Step 1: Understanding the condition.
We are told \(f(x) \in \mathbb{Q}\) iff \(f(x + 1) \notin \mathbb{Q}.\)
So rational and irrational values alternate for \(x\) and \(x+1.\)
Step 2: Check continuity.
The set of rational and irrational numbers are both dense in \(\mathbb{R}.\)
Hence, such alternating behavior is impossible for a continuous function —
it would require discontinuous jumps between rational and irrational values.
Step 3: Conclusion.
No continuous function satisfies the given property.
Hence, the number of such functions is \(\boxed{0}.\)