hadda ka fiirso set-ka \(Y = \{y \in \mathbb{Q} \mid y > 0,\; y^2 > 2\}\) . Sida iska cad, \(Y \neq \varnothing\) . Intaa waxaa dheer, \(Y\) waxaa ku xaddidan hoos oo aan la xaddidin kor, taasi waa \(Y \subset (0, +\infty)\) . Ha u ogolaan \(y \in Y\) . Waxaan tusinaynaa inuu jiro \(m \in \mathbb{N}\) si \(y - \frac{1}{m} \in Y\) . Waxaa la arkay in
\[\left(y - \frac{1}{m}\right)^2 = y^2 - \frac{2y}{m} + \frac{1}{m^2} > y^2 - \frac{2y}{m}.\]
Sidaas darteed \(y^2 - \frac{2y}{m} > 2 \iff \frac{1}{m} < \frac{y^2 - 2}{2y}\) . Tan iyo \(y^2 - 2 > 0\) iyo \(2y > 0\) , axiom-ka Archimedes wuxuu mar kale dammaanad qaadayaa jiritaanka a \(m \in \mathbb{N}\) . Sidaa darteed, \(\left(y - \frac{1}{m}\right)^2 > 2\) , sidaas darteedna \(y - \frac{1}{m} \in Y\) .
Ka soo qaad \(w = \sup X\) . Ma noqon karto taas \(w^2 < 2\) , sababtoo ah markaas \(w \in X\) , waxaana jiri doona \(n \in \mathbb{N}\) oo leh \(w + \frac{1}{n} \in X\) iyo \(w < w + \frac{1}{n}\) , taas oo ka hor imaanaysa xaqiiqda ah in \(w\) uu yahay xuduud sare oo ah \(X\) . Sidoo kale run ma noqon karto taas \(w^2 > 2\) , sababtoo ah markaas \(w \in Y\) , waxaana jiri doona \(m \in \mathbb{N}\) oo leh \(w - \frac{1}{m} \in Y\) iyo \(w - \frac{1}{m} < w\) , taas oo ka hor imaanaysa xaqiiqda ah in \(w\) uu yahay xuduudda ugu yar ee sare ee \(X\) .
Aan hadda u qaadanno in \(v = \inf Y\) . Ma noqon karto run in \(v^2 > 2\) , sababtoo ah markaas \(v \in Y\) , iyo \(v\) ma noqon doonaan xuduud hoose oo ah \(Y\) . Sidoo kale, ma noqon karto run in \(v^2 < 2\) , sababtoo ah markaas \(v \in X\) , sidaas darteed \(v\) ma noqon doonto xuduudda ugu weyn ee hoose ee \(Y\) .
Suurtagalnimada kaliya ee hartay waa \(w^2 = 2\) iyo \(v^2 = 2\) . Si kastaba ha ahaatee, ma jiro tiro macquul ah oo labajibbaarankeedu la mid yahay \(2\) . Sidaa darteed \(w = \sup X\) kuma jiri karo \(\mathbb{Q}\) . Sababtaas awgeed, \(v = \inf Y\) kuma jiri karo \(\mathbb{Q}\) .
Sidaa darteed, \(\mathbb{Q}\) ma aha goob si buuxda loo nidaamiyay. \(\square\)