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Irrational power of an irrational number

  1. Mar 18, 2012 #1
    What kind of number is sqrt(2)^sqrt(2)?

    I have noted sqrt(2)^sqrt(2) = 2^(sqrt(2)/2) = 2^(1/sqrt(2)), i.e. a rational number to an irrational power.

    Now, 1/sqrt(2) is less than 1, but greater than zero. So, given that 2^x is an increasing function, 2^(1/sqrt(2)) is less than 2^1, but greater than 2^0.

    Also, I tried the following. Let a,b be positive integers such that 2^(1/sqrt(2)) = a/b. What I want to do is reach a contradicition for otherwise I would assume the number in question is rational.
     
  2. jcsd
  3. Mar 18, 2012 #2

    HallsofIvy

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    Your last statement is poor logic. If you can show [itex]2^{1/\sqrt{2}}= a/b[/itex] then, yes, by definition, it is rational. If you cannot show that you don't know what it is. The fact that you cannot show something doesn't mean it isn't true.
     
  4. Mar 18, 2012 #3
    Thanks for replying. I see what you mean by poor logic. I feel silly. After posting the problem, I read a thread in which you answer the question of whether an irrational number to an irrational power can be rational. That makes my day as I thought I had to know what kind of number sqrt(2)^sqrt(2) is.
     
  5. Mar 19, 2012 #4
    Anyways, the question remains: what kind of number is sqrt(2)^sqrt(2)?

    1 < sqrt(2) < 2

    Therefore,

    1/2 < sqrt(2)/2 = 1/sqrt(2) < 1

    Now, 1/sqrt(2) is irrational, i.e. some infinte nonrepeating decimal.

    It has the form

    0 + digit number/10 + digit number/100 + digit number/1000 + ...

    Then, sqrt(2)^ sqrt(2) = 2^(1/ sqrt(2)) = 2^0 * 2^(digit number/10) *
    2^(digit number/100) * 2^(digit number/1000) * ...

    Now, what kind of number is 2^(digit number/(10^n)) for every positive

    integer n?

    First, 10^n > digit number. When digit number = 2, digit number/10^n

    takes the form 1/(5*10^(n-1)).

    Now, let 2^(1/(5*10^(n-1))) = a/b, with a, b positive integers (necessarily)

    greater than 1:

    2 = (a/b)^(5*10^(n-1))
    2 * b^(5*10^(n-1)) = a^(5*10^(n-1))

    But the last statement seems to be a contradiction. Both sides of the

    equation represent nonprime natural numbers greater than one. Every

    nonprime natural number greater than one can be expressed as a unique

    product of primes. The right-hand side of the equation has 2 as factor a

    different number of times from the left-hand side. According to this,

    2^(1/(5*10^(n-1))) is an irrational number. Correct me if I am wrong.


    At last, what I suspect, but have not proved of course, is that 2^(1/sqrt(2))

    is the product of infinitely many irrational numbers. But, does this

    approach lead anywhere?
     
  6. Mar 20, 2012 #5
    A way to prove that sqrt(2) is an irrational number is the following.

    Let a, b be natural numbers greater than 1. Now, I plug

    sqrt(2) = a/b

    As you will see, that statement leads to a contradiction. Square both sides of the equation:

    2 = (a^2)/(b^2)

    2(b^2) = a^2

    The fundamental theorem of arithmetic: every nonprime number greater than one can be expressed as a unique product of primes.

    Now, clearly, the right-hand side of the equation has the factor 2 a different number of times from that of the left-hand side. Namely, on the right-hand side, 2 appears as a factor an even number of times. But on the left-hand side, it appears as factor an odd number of times.

    What went wrong? It happens to be that sqrt(2) is not a rational number (a/b). It is an irrational number.

    I have to say that I took this proof from Varberg and Fleming's Algebra and Trigonometry with Analytic Geometry: A problem-solving approach (the Spanish language version).

    I have used the same logic above to show that 2^(1/(5*10^(n-1))) is an irrational number. Maybe I omitted some details. So I will be more explicit:

    Let a, b, n be positive integers. Then, plug

    2^(1/(5*10^(n-1))) = a/b

    Now, 0 < 1/(5*10^(n-1)) < 1

    Given that 2^x, with x real, is an increasing function,

    2^0 < 2^(1/(5*10^(n-1))) < 2^1

    So, 2^(1/(5*10^(n-1))) lies somewhere between 2 and 1. Being that the case, it is not an integer. So b can't be equal to 1. Also, a is not equal to b; it must be greater than it. That's why a is not equal to 1.

    Going back, I raise the following equation to the power of 5*10^(n-1):

    2^(1/(5*10^(n-1))) = a/b

    2 = (a/b)^(5*10^(n-1))

    2*b^(5*10^(n-1)) = a^(5*10^(n-1))

    On the left-hand side, there may be only one factor 2. And, if tha't's the case, the right-hand side also must have one factor 2, but that's not the case.

    Now, the other possibility is that, on the left-hand side, 2 appears as factor a multiple-of -five-plus-one number of times, whereas on the right-hand side it appears a multiple-of-five number of times. Now, it seems to me that that's a contradiction.

    The conclusion is that 2^(1/(5*10^(n-1))) is an irrational number.

    I want to prove in the same fashion (on a case by case basis) what kinds of numbers the expression 2^(digit number/(10^(n))) yields (rational, irrational?). But, does this program lead anywhere concerning the question of what kind of number is sqrt(2)^sqrt(2)?

    Remember,

    sqrt(2)^sqrt(2) = 2^(1/sqrt(2)) = 2^0 * 2^(digit number/10) * 2^(digit number/100) * 2^(digit number/1000) * ...

    I would also be grateful if someone can assess the reasoning I have followed on this post.
     
  7. Mar 20, 2012 #6

    Dick

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    You aren't going to be able to show a number is irrational by showing it's a limit of rational or irrational numbers. sqrt(2)^sqrt(2) is irrational. In fact, it's transcendental. It's a easy consequence of the Gelfond-Schneider theorem. I'm not sure there is any elementary way to show this.
     
    Last edited: Mar 20, 2012
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