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Proof: x is irrational => sqrt(x) is irrational

  1. Aug 29, 2005 #1
    Ok so I am to prove: If x is irrational, then [tex]\sqrt{x}[/tex] is irrational. So I started by trying to prove the contrapositive: If [tex]\sqrt{x}[/tex] is rational, then x is rational.

    So then [tex]\sqrt{x} = \frac{m}{n}[/tex] For integers m and n, n[tex]\neq[/tex]0

    Then square both sides. [tex]x = \frac{m^2}{n^2}[/tex]

    This is clearly rational because m^2 and n^2 are integers.

    Now, is this a satisfactory proof? I am sure it is, it just seems as though it was too easy. Did my teacher ask it because it shows how proving the contrapositive can sometimes make life easy? Thanks.
  2. jcsd
  3. Aug 29, 2005 #2


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    Yes, that's correct, and it is an easy proof. It boils down to this: Every square of a rational number is rational, and so these rational squares are the only numbers which have rational square roots.
  4. Aug 29, 2005 #3


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    Yep! Ideally, when you're posed with the problem of proving a statement, looking at its contrapositive should become second nature! With luck, you'll get to the point where you barely even notice the difference between A→B and ~B→~A!
  5. Nov 5, 2010 #4
    Actually that is incorrect.
    The negation of "irrational" is simply "not irrational". For a number to be "not irrational" has 2 cases. The number must be either complex (including i) or rational. Thus your statement of what the contrapositive is is not logically equivalent. This proof must be done by contradiction not by contrapositive.
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