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Proving or Disproving rational raised to rational is rational number

  1. Oct 20, 2005 #1
    Im trying to either prove or disprove that if a and b are rational numbers, then a^b is also rational. I tried doing it with a contradiction, but i cant seem to correctly arrive at a solution. this is how i started the problem

    defn of rational number: a,b = {m/n: m,n are all nonzero integers}
    1. a^b is irrational (hypothesis/assumption)
    2. b^b is irrational (from 1)
    3. (m/n)^(m/n) (from defn. of rational number)
    4. [m^(m/n)]/[(n^(m/n)] (algebra)

    i'm stuck right here. i need to prove that an integer raised to a rational number is either rational or irrational. any inputs will be really helpful. thank you
     
  2. jcsd
  3. Oct 20, 2005 #2
    Use a counter-example. Have you considered [tex]\sqrt{2}[/tex]?
     
  4. Oct 20, 2005 #3
    i dont understand what you mean by "consider square-root of two". can you be more specific?
     
  5. Oct 20, 2005 #4

    Tide

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    And then there is [itex]1^1[/itex].
     
  6. Oct 20, 2005 #5

    Tide

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    He means [itex]2^{1/2}[/itex].
     
  7. Oct 20, 2005 #6
    oh i see. 2^(1/2) is irrational number, which disproves the above statement. however like Tide said, 1^1 is a rational number. but since there's a one statement that made it false, it makes the entire statement false, right?
     
  8. Oct 20, 2005 #7

    Tide

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    All it takes is a single counterexample to disprove a theorem!
     
  9. Oct 20, 2005 #8
    thank you very much for all your help Tide and devious. :)
     
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