Proof Involving Imaginary Number: Where's the wrong one? Here are the propositions:
$$i=\sqrt{-1}$$
$$i^2=-1$$
$$(i)(i)=-1$$
$$\sqrt{-1}\sqrt{-1}=-1$$
$$\sqrt{(-1)(-1)}=-1$$
$$\sqrt{1}=-1$$
There's an error in the propositions above. I think it's in the fifth line where $\sqrt{(-1)(-1)}=-1$.
Are $\sqrt{ab}=\sqrt{a}\sqrt{b}$ and $a=(\sqrt{a})(\sqrt{a})$ different from each other?
 A: 1) The rule "$\sqrt{a}\sqrt{b}= \sqrt{ab}$" is not true for a and b non-real.
2) As Elio Joseph said, defining "i" as "$\sqrt{-1}$ is not rigorous.  Better is to define the complex numbers to b pairs of real numbers with addition defined by (a, b)+ (c, d)= (a+ c, b+ d) and multiplication by (a, b)x(c, d)= (ac- bd, ad+ bc).  We think of the real numbers as "embedded" in the complex numbers by a<==> (a, 0).  The "imaginary numbers" are the set of pairs (0, b).  And, of course, "i" is (0, 1):  (0, 1)x(0, 1)= (0x0- 1x1, 0x1+ 1x0)= (-1, 0) which is what we associate with "-1".  
A: Both 5 and 6 are wrong, assuming the usual choice of positive square root. In order to make 
$$f(x)=\sqrt{x} $$
a bonafide function, mathematicians have defined it to output the positive square root. By this logic, both 5 and 6 are wrong, as they select the negative square root. 
A: The error is that you can't say that $$i=\sqrt{-1}$$
because the function $\sqrt.$ is not defined on $\mathbb C$ properly. 
You can't assigned a unique value for $\sqrt{z}$ if $z\in \mathbb C$ like you do for real positive numbers.
Therefore, writing $\sqrt{-1}$ instead of $i$ may and will lead to wrong results.
