# Why is √2 irrational?

- **Definitions**
  - A **rational number** can be expressed as `a/b` for integers `a` and `b ≠ 0`. {#rational}
  - A fraction is in **lowest terms** when its numerator and denominator share no integer factor greater than one. {#lowest-terms}
  - An integer is **even** when it equals `2k` for some integer `k`. {#even}
- If an integer's square is even, the integer itself is even. {#parity-lemma}
  - 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](#rational). {#assumption}
  - Choose positive integers `a` and `b` such that `√2 = a/b` in [lowest terms](#lowest-terms). {#fraction}
    - 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`. {#equation}
  - The equation makes `a²` [even](#even).
  - The [parity lemma](#parity-lemma) makes `a` even.
  - Therefore `a = 2k` for some integer `k`.
- Substituting `a = 2k` into the [equation](#equation) gives `b² = 2k²`. {#both-even}
  - Substitution first gives `4k² = 2b²`.
  - Dividing both sides by two gives `2k² = b²`.
  - The resulting equation makes `b²` even.
  - The [parity lemma](#parity-lemma) makes `b` even.
- Both `a` and `b` are divisible by two. {#contradiction}
  - Their shared factor contradicts the [choice of a fraction in lowest terms](#fraction).
  - Therefore the [rationality assumption](#assumption) is false.
- Consequently, √2 is irrational.
  - The contradiction rules out every integer fraction representing √2 rather than only one attempted fraction.