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If a, b are irrational, then is ##a^b## irrational?

  1. Jan 13, 2017 #1
    1. The problem statement, all variables and given/known data
    True or false and why: If a and b are irrational, then ##a^b## is irrational.

    2. Relevant equations
    None, but the relevant example provided in the text is the proof of irrationality of ##\sqrt{2}##

    3. The attempt at a solution
    Attempt proof by contradiction. Say ##a^b## is rational, then we can express
    ##a^b = \frac{c}{d}##
    for some integer c and d. Without loss of generality, we can choose c and d to express the fractions in lowest terms.
    Then, ##a^b d = c## which implies that ##a^b## must be an integer. This seems to me like a suspicious statement. The problem reduces to whether or not we can have ##a^b = c## where c is an integer. How would I show this possible/ impossible, if indeed this is the correct path to a solution?
     
  2. jcsd
  3. Jan 13, 2017 #2

    Mark44

    Staff: Mentor

    Can you raise e (which is irrational) to some irrational power, but still get a rational number?
     
  4. Jan 13, 2017 #3
    Yes, if we raise it to the natural log of some rational number. ## e^ {\ln{ \frac{c}{d} }} = \frac{c}{d} ##.

    Applying this to the original statement, we would have ## b = \log_{a} \frac{c}{d} ##. Do we have to prove that ## \log_{a} \frac{c}{d} ## is irrational to finish this? My attempt to do so: assume it is rational. Then ## \log_{a} \frac{c}{d} = \frac{e}{f} ## where e, f are integers such that fraction is in lowest terms. Then ## \frac{c}{d} = a^{\frac{e}{f}} ##. I'm tempted to say that raising an irrational number to a rational power (other than 0) will result in an irrational and hence a contradiction, but I'm not sure how to prove. Where do I go from here?
     
  5. Jan 14, 2017 #4

    Mark44

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    More simply ##e^{\ln(2)} = 2##. Surely ##\ln(2)## is irrational...
     
  6. Jan 14, 2017 #5

    Math_QED

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    As your text refers to the irrationality of ##\sqrt{2}##

    Consider: ##\sqrt{2}^{\sqrt{2}}##. If this is rational, we are done. Otherwise, when it is irrational, we have ##(\sqrt{2}^{\sqrt{2}})^{\sqrt{2}} = \sqrt{2}^{2} = 2##. Thus, surely any of these two numbers will be the number you seek. Note that this proof does not show which one (But I believe it's the second one)
     
  7. Jan 14, 2017 #6

    Math_QED

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    Why is ln(2) irrational?
     
  8. Jan 14, 2017 #7

    fresh_42

    Staff: Mentor

    Just as a remark on the subject in case you're interested in: Theorem of Gelfond-Schneider

    If ##a## and ##b## are algebraic numbers (e.g. rational) with ##a \notin \{0,1\}## and ## b## irrational, then any value of ##a^b## is a transcendental number (which are irrational).
     
  9. Jan 14, 2017 #8

    fresh_42

    Staff: Mentor

  10. Jan 14, 2017 #9
    My favorite example of this (maybe a stretch because it uses imaginary numbers) is simply @Math_QED s profile picture.

    ##e^{i\pi}=-1##

    Even though it is imaginary, I still believe that ##i\pi## is irrational, because it is i*3.14159265.... (it still goes on forever).
     
  11. Jan 14, 2017 #10

    mfb

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    2016 Award

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    As the original question is answered, but the flaw in the attempt didn't get pointed out:
    It does not. Ignoring the details of the exponent, imagine ##a^b = \frac 5 3##. Therefore, ##a^b \cdot 3 = 5##. Does that imply ##a^b## has to be an integer?
     
  12. Jan 14, 2017 #11

    Mark44

    Staff: Mentor

    The irrational numbers are a subset of the reals. ##i\pi## is a pure-real complex number.
     
  13. Jan 14, 2017 #12

    Mark44

    Staff: Mentor

  14. Jan 14, 2017 #13

    mfb

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  15. Jan 14, 2017 #14

    Mark44

    Staff: Mentor

    I wasn't giving it as another possible counterexample, but rather in answer to Isaac's question. Proving that ##\log_2(3)## is irrational is pretty easy, but proving that ##\ln(2)## is irrational would be quite a bit harder.
     
  16. Jan 14, 2017 #15

    fresh_42

    Staff: Mentor

    This is also interesting in this context (and the first time I heard of / read about it):
    https://en.wikipedia.org/wiki/Transcendental_number_theory
     
  17. Jan 14, 2017 #16

    LCKurtz

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    Gold Member

    Nice! Best argument of the bunch because it only uses the irrationality of ##\sqrt 2##.
     
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