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I tried to solve this problem, but check this out: $$\sqrt{x} = -1$$ $$x^\frac{1}{2}=-1$$ $$\frac{1}{2}\ln x = \ln (-1) = \pi i$$ $$\ln x = 2 \pi i$$ $$x = e^{2\pi i}$$

And if we plug this into the first equation: $$\sqrt{e^{2\pi i}} = \sqrt{(-1)^2} = \sqrt{1} = 1$$

This makes a paradox and there is actually no solution to the equation $\sqrt{x} = -1$ !

What is the value of the $x$ which satisfies the first equation? If there isn't, why can't we make an another imaginary number like $i$, where $i = \sqrt{-1}$?

FoiledIt24
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프로형
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  • One thing to consider is that when a square root is taken, there are branches; this is why most formulas involving the square root have the form $y=x^2\to \pm\sqrt y=x$. So either $\sqrt y=x$ or $-\sqrt y=x$. In this case, we would have $-\sqrt 1=-1$. – abiessu Feb 23 '21 at 21:22
  • In Real Analysis, which adheres to the convention that $\sqrt{|x|}$ must be non-negative, there is no solution. In Complex Analysis, to the best of my knowledge, there is no analogous universal convention, so the comment of @abiessu pertains. – user2661923 Feb 23 '21 at 21:31
  • Does this answer your question? Why $\sqrt{-1 \cdot {-1}} \neq \sqrt{-1}^2$? – vitamin d Feb 23 '21 at 21:34
  • But can't we just make another imaginary number for x when sqrt(x) = -1, like when x = k (where k is an imaginary number), sqrt(k) = -1? We have made number like i = sqrt(-1)... – 프로형 Feb 23 '21 at 21:44
  • "But can't we just make another imaginary number for x when sqrt(x) = -1," No, you've got it backwards. In Complex Analysis, if $\sqrt{z} = a,~$ then $z$ must be the specific value $(a^2).$ However, if (for example) you have the equation $(z^2) = b$, and you identify the specific value $z_1$ where $(z_1)^2 = b$, then you have the 2nd root of the equation $(z^2) = b$, namely $z = -(z_1).$ – user2661923 Feb 24 '21 at 05:13

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$\sqrt{x}$ is defined as the non-negative square root aka the principle square root. This is due to the fact that the each number has two square roots. For example, $4$ has $2$ and $-2$. However, for $\sqrt{x}$ to be a function, it must map every $x$-value to only one $y$, thus only one of the $2$ possible values of the square root may be chosen to have $\sqrt{x}$ be a function. So, the positive value is chosen by convention. This thus means that it is impossible for $\sqrt{x}$ to be negative by definition.

This problem is similar to why $|x|$ cannot be negative. The $|x|$ is defined as the distance to $0$, and distances are non-negative, thus $|x|$ cannot be negative by definition.

Some Guy
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  • Thanks ${}{}{}$ – Some Guy Feb 23 '21 at 21:42
  • But can't we just make another imaginary number for x when abs(x) = -1, like when x = k (where k is an imaginary number), abs(k) = -1? We have made number like i = sqrt(-1)... – 프로형 Feb 23 '21 at 21:47
  • That's like saying that the solution to $2+2$ is the number that is equal to $2+2$. It isn't of much use to define a number as the solution to a problem you don't know how to solve – Some Guy Feb 23 '21 at 21:49
  • Also $i$ is much much more useful than the $k$ you propose, you can look up ways complex numbers are used @프로형 – Some Guy Feb 23 '21 at 21:49