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

  • Thread starter The UPC P
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  • #1
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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?
 

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  • #2
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m and n are integers.

log2(i) = m/n
2^(m/n) = i
2^m = i^n
What property of exponents justifies the step above?
The UPC P said:
2^0 = i^4 = 1
So you're saying that ##\log_2(i) = \frac 0 4 = 0##? That's equivalent to saying that ##i = 2^0##.
The UPC P said:
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?
 
  • #3
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##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.
 
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  • #4
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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?
 
  • #5
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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
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.

The UPC P said:
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
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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.
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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