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View Full Version : 1 = -1 ... where is the error


ex-xian
Nov18-05, 10:39 PM
I've tried and tried and I cannot find the error in the reasoning below. It's probably something simple and I'll feel like an idiot when someone explains it.
i = i
sqrt(-1) = sqrt(-1)
sqrt(1/-1) = sqrt (-1/1)
sqrt(1)/sqrt(-1) = sqrt(-1)/sqrt(1)
sqrt(1) * sqrt(1) = sqrt(-1) * sqrt(-1)
[sqrt(1)]^2 = [sqrt(-1)]^2
1 = -1

Does it have something to do with sqrt(-1/1) = sqrt(-1)/sqrt(1)? Does complex numbers not obey this property?

1800bigk
Nov18-05, 11:03 PM
doesnt [sqrt(a)]^2 = |a| ? maybe thats just in the reals

vsage
Nov19-05, 12:03 AM
\frac{\sqrt{1}}{\sqrt{-1}} pretty much sums up what's wrong with the train of thought. This is a good reason why when you divide a real number by an imaginary number that you must first multiply by the conjugate on the numerator and denominator. I can't nail down a good reason other than that.

I should note that this inconsitency does not exist if you do the division with sqrt(1) and sqrt(-1) in polar form.

matt grime
Nov19-05, 09:06 AM
square rooting is 1 to 2, so you need to pick a choice of square root. You've not done so consitently.