|I was taught that sqrt(9) was equal to +/- (read: plus minus) 3 |
Definition of sqrt: sqrt is a function such that
1. If x>=0, sqrt(x) is a non-negative number such that sqrt(x)^2 = x.
2. If x<0, sqrt(x) is undefined.
You could define a multifunction f:C
) such that: for all y in f(x), y^2 = x <edit>and there exist no other complex numbers that meet this condition</edit> (f(x) is a set). But then obviously f != sqrt.
|...which is why I thought x had to be imaginary. But apparently there's not even an imaginary solution?|
|Imaginary numbers are when you have a square root of a negative number, not when the square root of a number equals a negative number.|
(-sqrt(3)i)^2 = -3 (i^2) = -3 (-1) = 3
|And was wondering why? (The later explanation fails to convince me)|
Like that page says, there's a
distinction between a
square root, which is just any number whose square is equal to the given number, and the
square root, which is a mathematical function with a precise definition. Because the
square root of x (or sqrt(x), or x^0.5) is a function, it is one number and one number alone. There's no technical reason why the image of sqrt is [0;+inf.). It's just so by convention. Whatever convention you choose, exactly one of these equations has no solution (provided that both sqrt follow the same convention):