prove that : $ i = \sqrt {-1}\ $ i have a pretty nasty question.
i was glancing through a few olympiad papers and stumbled upon this question:
prove that
$
i = \sqrt {-1}\
$.
i tried the conventional methods namely euler's formula  but could not figure what to do next.
how do you actually proceed further? i have checked a few books and a proof is not available.are there any proof available for this 
 A: The square root is typically only defined for real numbers. This has a number of reasons, for instance, a number of simplification rules for the square root only work if the argument is nonnegative, such as
$$\sqrt{ab} = \sqrt{a}\sqrt{b}$$
If we assume this also holds for negative numbers, we get:
$$1 = \sqrt{1} = \sqrt{(-1)\cdot(-1)} = \sqrt{-1}\cdot \sqrt{-1} = \left(\sqrt{-1}\right)^2 = -1$$
This is obviously false. Another thing is that $\sqrt{\ \ \ }$ is defined as always returning the non-negative root when there are two, i.e. $\sqrt{9}=3$ and not $-3$, even though $-3$ is also a solution of $x^2=9$. Complex numbers, however, are not ordered; therefore the definition gets a bit tricky. Is $\sqrt{-2i}=1-i$ or $i-1$?
Furthermore, as a consequence of this, there is no way of distinguishing $i$ and $-i$ except on a formal level. You say that $i$ is, formally, some number that, when squared, yields $-1$; but $-i$ also fulfils that requirement, so if you mail-order the $i$ number, i.e. a number that when squared gives you $-1$, I could take it out, negate it and put it back in and you would never be able to tell the difference.
So, to summarise: defining square roots of negative numbers is probably not a good idea.
A: Something similar was discussed here:
What's the thing with i
Afaik. $i$ is usually +defined as $i^2 = -1$, which is different from $i = \sqrt{-1}$.
A: You may want to show that $i$ solves the equation $z^2 = -1$. 
The square root of x is typically defined as the (positive) solution of the equation $z^2 = x$. If we are talking about roots in $\mathbb C$, one usually avoids to write $i =  \sqrt{-1}$, because there are multiple solutions to this (consider $(-i)^2 = (-1)^2 i^2 = -1$).
