Re: Rational/Irrational Numbers


In considering the positive rationals less than some irrational p_i,
for each p_j < p_i the set of rationals Q_j: q E Q+, q < j is always a
proper subset of the set of rationals Q_i, those rationals less than
p_i. So, Q_i setminus Q_j is always non-empty, for each p_j < p_i in
P+, the set of positive irrationals. The set difference is always
non-empty, there is for each p_j < p_i at least one distinct rational
q_i such that p_j < q_i < p_i. Simlarly, yet in a contradicting
manner, for each q_i there is an irrational p_h such that p_j < q_i <
p_h < p_i, so q_i is not a distinct rational over each p_h < p_i, and
there exists q_h, and around it goes.

I want to use that kind of notion to see why Q_i setminus the union of
Q_j for p_j < p_i is or is not empty.

If you would claim it empty, exchange rationals with irrationals. Then
there wouldn't be a distinct irrational for each rational, by that
argument. Where there is exchange them back and it's not.

For the difference of Q_i and union of the Q_j's for p_j < p_i to be
empty, as the union is a subset of Q_i they would be equal. Where the
union contains all the elements, how did that last one get in there,
and where it is, how is that p_j less than p_i?

The union can't contain an element not in at least one of those sets
which comprise it by their elements, unless you're looking at something
like a Russell set.



So, for no positive irrational p_h < p_i is it so that the set of
positive rationals less than p_h, Q_h, is equal to the set of rationals
less than p_i, Q_i. So, Q_h < Q_i, the (positive) rationals less than
p_h is a proper subset of Q_i, so Q_i setminus Q_h is non-empty, and
for each irrational p_i there's an element of the difference of Q_i and
the union of the Q_h's for p_h < p_i such that q_i is related to p_i
and not to any other p_j =/= p_i.

That Q_i, setminus the union of the Q_h's, is non-empty, is true. Do
you agree? If not there's a p_h >= p_i, yet each p_h < p_i, or a q_h
in a Q_h greater than p_h, yet
each q_h is less than p_h.

So, here's the question: is the intersection of (Q_i setminus Q_h) for
each p_h < p_i empty? Again, is the difference Q_i setminus the (union
of the Q_h)'s for each p_h < p_i empty?

There's a rational between any two irrationals.

So, the union of the Q_h's for p_h < p_i is equal to Q_i, else Q_i \
Union Q_h is non-empty and there's a distinct rational element of that
difference for each irrational p_i, and thus an injection from the
irrationals into the rationals and thus a surjection from the rationals
onto the irrationals. (I think infinite sets are equivalent so that's
not a contradiction, for Union Q_h < Q_i.)

Where the union of the Q_h's contains an (at least one) element not in
any of the Q_h's, for there to not be a distinct rational for each
irrational, that's a contradiction., the set is defined by its
elements, and p_h < p_i, not p_h = p_i.

Yet, for any q_i, there exists a p_j such that q_i < p_j < p_i, then
there wouldn't be a rational between any two irrationals,
contradiction.

The difference Q_i \ Q_h is non-empty for each p_h < p_i, and that
difference Delta(Q_i,Q_h) is a superset of Delta(Q_i,Q_j) for p_h < p_j
< p_i. So the intersection is only empty when p_h = p_i, but p_h <
p_i, the intersection is non-empty else contradiction.

Transpose p and q, (positive) P and Q, irrationals and rationals, it's
the same system.

It seems there are contradictory true statements, so one of them is
false.

Again, it seems like the Russell paradox. Where it is true that the
rationals and irrationals are each dense in the reals, it can not be
that of the set of irrationals [0,p_i) that one of them equals p_i,
because they are defined to be not. It's reminiscent of the ordinals
where the set of ordinals up to but not including n has order type n,
here the union of the Q_h's for p_h < p_i equals Q_i, else infinite
sets are equivalent, yet then p_h = p_i, a contradiction.


So, where's the contradiction?

If the union of the Q_h's for p_h < pi is equal to Q_i, else there
exists a distinct rational for each irrational, then the union of the
P_h's for q_h < q_i is equal to P_i, and there doesn't exist a distinct
irrational for each rational. Then exchange rationals and irrationals
and it's neither.



Ross

.

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