Cantor's diagonal
argument
shows that for any countable list of numbers, one can
construct a number not on that list. Cantor used this
argument to show, for example, that there are
transcendental numbers, since one may describe a way to
list all the polynomials with integer coefficients and
their roots and hence to list all the algebraic numbers.
One sometimes hears it asserted that Cantor's proof of the
existence of transcendental numbers is a "pure-existence"
proof, showing merely that transcendental numbers exist,
but not that any particular number is transcendental. But
that view is not correct, for the argument is completely
constructive: one may explicitly describe an enumeration of
the algebraic numbers and the diagonal procedure produces a
definite real number that is not algebraic. (I once saw an
article, I think in one of the MAA volumes, with the title
something like "0.24543... is transcendental", where they
implemented Cantor's actual algorithm.)
The relevance to this question is that Cantor
diagonalization also can be used to prove that specific
real numbers are not rational, by producing real numbers
that are explicitly different from every rational number.
Specifically, we may enumerate the rational numbers as
$q_k$ in any of the usual effective manners, and define a
real number $z$ so that the $k$-th digit of $z$ is $4$,
say, if $r_k$ does not have $k$-th digit $4$, and otherwise
the $k$-th digit of $z$ is $5$. It now follows by
construction that $z\neq r_k$ for each $k$, which is what it means for $z$ to be irrational.
Best Answer
One can adapt any of the proofs for the squart root of 2 to this problem. I provide one here:
By definition $\sqrt{24}$ is a root to $x^2 - 24 = 0$. By the Rational root test the only candidates for rational roots of this polynomial are $\pm 1, 2, 3, 4, 6, 8, 12, 24$. It is easy to check none of these square to $24$, so $x^2 - 24$ has no rational roots.