What exactly IS a square root? It's come to my attention that I don't actually understand what a square root really is (the operation). The only way I know of to take square roots (or nth root, for that matter) it to know the answer! Obviously square root can be rewritten as $x^{1/2}$ , but how does one actually multiply something by itself half a time? 
How do calculators perform the operation?

Let $n$ be an integer that is not the square of any integer. Two versions of the question have already been answered, namely


*

*What is the definition of $\sqrt{n}$, i.e. what does it mean for a given number to be a square root (or the positive square root) of $n$?


*How can we approximate $\sqrt{n}$, i.e. how can we find rational numbers (or decimals) whose square is close to $\sqrt{n}$?

I will try to answer yet another version of the question that I think is lurking in the "is".



*What kind of object is it whose square is $n$?


By our assumption such an object cannot be an integer.  Also, it cannot be a rational number.  So we need some new type of number.  Here are a few approaches:

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*Dedekind cuts.  If we define real numbers as Dedekind cuts, then we can simply define $\sqrt{n}$ as the set $\{a \in \mathbb{Q} : a^2 < n\}$.  It takes some argument to see why it makes sense to consider such a set as a kind of number, however.


*Cauchy sequences. An algorithm for computing square roots gives a sequence of rational numbers (which we can think of as better and better approximations for $\sqrt{n}$, whatever that is.)  Such a sequence is a Cauchy sequence, meaning roughly that it looks like it should be converging to something, even if we don't know what that thing is yet.  The sequences given by different algorithms might not be equal, but they are equivalent in the sense that they look like they must be converging to the same thing (again this can be made precise) and so we can uniquely define $\sqrt{n}$ as this equivalence class of Cauchy sequences.


*Algebraic field extensions.  Dedekind cuts and Cauchy sequences are overkill for defining $\sqrt{n}$ in the sense that they also add uncountably many other numbers to $\mathbb{Q}$, which we might not be interested in.  A more economical way to add a square root of $n$ to $\mathbb{Q}$ is to add an indeterminate $x$ to get the ring of polynomials in $x$ with rational coefficients, and then take a quotient of this ring by setting the polynomial $x^2-n$ equal to zero, giving us a field containing $\mathbb{Q}$ and also containing an element $x$ that acts like a square root of $n$.
