If \sqrt{2} were the image of a rational r, squaring gives r^2 = 2 in \mathbb{Q}. Writing r as the quotient of its numerator and denominator and clearing denominators yields \mathrm{num}(r)^2 = 2\,\mathrm{den}(r)^2, so taking absolute values gives a natural-number solution of p^2 = 2q^2 with q = \mathrm{den}(r) \neq 0.

Use Theorem 3.3 to reduce to the natural-number form. Suppose p^2 = 2q^2 with q > 0. Then 2 \mid p^2, so 2 \mid p since 2 is prime; write p = 2r. Substituting gives q^2 = 2r^2, a new solution whose first component q is strictly smaller than p. By strong induction no solution with q \neq 0 exists.

Use Theorem 3.3 to reduce to the natural-number form. Apply the exponent of 2 in the prime factorization to both sides of p^2 = 2q^2. The left side has even 2-adic valuation 2v_2(p), while the right side has odd valuation 1 + 2v_2(q), a contradiction.

This proof establishes the irrationality of \sqrt{2} Definition 3.1 directly, without the natural-number reduction. One has the identity (18+17\sqrt{2})^3 + (18-17\sqrt{2})^3 = 42^3, due to Jarvis and Meekin. If \sqrt{2} were a rational r, then 18+17r, 18-17r, and 42 would be nonzero rationals (nonzero because (\pm 18/17)^2 \neq 2) solving x^3 + y^3 = z^3, contradicting Fermat's Last Theorem for exponent 3. See this Math StackExchange answer.