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A counter-example to the Gelfond–Schneider theorem

  1. Mar 26, 2012 #1
    Gelfond–Schneider theorem can be seen here(http://en.wikipedia.org/wiki/Gelfond's_theorem) wiki.
    Given a simple calculation:
    where a<0;
    and let b be a fraction : u/v
    so there are 3 possible ways of u and v s' arrangement:
    I. u even, v odd
    II. u odd, v odd
    III. u odd, v even.
    in I, a^b>0
    in II, a^b<0
    in III, a^b is a complex number.​
    then the part of my study kicks in:
    So what happens when b is a transcedental number?
    since b is transcedental;
    it no longer can be described as a fraction.
    so the above 3 rules doesn't apply.
    So I pick one of the most common transcedental numbers for b : b = e
    and (-1) for a.
    so the equation is: (-1)^e
    it is known that e=1+1/1!+1/2!=1/3!+....
    so it becomes: (-1)^1*(-1)^1*(-1)^(1/2)*(-1)^(1/6).....
    and for all exponents starting from the 3rd one: the above (III) rule applys.
    so the answer to the equation is : -1*-1*i*i*i.....
    (so this is what I was working up to now)
    so there can be 2 conjectures:
    1. (-1)^e = 1 or -1 or i or -i
    2. the equation is undefined.
    but either way, we will just leave it alone for a moment.
    If the Gelfond–Schneider theorem is true:
    then (-e)^e must be transcendental.

    and (-e)^e is equal to [(-1)^e]*[e^e]
    still if the theorem is true; e^e must be transcendental.
    then assume 1 from the above conjecture is true:
    there are 2 complex answers of (-e)^e
    or assume 2 is true:
    (-e)^e is undefined
    so clearly either way, it is a paradox.

    By Victor Lu, 16
    Burnside High School, Christchurch, NZ
  2. jcsd
  3. Mar 26, 2012 #2


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    Science Advisor

    Hey n_kelthuzad and welcome to the forums.

    It might help you to realize the following identity:

    Consider the same c = a^b number. If a < 0 then we can write this number as:

    c = a^b = |a|^b x e^(ib) where |a| is the absolute value of a and y is known as the argument of the complex number which is given by calculating:

    The way you can interpret is that if you graph two functions corresponding to the real part of c and the imaginary part, then you'll get something that looks like a wave but it will either keep increasing or decreasing as the b value increases. I'm assuming also that b is a real number.

    If you are having trouble, then a normal sine or cosine wave is when a = -1. If -1 < a < 0 then the peak and the trough of the wave will go towards the x-axis as b goes to infinity. If a < -1 then the peak and the trough will go towards infinity.
  4. Mar 26, 2012 #3
    c = a^b = |a|^b x e^(ib) ?

    so if a= -1, b=2 c=1; then |-1|^2*e^(2i) = e^2i?
    i suppose that can only be e^2i∏ which equals to 1.
    and whats this formula called?
  5. Mar 26, 2012 #4


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    Science Advisor

    e^(ix) = cos(x) + i sin(x) where i is the square root of -1. In other words i = SQRT(-1) and i^2 = -1.

    Here is some more information:

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