Why $\sqrt{-1 \times -1} \neq \sqrt{-1}^2$? We know $$i^2=-1 $$then why does this happen?
$$
i^2 = \sqrt{-1}\times\sqrt{-1}
$$
$$
=\sqrt{-1\times-1}
$$
$$
=\sqrt{1}
$$
$$
= 1
$$
EDIT: I see this has been dealt with before but at least with this answer I'm not making the fundamental mistake of assuming an incorrect definition of $i^2$.
 A: Following Willie Wong's suggestion, I will put the comments I made before into an answer. The problem with this argument is that the formula $\sqrt{ab} = \sqrt{a} \sqrt{b}$ is valid for $a,b \geq 0$. In particular, it is not valid in the case when $a=b=-1$ so that the step $\sqrt{-1}\sqrt{-1} = \sqrt{(-1)(-1)}$ is incorrect.
A: Don't take me the wrong way, all of you are right from a certain point of view.
Let me show you the following. Let $a$ and $b$ be complex numbers. If $\sqrt a$ is a square root of $a$ and $\sqrt b$ is a square root of $b$, then $(\sqrt a\sqrt b)^2 = (\sqrt a)^2 (\sqrt b)^2 = ab$ by assumption, so indeed $\sqrt a\sqrt b$ is a square root of $ab$. Notice that it is some square root; there is no canonical choice over $\mathbb{C}$!
From this point of view, your arguments work until $i^2 = \sqrt1$. All you've shown is that $-1 = i^2$ is a square root of 1 and no one will disagree with that fact.
So what I'm trying to say is, if you want to write something like $\sqrt a\sqrt b = \sqrt{ab}$, which isn't the worst idea, you can't write something like $\sqrt 1 = 1$.
A: From $i^2=-1$ you cannot conclude that $i=\sqrt{-1}$, just like from $(-2)^2 = 4$ you cannot conclude that $-2=\sqrt 4$. The symbol $\sqrt a$ is by definition the positive square root of $a$ and is only defined for $a\ge0$.
It is said that even Euler got confused with $\sqrt{ab} = \sqrt{a}\,\sqrt{b}$. Or did he? See Euler's "mistake''? The radical product rule in historical perspective (Amer. Math. Monthly 114 (2007), no. 4, 273–285).
A: @Greg: $i^2=-1$ is a matter of definition.  It was introduced to handle a negative sign under the radical when solving polynomial equations.  The best example is $x^2 +1 =0$, which yields $x = \pm\sqrt{-1}$.  Using your manipulation above would yield $x^2 = 1$ which clearly does not solve the equation.
A: You may only apply $\sqrt{a} \sqrt{b} = \sqrt {ab}$ when $a,b \ge0$.
A: $\hspace{0.2in}$
The problem is this: the (polar) representation of a complex number depends on your choice of branch. Once you choose a branch of the square root, you cannot simultaneously represent $1$, and $-1$, because they are $\pi$ apart, so that $-1$ and $1$ will necessarily be in different branches of $z^{1/2}$. The choice of square root you make (if you want it to be well-defined) depends on the choice of branch you are working with. You are trying to combine numbers that live in different branches; it is as if you add $1+1$ and get $1-i\pi$, since 1 can also be represented as $i\pi$ ($~1$ can actually be represented as $ik\pi$, but when you operate, you are supposed to stay within a branch, and, in your case, you are not$~$).
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In complex numbers, from the perspective of polar representations, when you multiply $z_1 \dot z_2$, you multiply the respective lengths, and add the respective angles (but you have to make up for the fact that the sum of the angles may be larger than $2\pi$ (or whatever argument-system you are working with. So in this sense, when you multiply $i$ by itself, you multiply the length of $i$ by itself, and add the argument to itself; i.e., you double the argument. In complex variables, you have many possible polar representations for a given number; specifically, given $z=re^{it}$, then $z=re^{i(t+2\pi)}$ is also a valid representation. 
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You must then choose a specific representation for your $z$, specifically; you will need to specify the range of the argument you will be working with. So, say you work with the "standard" range of $[0,2\pi)$. Then the expression for $i$ (me) is as $i=1e^{i\pi/2}$, so that $i^2$= $(1)(1).e^{i(\pi/2+\pi/2)}=e^{i\pi}=-1$. But the way backwards from multiplying to taking roots is more complicated if your base is non-negative, and/or your exponent has non-zero imaginary part. When this last is the case, you define:
$z^{{1}/{2}} \:=e^{{\Large{log(z)}}},~$
where we define:
$\log(z):=\ln |z|+i arg(z)$
(This choice of definition has to see in part with wanting to have the complex log agree with the standard real $\log$--though this agreement is possible only for one choice of "branch", as we will see.)
 But because of the infinitely-many possible choicesfor the argument of a number, the $\log(z)$ itself--defined locally as the inverse of $e^z$  is somewhat-ambiguously-defined, since $e^z$ does not have a global inverse (since it is not $1-1$, for one thing, but $e^z$ does have local inverses, e.g., by using the inverse function theorem). So when we mention $\log$, we are referring just to one of (infinitely-) many possible local inverses of $e^z$ .Each possible local inverse to $e^z$ is called a "branch" of the $\log$. So once we choose a branch for the $\log$, which is a choice of an open set (technically, it is half-open) of width $2\pi$ from which we will choose the argument we will use. So, say we choose the standard branch $(0,2\pi)$, which we call Log(z). We then define :
$z^{1/2}:=e^{\large{Log(z)/2}}$
But, in this branch , $(-1)^{1/2}$ is not even defined, because the argument for $(-1)$ is $0$ , which is outside of the allowable values $(0,2\pi)$. So, in this sense, the expression $(-1)^{1/2}$ is not well-defined, i.e, does not really make sense. ($~$Note that this particular branch of $\log$ reduces to the standard one when you select an argument of $t=0~$).
A: Any non zero number has two distinct square roots. There's an algebraic statement which is always true : "a square root of $a$ times a square root of of $b$ equals a square root of $ab$", but this does not tell you which square root of $ab$ you get.
Now if $a$ and $b$ are positive, then the positive square root of $a$ (denoted $\sqrt{a}$) times the positive square root of $b$ (denoted $\sqrt{b}$) is a positive number. Thus, it's the positive square root of $ab$  (denoted $\sqrt{ab}$). Which yields
$$\forall a,b \ge 0, \ \sqrt{a} \sqrt{b} = \sqrt{ab}$$
In your calculation, because $i$ is a square root of $-1$, then $i^2$ is indeed a square root of $1$, but not the positive one.
A: There are many ways to show that your second equality is incorrect. Just for the sake of argument and a refreshing change of pace, suppose your point is true. That is, suppose $i^{2} = 1$. Then
\begin{align}
(x + i)(x-i) = x^2 - i^{2} = x^{2} - 1.
\end{align}
Use the Descartes Rule of signs to derive a contradiction. Hint: $i$ is not real. Can you finish the line of reasoning and derive an absurdity?
A: In general,
$\forall a,b \in \mathbb{R} ,  ab\geq  0, (\sqrt{ab} =\sqrt{ |{a}| }\sqrt{|b|}) $
