Another angle on this is to prove that no integer is the square of a ratio. All squares are the squares of integers. Then, since $12$ does not have an integer square root, its square root cannot be rational, either.
To show that no integer is the square of a ratio, suppose $(\frac{n}{m})^2 = k$ where $m, n$ and $k$ are integers, $n/m$ is in lowest terms, $m\neq 1$, and all are integers. But that situation is impossible.
$$(\frac{n}{m})^2 = k$$
$$\frac{n\cdot n}{m\cdot m} = k$$
If $n/m$ is in lowest terms, as we assumed, that means that $m$ does not divide $n$. This implies that $m$ and $n$ have completely distinct prime factors. Which implies that no multiple of $m$ divides any multiple of $n$, because multiples of a number are just combinations of its prime factors. Thus $\frac{n\cdot n}{m\cdot m}$ cannot be an integer, unless $m = 1$ which we ruled out.
Therefore, $\sqrt 12$ cannot be a ratio. It must either be an integer ($12$ is a square), or else irrational. Our goal is therefore to show that $12$ isn't a square.
Observe that $12$ factors into $2\cdot 2\cdot 3$. A square has prime factors that are all of even duplicity, so that these factors can be divided into two identical groups. There is no way to separate the factors $2\cdot 2\cdot 3$ into two identical groups because $3$ occurs only once. (By contrast, consider $36 = 2\cdot 2\cdot 3\cdot 3$ whose factors are each of duplicity 2, and so can be split into two groups $(2\cdot 3)(2\cdot 3) = 6\cdot 6$.)
Since $\sqrt 12$ isn't an integer, and no integer has a square root which is a ratio, $\sqrt 12$ must be irrational.