This is an example on how to design proofs. Another one is needed for definitions and axioms. This result uses the following: Definition of rational number. Definition of prime and coprime. Definition of square root. The square root of 2 is irrational, 2 ∉ Q {\displaystyle {\sqrt {2}}\notin \mathbb {Q} } This is a proof by contradiction, so we assume that 2 ∈ Q {\displaystyle {\sqrt {2}}\in \mathbb {Q} } and hence 2 = a b {\displaystyle {\sqrt {2}}={\frac {a}{b}}} for some a, b that are coprime. This implies that 2 = a 2 b 2 {\displaystyle 2={\frac {a^{2}}{b^{2}}}} . Rewriting this gives 2 b 2 = a 2 {\displaystyle 2b^{2}=a^{2}\!\,} . Since b 2 ∈ Z {\displaystyle b^{2}\in \mathbb {Z} } , we have that 2 | a 2 {\displaystyle 2|a^{2}} . Since 2 is prime, 2 must be one of the prime factors of a 2 {\displaystyle a^{2}} , which are also the prime factors of a {\displaystyle a} , thus, 2 | a {\displaystyle 2|a} . So we may substitute a with 2 k , k ∈ Z {\displaystyle 2k,k\in \mathbb {Z} } , and we have that 2 b 2 = 4 k 2 {\displaystyle 2b^{2}=4k^{2}\!\,} . Dividing both sides with 2 yields b 2 = 2 k 2 {\displaystyle b^{2}=2k^{2}\!\,} , and using similar arguments as above, we conclude that 2 | b {\displaystyle 2|b} . Here we have a contradiction; we assumed that a and b were coprime, but we have that 2 | a {\displaystyle 2|a} and 2 | b {\displaystyle 2|b} . Hence, the assumption was false, and 2 {\displaystyle {\sqrt {2}}} cannot be written as a rational number. Hence, it is irrational. As a generalization one can show that the square root of every prime number is irrational. Another way to prove the same result is to show that x 2 − 2 {\displaystyle x^{2}-2} is an irreducible polynomial in the field of rationals using Eisenstein's criterion.
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