Or, you could just use the dandy world of Complex Numbers :)

By definition i = sqrt(-1)

Oh, and it's also interesting to note that all numbers are complex, it's
just that the imaginary i value has a co-efficient of 0.

By introducing this number any equation where there are negative sqrts can
be solved.  (after all, if we didn't do that, the poor Julia and Mandelbrot
sets wouldn't exist, and we wouldn't want that now would we?)

;)

--
Lance.


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Larry Fontaine wrote in message <37D054E0.A679C841@isd.net>...
>>         May I point out that sqrt(-1) isn't defined? sqrt is only a
function
>> from the set of positive reals into itself... (I know, if you take the
>> complexes, you can find a number (well two actually, that's the problem)
>> whose square is -1, but since you couldn't know wich to choose you can't
>> define *the* square root)
>
>You're right, it can be (0,1) or (0,-1). And those multiplied would be 1,
not
>-1. But that happens with positive numbers too. The square root of 1 is 1
or -1.
>And -1 times 1 is -1, not 1. Which would seem to imply that whichever
number you
>choose as the square root, you must use the same one when you square it
back,
>which makes sense because A^2 means A*A, not A*-A. So I don't think that's
an
>issue.
>

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