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Proof that √2 is irrational

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  1. Assume that √2 is a rational number, meaning that there exists an integer a and an integer b such that a / b = √2.
  2. Then √2 can be written as an irreducible fraction (the fraction is reduced as much as possible) a / b such that a and b are coprime integers and (a / b)2 = 2.
  3. It follows that a2 / b2 = 2 and a2 = 2 b2.
  4. Therefore a2 is even because it is equal to 2 b2 which is obviously even.
  5. It follows that a must be even. (Odd numbers have odd squares and even numbers have even squares.)
  6. Because a is even, there exists an integer k that fulfills: a = 2k.
  7. We insert the last equation of (3) in (6): 2b2 = (2k)2 is equivalent to 2b2 = 4k2 is equivalent to b2 = 2k2.
  8. Because 2k2 is even it follows that b2 is also even which means that b is even because only even numbers have even squares.
  9. By (5) and (8) a and b are both even, which contradicts that a / b is irreducible as stated in (2).

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