Why is √2 irrational?

Definitions
- A rational number can be expressed as a/b for integers a and b ≠ 0.
- A fraction is in lowest terms when its numerator and denominator share no integer factor greater than one.
- An integer is even when it equals 2k for some integer k.
If an integer's square is even, the integer itself is even.
- We prove this by showing that an odd integer has an odd square.
  - An odd integer has the form 2k + 1 for some integer k.
  - Squaring gives (2k + 1)² = 4k² + 4k + 1.
  - Factoring gives 4k² + 4k + 1 = 2(2k² + 2k) + 1.
  - This expression is odd because it is one more than an even integer.
- Every integer is either even or odd.
- Therefore an integer with an even square cannot be odd.
Suppose for contradiction that √2 is rational [1.1].
- Choose positive integers a and b such that √2 = a/b in lowest terms [1.2].
  - Any common factors can be canceled without changing the value of a fraction.
  - Canceling common factors eventually yields lowest terms because the positive denominator decreases.
Squaring the chosen equality gives a² = 2b².
- The equation makes a² even [1.3].
- The parity lemma [2] makes a even.
- Therefore a = 2k for some integer k.
Substituting a = 2k into the equation [4] gives b² = 2k².
- Substitution first gives 4k² = 2b².
- Dividing both sides by two gives 2k² = b².
- The resulting equation makes b² even.
- The parity lemma [2] makes b even.
Both a and b are divisible by two.
- Their shared factor contradicts the choice of a fraction in lowest terms [3.1].
- Therefore the rationality assumption [3] is false.
Consequently, √2 is irrational.
- The contradiction rules out every integer fraction representing √2 rather than only one attempted fraction.