imaginary numbers: what am I missing

[daijobu]

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.  

[regentrude]

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.

[mom2bee]

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.  

 

[regentrude]

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.

 

[daijobu]

Thank you, ladies!

[Mike in SA]

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.

[daijobu]

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.

[Mike in SA]

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...