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Proof that log2(i) is rational but I think it is wrong

  1. Jan 29, 2016 #1
    m and n are integers.

    log2(i) = m/n
    2^(m/n) = i
    2^m = i^n
    2^0 = i^4 = 1

    so that means that log2(i) is rational because there are integers n and m so that log2(i) = m/n , they are m=0 and n=4.

    But what I do get about this proof is that it seems to imply that log2(i) = 0/4 = 0 while google says it is 2.26618007 i. So what is going on here? Is my proof wrong?
  2. jcsd
  3. Jan 29, 2016 #2


    Staff: Mentor

    What property of exponents justifies the step above?
    So you're saying that ##\log_2(i) = \frac 0 4 = 0##? That's equivalent to saying that ##i = 2^0##.
  4. Jan 29, 2016 #3


    Staff: Mentor

    ##log_2 i ## is not rational, so you start at a false assumption.
    The complex logarithm function does not obey the same laws as the real one. You may not mix both concepts just as you like.
    ##log_2 i = \frac{ln i}{ln 2}## and ##ln## ##i## is defined via the exponential function. i.e. ##ln ## ##i = x## means ##e^x = i## and therefore
    ##x = \frac{\pi}{2}i## for the main branch.
  5. Jan 30, 2016 #4
    Thanks for the help! Hoewever I am still having problems.
    I want to make a proof by contradiction if log2(i) is irrational so that is why I start with a false assumption. I now made a new proof but I still do not get it:

    log2(i) = m/n
    ln(i)/ln(2) = m/n
    ln(i) = ln(2)*m/n
    e^(ln(2)*m/n) = i
    e^(ln(2)*m/n)^n = i^n
    e^((ln(2)*m/n)*n) = i^n
    e^(ln(2)*m) = i^n
    e^ln(2)^m = i^n
    2^m = i^n

    So this still means that m=0 and n=4 works out even though there should not exist integers n and m for which log2(i) = m/n holds!

    However if I do

    2^m = i^n
    2^m^(1/n) = i^n^(1/n)
    2^(m*(1/n)) = i^(n*(1/n))
    2^(m/n) = i^1
    2^(m/n) = i

    Then it makes sense because 2^(m/n) can never be i since i is imaginary.

    So where is my mistake in my new proof by contradiction?
  6. Jan 30, 2016 #5


    Staff: Mentor

    You're using the ordinary properties of the logarithm (which is defined only for positive real numbers) when they don't apply. You need to be looking at the complex logarithm. See https://en.wikipedia.org/wiki/Complex_logarithm.

  7. Jan 30, 2016 #6


    User Avatar
    Science Advisor
    Gold Member

    And it is more complicated, in that , unless restricted, ln_2(i) is a set, not a single value. so one may ask if there exist a single value of ln_2(i) which is rational.
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