Re: Rational/Irrational Numbers


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.


i) rationals and irrationals are reals
ii) the reals are a set
iii) the reals are trichotomous
iv) the rationals and irrationals are each dense in the reals
a) between any two irrationals there is a rational
b) between any two rationals there is an irrational

Considering only the positive reals, consider a rational q_i (let q_i
be a positive rational). Is there a distinct irrational p_i for each
q_i? The answer would necessarily be yes for any who think the
cardinality of P is greater than, or equal to, the cardinality of Q.
So, let P_x be the set of irrationals less than q_x. Consider the
union of the P_h's for q_h < q_i. If the difference of P_i and that
union is empty, then there's a rational for which there is not a
distinct irrational. So, is the union of the P_h's for q_h < q_i not
equal to P_i? If so, then the union of the Q_h's for p_h < p_i is not
equal to Q_i, and there's a distinct rational for each irrational. If
not, there's not a distinct irrational for each rational, for every
q_i.

"Rational" and "irrational" could be replaced here with "algebraic" and
"transcendental", or even in a three-state case "rational", "algebraic
irrational", and "transcendental", or any other partition of the reals
into subsets dense in the reals.

Is there not a simple answer to this conundrum?

Infinite sets are equivalent.

Ross

.

Relevant Pages

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