Template:Proof

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{{PAGENAME}}|title=Proof that <math>\sqrt{2}</math> is irrational|
contents=Assume <math>\sqrt{2}</math> is ''p''/''q'', where ''p'', ''q'' ∈ ℤ<sub>>0</sub> and gcd(''p'', ''q'') = 1. Then ''p''<sup>2</sup>/''q''<sup>2</sup> = 2, and ''p''<sup>2</sup> = 2''q''<sup>2</sup>. Since ''p''<sup>2</sup> is even, and gcd(''p'', ''q'') = 1, gcd(''p''<sup>2</sup>, ''q''<sup>2</sup>) = 1, ''q''<sup>2</sup> must be odd. Hence ''p''<sup>2</sup> ≡ 2 (mod 4), which is impossible because the square of an integer must always be congruent to 0 or 1 (mod 4). 
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Proof that [math]\displaystyle{ \sqrt{2} }[/math] is irrational

Assume [math]\displaystyle{ \sqrt{2} }[/math] is p/q, where p, q ∈ ℤ_>0 and gcd(p, q) = 1. Then p²/q² = 2, and p² = 2q². Since p² is even, and gcd(p, q) = 1, gcd(p², q²) = 1, q² must be odd. Hence p² ≡ 2 (mod 4), which is impossible because the square of an integer must always be congruent to 0 or 1 (mod 4). [math]\displaystyle{ \square }[/math]

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