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Proof - irrational numbers

  1. Nov 1, 2008 #1
    1. The problem statement, all variables and given/known data

    Prove that if x^2 is irrational then x must be irrational.

    2. Relevant equations

    3. The attempt at a solution

    Maybe do proof by contradiction. I'm not really sure where to start.
  2. jcsd
  3. Nov 1, 2008 #2


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    Proof by contradiction sounds good. What if x is rational?
  4. Nov 1, 2008 #3


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    "Suppose x is rational. Then x= __________"
  5. Nov 1, 2008 #4
    so i let x= a/b

    then obviously x^2 = a^2/b^2

    im not sure how to continue to reach the contradiction
  6. Nov 1, 2008 #5


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    Assume x^2 is irrational and x is NOT irrational. You've shown that in such a case, x^2 is rational. That's a contradiction. x^2 can't be both rational and irrational. Therefore x must be irrational.
  7. Nov 1, 2008 #6


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    It is not enough just to say "let x= a/b" without saying what a and b are. A number is rational if and only if it can be expressed as a ratio of integers: a/b where a and b are integers (and b is not 0). If x is rational, the x= a/b where a and b are integers. You arrive at the fact that x2= a2/b2, also a ratio of integers. That itself contradicts the hypothesis, that x2 is irrational.
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