Four Proofs of the Irrationality of √2

Definitions and Aim

\[ \mathbb{Q} = \left\{\frac{p}{q} : p,q \in \mathbb{Z},\; q\neq0 \right\}, \qquad \mathbb{I} = \mathbb{R}\setminus\mathbb{Q}. \] Irrational: \(x\notin\mathbb{Q}\). Goal: \[ \sqrt{2}\in\mathbb{I} \;\;\Leftrightarrow\;\; \neg\exists(p,q)\in\mathbb{Z}^2_{>0}\; [\gcd(p,q)=1\wedge p^2=2q^2]. \]

Proof 1 — Contradiction in Parity

Assume \(\exists(p,q)\in\mathbb{Z}^2_{>0}\) such that \(\gcd(p,q)=1\) and \(p^2=2q^2\). \[ p^2\text{ even }\Rightarrow p\text{ even }\Rightarrow p=2k\Rightarrow 4k^2=2q^2\Rightarrow q^2=2k^2\Rightarrow q\text{ even.} \] Thus both \(p,q\) even ⇒ contradiction with coprimality. \(\therefore\ \sqrt{2}\notin\mathbb{Q}\).

Proof 2 — Modulo 4 Argument

\(\forall x\in\mathbb{Z},\ x^2\equiv0,1\pmod4.\) If \(p^2=2q^2\): \[ p^2\equiv0,1\pmod4,\qquad 2q^2\equiv0,2\pmod4. \] \(2\not\equiv0,1\pmod4\Rightarrow\) contradiction unless both even ⇒ \(\bot\). Hence \(\sqrt{2}\notin\mathbb{Q}\).

Proof 3 — Infinite Descent

Suppose minimal \(q>0\) satisfies \(p^2=2q^2\). Then \(p\) even ⇒ \(p=2k\) gives \(q^2=2k^2\). Hence smaller solution \((k,q/2)\) exists ⇒ violates minimality. \[ \therefore\text{no integer pair }(p,q)\text{ with }p^2=2q^2. \]

Proof 4 — Continued Fraction

\(r\in\mathbb{Q}\Leftrightarrow\) finite continued fraction. For \(\sqrt{2}\): \[ \sqrt{2}=1+\frac{1}{2+\frac{1}{2+\frac{1}{2+\cdots}}} = [1;\overline{2}], \] an infinite periodic expansion ⇒ non-rational. \[ \boxed{\sqrt{2}\in\mathbb{I}}. \]

Summary

Four views, one conclusion: \[ \forall(p,q)\in\mathbb{Z}_{>0}^2,\; p^2=2q^2 \Rightarrow \bot, \qquad \therefore\ \sqrt{2}\notin\mathbb{Q}. \]