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Question about simple analysis proof

  1. Jan 10, 2012 #1
    I saw a proof on a textbook, but I don't understand why m and n should not both be even?

    Prove p^2 =2 then p is irrational

    (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.

    Then (1) implies

    (2) m^2=2n^2

    This shows that 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  is even, which implies that n is even.

    The assumption that (1) holds thus 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.

    Thanks in advance
    Last edited: Jan 10, 2012
  2. jcsd
  3. Jan 10, 2012 #2
    We can choose m and n not both even.

    Indeed, let's say that [itex]p=\frac{12}{8}[/itex] then both the numerator as the denominator are even. But we can simplify it as [itex]p=\frac{3}{2}[/itex], not not both are even.

    So we can always write p=m/n with not both m and n even. If m and n were both even, then we can just simplify the fraction.
  4. Jan 10, 2012 #3
    thanks for the reply, but I still don't understand why is m and n both are even a conclusive evidence that p is not a rational?
  5. Jan 10, 2012 #4
    We assumed p to be rational. We could write p=m/n with m and n not both even. But then we reasoned further and we found that m and n were both even. This is a contradiction, so the hypothesis that p be rational was false. Thus p is irrational.
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