daijobu Posted April 13, 2015 Share Posted April 13, 2015 From dd13: i^2 = -1 i = sqrt(-1) i^2 = i*i = sqrt(-1) * sqrt(-1) = sqrt([-1]*[-1]) = sqrt(1) = 1 That is, the product of the square root of negative 1 and square root of negative 1 is the square root of the product (-1)*(-1) which is the square root of 1. So i^2 = 1. I know I'm missing something here. Quote Link to comment Share on other sites More sharing options...
regentrude Posted April 13, 2015 Share Posted April 13, 2015 The problem is that the square root function is only defined for non-negative numbers under the square root. Sqrt(-1) really means i * sqrt(1). The property that the sqrt of a product is the product of the sqrts does not apply for negative arguments. 2 Quote Link to comment Share on other sites More sharing options...
mom2bee Posted April 13, 2015 Share Posted April 13, 2015 You are forgetting that square roots do NOT apply to negative numbers. Its for all integers a, such a >= 0. From dd13: i^2 = -1 i = sqrt(-1) i^2 = i*i = sqrt(-1) * sqrt(-1) = sqrt([-1]*[-1]) = sqrt(1) = 1 That is, the product of the square root of negative 1 and square root of negative 1 is the square root of the product (-1)*(-1) which is the square root of 1. So i^2 = 1. I know I'm missing something here. 1 Quote Link to comment Share on other sites More sharing options...
regentrude Posted April 13, 2015 Share Posted April 13, 2015 You are forgetting that square roots do NOT apply to negative numbers. Its for all integers a, such a >= 0. Not quite correct; it's not all integers, but all non-negative numbers. The square root is defined for positive non-integers. 2 Quote Link to comment Share on other sites More sharing options...
daijobu Posted April 13, 2015 Author Share Posted April 13, 2015 Thank you, ladies! Quote Link to comment Share on other sites More sharing options...
Mike in SA Posted April 13, 2015 Share Posted April 13, 2015 OK, to be even more pedantic. :) (this is a classic problem in introductory abstract algebra) If you consider sqrt(-1) = i as a point in the complex plane (0, 1i), then, yes, you can do a square root of a negative number. It is defined and completely valid. However, square roots are not always distributive in the complex plane. That is, sqrt(-1)*sqrt(-1) <> sqrt(-1 * -1). That property holds for real numbers, but not for complex numbers with a nonreal component. 4 Quote Link to comment Share on other sites More sharing options...
daijobu Posted April 14, 2015 Author Share Posted April 14, 2015 OK, to be even more pedantic. :) (this is a classic problem in introductory abstract algebra) If you consider sqrt(-1) = i as a point in the complex plane (0, 1i), then, yes, you can do a square root of a negative number. It is defined and completely valid. However, square roots are not always distributive in the complex plane. That is, sqrt(-1)*sqrt(-1) <> sqrt(-1 * -1). That property holds for real numbers, but not for complex numbers with a nonreal component. And thank you, sir. This does in fact help to clarify. Quote Link to comment Share on other sites More sharing options...
Mike in SA Posted April 14, 2015 Share Posted April 14, 2015 And thank you, sir. This does in fact help to clarify. Not a problem. For some reason, many texts treat complex numbers as a funny extension of the reals, when they are in fact 2-dimensional, and have an unusual definition of multiplication [ (a,b)*(c,d) = (ac-bd, ad+bc) ]. I really wish the textbooks would be clearer... 1 Quote Link to comment Share on other sites More sharing options...
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