Are there any irrational numbers that have a difference of a rational number?

If $q$ is a rational number, and a is an irrational number, then $q+a$ is irrational, $(q+a)-a=q$.

This is a surprisingly tricky question, if you discount the answers already given. The set of irrational numbers can be further split into

• algebraic numbers (zeros of polynomials with integer coefficients, such as $\sqrt{7}$, $\root3\of2$ that are zeros of $x^2-7$ and $x^3-2$ respectively), and
• transcendental numbers - the rest of them. Famous known transcendentals include $e,\pi$, $\log n$ for an integer $n>1$. In general it is very difficult to prove that a number given by some formula is transcendental. The odds are in favor of a number being transcendental unless it is "obviously" algebraic (such as $\sin(\pi/4)=\sqrt2/2$ that happens to be algebraic). That is, unless the formula only involves rationals and root extractions.

What can be said in general is the following

• The difference between two algebraic numbers is irrational, unless it is of the type described in other answers. The methods needed to identify, when this may be the case involve the theory of field extensions. Algebraic number theory in particular. See questions carrying that tag.
• The difference between an algebraic and a transcendental is ALWAYS transcendental, hence also irrational. So no cool examples like $\pi^{7/5}-\sqrt{131}$ can possibly work. Such a difference is automatically irrational.
• The difference between two transcendentals? Who knows? I'm not aware of any non-trivial examples.

$\hspace{20mm} \sqrt{2}-\sqrt{2}=0$