Consider the algebraic equation

xⁿ + a₁x^n-1 + … + a_n-1xⁿ + a_n = 0

in which the coefficients a_i are integers.

Lemma: Such algebraic equations cannot have a root that is rational but not an integer.

Proof:

Suppose there is a root x=p/q in which p and q have no common factor. If the equation is multiplied by q^n-1 the equation can be put into the form

a₁p^n-1 + a₂p^n-2q + … + a_nq^n-1 = −p^n-1/q

The left-hand side of the above equation can only be an integer, but the right-hand side is an integer only if q=1.

Application:

The cube root of 2 cannot be rational since it is a solution to the algebraic equation x³−2=0 and it cannot be an integer.

To show that an expression, such as √3−√2, is not rational it is only a matter of finding an algebraic equation with integer coefficients for which it is a root. For the case of x=√3−√2, it immdiately follows that x² is 3−2√6+2 and hence

x² − 5 = −2√6
and therefore
(x² − 5)² = x⁴ − 10x² + 25 = 24
and hence
x⁴ − 10x² + 1 = 0

This is an algebraic equation with integer coefficient for which √3−√2 is a root. There is no integer equal to √3−√2. Therefore √3−√2 cannot be rational.

The failure of some expression to be a real rational number should not be construed to mean that it is a real irrational number. For example, the solution to

x² + 1 = 0

cannot be a real rational number. It is of course a complex number, of an integral nature.