Re: Rational numbers, irrational numbers: each dense in real numbers

Ross wrote:

In ZFC, with standard definitions of the real,
rational, and
irrational numbers, let p_i be an irrational number
between zero and
one for i from a suitably large well-ordered index
set X. With the
well-ordering of the index set, let the i'th element
p_{i+1} be an
irrational number between zero and p_i, where i+1 is
the least element
of the well-ordering X_i setminus i, that is defined
to equal X_{i
+1}. There are uncountably many irrational numbers
less than each p_i
and greater than zero, else the irrational numbers
are countable (or
the real numbers are not standard). Define P to be
comprised of the
p_i's. There exists a rational number q_i between
p_i and p_{i+1},
else the rational numbers are not dense in the reals
thus that between
any two irrational numbers there is a rational
number. For each of
the irrational p_i's, there thus exists at least one
unique rational
q_i between p_i and p_{i+1}, and infinitely many.
Let the ordered
pair (p_i, q_i) be an element of a function, as a
set, from P to Q.
If there is an uncountable set P of irrational
numbers in (0,1), then
there is a 1-to-1 function defined by the set {(p_i,
q_i), i E X} from
uncountable P to a subset of Q the rational numbers,
and there is thus
an injection from an uncountable set of irrational
numbers to a subset
of the rational numbers, a subset of a countable set
is countable.

Contradiction ensues, from that an uncountable set
injects into a
countable set. Which presupposition is false?
Perhaps it is so that
the irrationals can not be well-ordered in their
normal ordering, but,
via separation, for any given subset of the
irrational numbers in (0,
p_i) there exists (0, p_{i+1}) for any p_i such that
0 < p_{i+1} <
p_i, or the irrationals are not dense in the reals.
Perhaps it is so
that there are no uncountable subsets of the
irrationals in (0,1), but
then the irrationals wouldn't be uncountable. The
rationals are dense
in the reals so between any distinct p_i and p_{i+1}
there exists a
q_i, or p_i = p_{i+1} and p_i =/= p_{i+1}. In ZFC
there exists a
suitably large well-ordered index set.

Quantify over sets, ZF(C) and/or the standard
definitions of the real,
rational, and irrational numbers are thus
inconsistent.

Ross

--
Finlayson Consulting


so ok.

between every pair of irrationals lies a rational.

and between every pair of rationals lies an irrational.

seems like the set 2n and 2n + 1 ( even and odd integers)

so implies equal density and both countable....

or both uncountable...

however

assume they are both uncountable.

then if the rationals are uncountable ; so are the integers...

that cant be right of course.

so assume irrationals are countable too.

but then there has to be a mapping from all rationals to all irrationals...

this means we need a function f(a,b) -> gives all irrationals for integer a and b.

f(a,b) does not exist.

even f(a,b,c) wont do.

for instance f(a,b,c) = a sqrt(2) + b sqrt(3) + c sqrt(5)

does not give all irrationals.

and if f was different then we would have left out the numbers of type a sqrt(2) + b sqrt(3) + c sqrt(5).

regards
tommy1729
.

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