What exactly IS a square root?

How does one actually multiply something by itself half a time ?

Zen Buddhism has a similar question: What is the sound of one hand clapping ? When my father told me, in passing, one day, that $\sqrt x=x^{1/2}$ I had pretty much the same reaction. But then I started thinking to myself: What is the fundamental property of an n-th root, $\sqrt[\Large^n]x$ ? It is basically the number which, when multiplied n times with itself, yields the desired value. At the same time, $x^{1/n}$ multiplied n times with itself, also yields the same value, since $\big(x^a\big)^b=x^{ab}$.

You appear to understand what a square root is just fine; $\sqrt{x}$ is any number with a square equal to $x$.

If you're asking “how do I compute that?”, well there's several possible algorithms for that. As @TrevorWilson points out, Wikipedia has an entire page on the subject.

One method is to pick a number, square it, and see if the answer is too big or too small. Once you have one number that's too big and another that's too small, you know the exact solution is somewhere between the two. You can then recursively subdivide the range into smaller and smaller chunks.

Alternatively, a computer might use Newton's method, which is good for solving all sorts of equations. $\sqrt{c}$ is the solution to $x^2 - c = 0$, which we can solve by

$$x_{n+1} = x_n - \frac{x_n^2 - c}{2x_n}$$

That is, given that $x_n$ is a (possibly very poor) estimate of $\sqrt{c}$, then $x_{n+1}$ is a better estimate. Repeat until you have sufficient accuracy.

"how does one actually multiply something by itself half a time?"

You answer that indirectly, from the definition of the square. The square of the square root of a number is that same number, or $$\left(\sqrt x\right)^2=x.$$

It turns out that this definition is consistent with a fractional exponent. For integer exponents, the following rule holds:

$$\left(x^a\right)^b=x^{ab}.$$

Then it is natural to write

$$\sqrt x=x^{1/2},$$ as this gives

$$\left(x^{1/2}\right)^2=x^{2/2}=x^1=x.$$

Assuming you didn't know how to compute a square root, you could do that by trials and errors. Let us take $\sqrt 2$ as an example.

$1^2=1<2$

$2^2=4>2$

Try the first decimal:

$1.1^2=1.21<2$

$1.2^2=1.44<2$

$1.3^2=1.69<2$

$1.4^2=1.96<2$

$1.5^2=2.25>2$

Try the second decimal:

$1.41^2=1.9881<2$

$1.42^2=2.0164>2$

$\cdots$

Some calculators still use this method. Others use the iterative Newton method. There is also the logarithm/antilogarithm approach, but this is a more advanced concept.

The method of Heron (a special case of Newton's method) was known early in history. It is based on the observation that if you know an approximation of the square root $a\approx\sqrt x$, then $\dfrac x a$ is also an approximation of the square root, and their mean $\dfrac12\left(a+\dfrac xa\right)$ is even better.

$$1\to\frac12\left(1+\frac21\right)=\frac32=1.5$$ $$\frac32\to\frac12\left(\frac32+\frac43\right)=\frac{17}{12}=1.41666\cdots$$ $$\frac{17}{12}\to\frac12\left(\frac{17}{12}+\frac{24}{17}\right)=\frac{577}{408}=1.414215\dots$$

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

1. 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 $n$?

2. 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".

3. 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:

1. 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.

2. 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.

3. 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 $n$.