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.

This is Theorem 2.1 of Keith Conrad's notes The congruent number problem, proved by Fermat's method of descent. Write the sides of a rational right triangle with area 1 as a/d, b/d, c/d with positive integers a, b, c, d, so that a^2 + b^2 = c^2 and ab = 2d^2. The greatest common divisor g of a and b satisfies g^2 \mid c^2 and g^2 \mid 2d^2, hence g \mid c and g \mid d, and dividing by g gives a solution with a and b coprime. Then exactly one of a and b is even, say a, and a = 2k^2, b = \ell^2 with \ell odd, so c is odd too. The coprime integers (c+b)/2 and (c-b)/2 multiply to k^4, so they are r^4 and s^4, and \ell^2 = r^4 - s^4 = (r^2+s^2)(r^2-s^2) with coprime factors. Hence r^2 + s^2 = t^2 and r^2 - s^2 = u^2 with t and u odd and coprime, and u^2 \equiv 1 \pmod{4} forces r to be odd and s to be even. Now a' = (t+u)/2, b' = (t-u)/2, c' = r, d' = s/2 is another solution with a' and b' coprime, and c' = r \le r^4 < r^4 + s^4 = c. By descent there is no solution.

This proof establishes the irrationality of \sqrt{2} Definition 3.1 directly, without the natural-number reduction. If \sqrt{2} were rational, then \sqrt{2}, \sqrt{2} and 2 would be the sides of a rational right triangle with area 1, contradicting Theorem 3.8. This is the remark after Theorem 2.1 in Conrad's notes; see also this Math StackExchange answer.