dj3vande said:
In a mathematical context, it's quite common to hear mathematicians
making claims like "the reals are a superset of the rationals".
Except they're not; the rationals are equivalence classes over ordered
pairs of integers, and the reals are sets of rationals[1], so no
rational number can possibly be a real number. This doesn't keep it
from being understood as "the rationals are isomorphic to a subset of
the reals", for both the person making the claim and the person hearing
it.
[1] Sometimes. There are several equivalent constructions, but these
are the ones that I saw in my math courses.
But, as far as I know, it's impossible to construct the reals in a
way that makes them look the same as the rationals, which is my
point here.
Take the set of real numbers, remove the ones in the subset isomorphic to
Q, and replace them with the true Q.
If you're working at the level where it makes sense to want to do that,
then by going ahead and doing it you'll break things.
F'rexample, if you've constructed R with Dedekind cuts, you've defined
real numbers as sets of rationals with some convenient properties, and
real addition ends up being defined as
A + B := {a+b|a in A, b in B}
If you replace the real zero {x in Q: x < 0} with the rational zero
[(0,1)]={(a,b): a,b in Z, a=0, b!=0}, then trying to add something like
0+sqrt(2) gives you... well, something that you can't evaluate with
either rational addition or real addition.
What you end up doing is proving that there's a sensible embedding,
i.e. the real number {x in Q: x < q} has all the properties you would
expect the rational number q to have.
Once you've done that, you then freely go ahead and treat the rationals
as a subset of the reals, with following the embedding bijection
implied when you say things like "Take pi and multiply it by a rational
number" instead of having to specify "Take pi and multiply it by
embed(q) where q is a rational number".
dave