Rational exponents (was: Math Discussion)

AI Thread Summary
The discussion centers on the expression (-64)^(3/2) and the absence of a real solution. Participants explain that the square root of a negative number results in an imaginary number, specifically involving the imaginary unit "i," where i^2 = -1. It is clarified that while real numbers can be negative, the operation of taking the square root of a negative number leads to an imaginary result. The conversation emphasizes the distinction between real and imaginary numbers in this context. Ultimately, there is no real answer to the problem due to the nature of square roots involving negative values.
goolalklk
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Homework Statement


(-64)^(3/2)

Homework Equations


None.

The Attempt at a Solution


There is no answer that can be reached and it is supposed not be a real number. I was wondering why that is. How is it that there is no "real" answer to this problem?
 
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goolalklk said:

Homework Statement


(-64)^(3/2)

Homework Equations


None.

The Attempt at a Solution


There is no answer that can be reached and it is supposed not be a real number. I was wondering why that is. How is it that there is no "real" answer to this problem?
##(-64)^{3/2} = [(-64)^{1/2}]^3##
Does that answer your question?
 
Alternatively: (−64)3/2 = √(-64)3

You're right in that there is no real solution. The square root of a negative number is an imaginary number, in which case you must use "i" to express √(-1).

i2 = -1 and √(-1) = i.
 
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When it says "there is no real solution to the problem" it means there is no real number. Do you understand the difference between "real numbers" and "imaginary numbers"?
 
goolalklk said:
There is no answer that can be reached and it is supposed not be a real number. I was wondering why that is. How is it that there is no "real" answer to this problem?

The negative sign makes it so that the stated number isn't real.
 
FeDeX_LaTeX said:
The negative sign makes it so that the stated number isn't real.
Not exactly.

Real numbers can be positive or negative. However, taking the square root of a negative number does not give a real number result. Instead, imaginary numbers were invented to overcome this problem. The imaginary unit i is defined such that i2 = -1.
 
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Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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