Yes, idiot. In the context of the reals, that is. *sigh*
Man, these lectures are dealing with real analysis.
> In spite of this, you will get thousands of [reasonable people, > knowing their stuff] (Klyver and "Me" and Burse included) talking > about Q as if it is not part of R.
The reason for this is that if we "build up" the number systems, staring just with IN (i.e. the Peano axioms) there's indeed a set Q which (at that point) is NOT "Part of IR, especially since at that point IR has not yet be defined.