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 Math Emeritus Sci Advisor Thanks PF Gold P: 39,552 This is an odd situation. Actually, there is a real number solution to $x^{0.5}= -1$, but you have to extend to the complex number system to find it! In the real number system, the square root function is "single valued" (as are all function in the real numbers) and we define $\sqrt{x}= x^{0.5}$ to be the non-negative number, a, such that $a^2= x$. By that definition, a square root is never negative and we cannot have real x such that $x^{0.5}= -1$. However, in the complex number system, for a number of reasons, including the fact that the complex numbers cannot be made into an ordered field, we must allow "multivalued" functions. In the complex number system, $\sqrt{x}= x^{0.5}$ is any number, a, satisifying $a^2= x$. In the complex number system, since $1^2= 1$ and $(-1)^2= 1$, $\sqrt{-1}$ is both 1 and -1. Thus, $\sqrt{x}= x^{0.5}= -1$, in the complex number system, is satisfied by x= 1, which happens to be a real number. (Although x= 1 is a real number, x= 1 would NOT be a solution to that equation in the real number system. In the real number system $1^{0.5}= 1$, only.)