# Does P not NP imply NP COMPLETE disjoint from RP?

According to Wikipedia https://en.wikipedia.org/wiki/RP_(complexity), $$P \ne NP$$ implies that $$RP$$ is a strict subset of $$NP$$. Does anybody have a reference? Furthermore, am I correct that if this indeed the case, then $$NP-COMPLETE \cap RP = \emptyset$$ since we can use $$NP$$ completeness to solve all other $$NP$$ problems?