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Proof: One more irrationality proof

  1. Aug 30, 2005 #1
    Proof: One more irrationality proof :)

    Ok, for this one I cannot even start the proof because I do not even know what I am trying to prove :mad:

    The question states:

    Prove of give a counterexample: there do not exist irrational numbers x and y such that x^y is rational.

    Ok, lets knock out the counterexample, because I think that there is definitely not one. And now it is a proof, and the statement is an implication. But which way is the implication is what I cannot figure out.

    Is it: If x and y are irrational, then x^y is irrational? Or, If x and y are rational, then x^y is rational?

  2. jcsd
  3. Aug 30, 2005 #2


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    The statement is not true. Think about logs.
  4. Aug 30, 2005 #3
    Ohh, well that makes it so that I do not even have to prove it. Sweet! I will mess around with that. Thanks.

    However, what if it were ture. Then how would I write the implication so that I could prove it?
  5. Aug 30, 2005 #4


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    That says:

    It is false that there exists irrational numbers x and y such that x^y is rational.

    right? How does negation propagate through quantifiers?

    P.S. it's a simple matter to have figured out how to find the counterexample yourself. You were probably limiting your options!

    Note that the counterexample consists of three parts:

    An irrational number x
    An irrational nubmer y
    A rational number x^y

    You probably tricked yourself into thinking you need to guess irrational x and y, hoping to get a rational x^y.

    It's far easier to guess a rational x^y and an irrational x, and hope for an irrational y. :smile:
    Last edited: Aug 30, 2005
  6. Aug 30, 2005 #5
    Then it would be: It is true that for all irrational numbers x and y that x^y is irrational. (Which is going to be false because there is a counter example.

    So then the implication would be: If x and y are irrational, then x^y is irrational. Correct? Thanks.

    So when I am looking for a counterexample I should think about finding a counterexample of the contrapositive too, or break it into parts as you have.

    So there are a infinite counterexamples. When you say it the other way around it is easy :smile:

    [tex]\pi^{log_{\pi}42} = 42[/tex]

    Whenever I get a chance to make things whatever I want, I am supposed to make them cool numbers right? :smile:

    Last edited: Aug 30, 2005
  7. Aug 31, 2005 #6

    matt grime

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    you proved thatlog of pi in base 42 is irrational did you? seems like oyu're assuming the answer.

    anyway, here is *the* counter example.

    conjsider sqrt(2), raise it to the power sqrt(2) what do you get? rational or irrational, if it's rational you've got a counter example, and even if not what can you do now to get a coutner example?
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