We all, by definition, know that $ i^2 = -1 $, but does that imply (or is it even equivalent ?) that $\sqrt{-1} = \pm i$ ? I often see debates on internet where people are arguing over this last equality, most people arguing that the square root operation is only defined for positive numbers.
I am not familiar with mathematics, so to me, both equations are equivalent, if not, could someone explain in easy terms ?
I think the usual definition of $\sqrt{-1}$ (or more generally $\sqrt{z}$, where $z$ is some number) is a number $y$ whose square is $-1$ (or whose square is $z$). This is inherently ambiguous, since every nonzero complex number $z$ has two distinct numbers $y$ whose square is $z$. When $z$ is a positive real number, usually when people write $\sqrt{z}$, they will mean the unique positive square root.
This naturally leads to the question of how you make sense out of complex numbers in the first place. You would have to resolve difficulties like if you have square roots of both negative and positive numbers, does it really make sense to add them, divide by them etc. In other words, suppose you're comfortable with the field of real numbers, and you want to embed $\mathbb{R}$ in a larger field in which $-1$ does have a square root. If you're familiar with abstract algebra, you can construct the field of complex numbers, and thereby the square roots of $-1$, as follows:
Let $f(X) = X^2 + 1 \in \mathbb{R}[X]$, and let $\mathfrak m$ be the ideal of $\mathbb{R}[X]$ generated by $(X^2 + 1)$. Since $f(X)$ is irreducible in $\mathbb{R}[X]$, $\mathbb{R}[X]/\mathfrak m$ is a field. You can define the set of complex numbers $\mathbb{C}$ to be the field $\mathbb{R}[X]/\mathfrak m$.
In $\mathbb{C}$ (as I have defined it here), the element (coset) $X + \mathfrak m$ will be what you think of as the number $i$. Indeed, $\mathbb{C}$ is a field containing $\mathbb{R}$ (formally, in place of each real number $y$, you write $y + \mathfrak m$), and here if you multiply $X + \mathfrak m$ by itself, you get $(X+\mathfrak m)^2 = X^2 + \mathfrak m = -1 + \mathfrak m$, and $-1 + \mathfrak m$ is now what you think of as the number $-1$.
If you aren't familiar with abstract algebra, you can replace this construction of $\mathbb{C}$ with a rather identical construction: define $\mathbb{C}$ to be the plane $\mathbb{R}^2$ with certain operations of addition and multiplication. This is a well known construction, you can find it in any good calculus textbook.
There is no canonical definition of $i$. Suppose you fix a definition, such that $i^2 = -1$, then we also have $$(-i)^2 = (-1)^2i^2 = 1\cdot -1 = -1$$ Thus, in any reasonable algebraic system of characteristic not 2, (in particular, in any field of characteristic not 2), if there is a single element whose square is $-1$, then there must also be another one, and there is no mathematical way of telling them apart. This is unlike the situation with positive real numbers, since even though there are two square roots of 2, only one of them is positive. Thus, we can define $\sqrt{2}$ to be the positive number whose square is 2.
Your question is a philosophical one. You have a symbol $i$, and you want to associate it to something that doesn't exist ($\sqrt{-1}$ isn't a priori defined). How does one do this? Well, you can't, but the way people often deal with this is to not define $i$. You simply use it as a symbol, which can be added and multiplied by other numbers and itself, satisfying the relation $i^2 = -1$.
Then, one defines the complex numbers to be the set of formal sums $a+bi$, where $a,b\in\mathbb{R}$ and $i^2 = -1$. (Formal in the sense that these sums don't a priori lie in any algebraic system, they have no "meaning" a priori, since after all $i$ isn't defined).
However, using the typical rules for adding/subtracting/multiplying/dividing, its easy to see that you can add and subtract any two such sums to get another such sum: $$(a+bi) + (c+di) = (a+c) + (b+d)i$$ Using the relation that $i^2 = -1$, you can also multiply such sums: $$(a+bi)(c+di) = ac + adi + bci + bdi^2 = (ac-bd) + (ad+bc)i$$ You can also divide any $a+bi$ by any $c+di\ne 0$ as follows: $$\frac{a+bi}{c+di} = \frac{(a+bi)(c-di)}{(c+di)(c-di)} = \frac{(ac+bd) + (bc - ad)i}{c^2+d^2} = \frac{ac+bd}{c^2+d^2} + \frac{bc-ad}{c^2+d^2}i$$ In doing all this, we have essentially shown that this set $\{a+bi : a,b\in\mathbb{R}\}$ forms what is called a "field". However, we still haven't really defined $i$ (ie, we haven't associated it to anything we are already familiar with. We simply called $i$ into existence, and decreed that it satisfies certain properties, including $i^2 = -1$). However, in doing this, we have implicitly given $i$ a meaning - in this set of formal sums $\{a+bi : a,b\in\mathbb{R}\}$, $i$ is simply itself!
Lastly, in more advanced language, you can define:
$$\mathbb{C} := \mathbb{R}[x]/(x^2+1)$$ Where the right hand side is the quotient of the polynomial ring $\mathbb{R}[x]$ by the principal ideal generated by $x^2+1$. In this situation, we may identify $i$ with the image of $x$ in this ring, though it would be just as okay to identify $i$ with (the image of) $-x$.
No, $i$ can NOT be defined as square root of -1. It would be a very poor definition. Square root? Of negative number? What is it? How many different roots are there? Only one? Or may be two? May be more than two? Why not zero then?
(Note: there are Quaternions - quite useful number system which resembles complex numbers, but there are three different "$i$'s": $i, j$ and $k$.)
Correct definition goes other way round. It defines "Complex Numbers" first: not yet numbers, but some objects (pairs of real numbers, or vectors or special matrices), than operations (+, -, *, /. And so on!) on these objects. It becomes clear that these objects "extend" usual real numbers: some subset of the set of Complex Numbers behaves exactly as set of real numbers. And there exists a complex number with a very special property: it's square is equal to -1 (correction: it is equal to a compex number which corresponds to a real number -1). Well, there are two of them, and one of them we call $i$.
