5 ms·

Errr in the set of numbers that complete the rationals wrt to standard metric? I guess you have to show it’s not rational so you probably did a proof of upper/l

• ShroudedNight 9y ago
  From what I remember, it's simpler than that: Assume sqrt(2) is rational and has the reduced form (x/y). Thus we are assuming that x and y are integers and that gcd(x,y) is 1. You square it and get x^2/y^2 which are somehow simultaneously coprime integers and also reducible to 2/1...
• alew1 9y ago
  I think here they just meant, show that no rational number is the square root of 2. Not rational = not expressible as a ratio. You assume that sqrt(2) = p/q for some integers p and q, then derive a contradiction.
• greenleafjacob 9y ago
  Here's the proof from Baby Rudin: > We now show that the equation (1) p^2=2 is not satisfied by any rational p. If there were such a p, we could write p=m/n where m and n are integers that are not both even. Let us assume this is done. Then (1) implies (2) m^2=2n^2. This shows m^2 is even. Hence m is even (if m were odd, m^2 would be odd), and so m^2 is divisible by 4. It follows that the right side of (2) is divisible by 4, so that n^2 is even, which implies that n is even. > The assumption that (1) holds leads to the conclusion that both m and n are even, contrary to our choice of m and n. Hence (1) is impossible for rational p.