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

The answer is no, because if so, it will contradict the fact that the set of rational numbers is countable, while the set of complex numbers is uncountable.In summary, the conversation discusses the relationship between m and n as integers in the equation log2(i) = m/n. It is shown that log2(i) is rational because of the existence of integers m=0 and n=4. However, this proof contradicts the fact that log2(i) is not a rational number, as it is defined by the complex logarithm which does not obey the same laws as the real one. A proof by contradiction is attempted, but it also uses the ordinary properties of the logarithm which do not apply in this case
  • #1
The UPC P
9
0
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
The UPC P said:
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
##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
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
The UPC P said:
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
Mark44 said:
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.
 

1. What is the proof that log2(i) is rational?

The proof involves using the definition of a rational number and the properties of logarithms. Let's start by writing log2(i) in exponential form: 2x = i. We know that 20 = 1, so x must be equal to 0. Therefore, log2(i) = 0, which is a rational number.

2. Can you explain why log2(i) is rational but some people think it is wrong?

Some people may think log2(i) is wrong because they are not familiar with the properties of logarithms and how they relate to rational numbers. They may also have misconceptions about the value of i, which is the imaginary unit and can be confusing to understand.

3. How does the definition of a rational number apply to log2(i)?

A rational number is any number that can be expressed as a ratio of two integers. In the case of log2(i), we can express it as 0/1, which is a ratio of two integers and therefore falls under the definition of a rational number.

4. Are there any other ways to prove that log2(i) is rational?

Yes, there are other ways to prove the rationality of log2(i). One way is to use the fact that the logarithm of any number can be written as a sum of logarithms of its prime factors. Since 2 and i are both prime numbers, log2(i) can be written as log2(2) + log2(i), which simplifies to 1 + 0 = 1, a rational number.

5. How does the rationality of log2(i) impact its applications in science and mathematics?

The fact that log2(i) is rational does not have a significant impact on its applications in science and mathematics. This is because the value of log2(i) is rarely used in its exact form, and is often rounded to a certain number of decimal places. However, understanding the rationality of log2(i) can help in understanding the properties and applications of logarithms in general.

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