We aim to prove that $\sqrt{6}$ is irrational. Consider the polynomial $p(x) = x^2 - 6$ Note that $p(\sqrt{6}) = (\sqrt{6})^2 - 6 = 0, $ hence $\sqrt{6}$ is a root of $p(x)$.
Now, consider the Rational Root Theorem, which states that any rational root of the polynomial equation $a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 = 0$, with integer coefficients, must be of the form $ \frac{p}{q}$, where (p) is a factor of the constant term $a_0$ and $q$ is a factor of the leading coefficient $a_n$.
For our polynomial $p(x) = x^2 - 6$, the set of all possible rational roots is 1, -1, 2, -2, 3, -3, 6, -6. However, $\sqrt{6}$ is not included in this set. Since $\sqrt{6}$ is a root of the equation but is not in the set of possible rational roots, it must not be rational.
Therefore, we conclude that $\sqrt{6}$ is irrational.
