Re: infinitely many nn's = infinite nn's?

In article <45F466BD.9060809@xxxxxxxxxxxxx>,
Phil <toob-headman@xxxxxxxxxxxxx> wrote:

Virgil wrote:

[snip]

By the way, a proof that the irrationals are not uncountable is
already established by the observation that there is no pair of
irrational numbers which is not separated by a terminating rational
number.


That's a good one. So, the fact that the rationals are dense implies the
reals are countable. I wonder how all the mathematicians missed that
theorem.

Well, it's intuititively plausible. If there's a rational between
every two irrationals, it would seem that the rationals and the
irrationals are equinumerous.


Possibly at first glance, but that assumes that the number of rationals
between two irrationals is the same as the number of irrationals between
two rationals, which is distinctly not the case.

Well, I'm pretty certain that there are infinitely many more irrationals
than rationals, and if that's true, then the ratio of irrationals to
rationals MUST be infinite. That would be consistent with your comment
above, which (correct me if I am wrong) states that the number of
irrationals between two rationals (x distance apart) COULD be infinitely
greater than the reverse, the number of rationals between two
irrationals which are also x distance apart.

While it does not quite state it, it implies it.

However, that can only be true if there can be several sequential
irrationals, with no rational in sight.

Not so, however much it seems so. There are never two or more reals "in
a row" without others between them, regardless of their rationality or
lack therof.

In fact, there must be
infinitely many irrationals "in a row" with no rational in sight.

Again, no. if there is another rational between any two rationals and
another irrational between any two irrationals, as their density
requires, one can never have tow or more of either "in a row", at least
if that requires no others between them.

Now, I
personally believe that to be the case, but it sure contradicts the
"proof" that between ANY two irrationals, there is a rational. I am
placing the word "proof" in quotes because my guess is that this proof
contains a flaw.

No flaw. One can actually prove the density of positive rational quite
easily:

Note that a/b < c/d, where a,b,c and d are positive integers, if and
only if a*d < b*c.

Let a, b, c and d be naturals such that a/b < c/d as rationals, then
with a little algebra, one can prove a/b < (a+c)/(b+d) < c/d.
Similarly for negative rationals.
And between any rational and zero is half that rational.




This proof really is a way of saying that the ratio of
irrationals to rationals is finite, because there is no such thing as
several sequential irrationals.

Then it is not a proof, as that "ratio" cannot be finite.

I agree (I've seen them but can't
remember them, and may not have understood them even then!) that proofs
of the contrary exist, but the existence of one proof does not, in and
of itself, "prove" that another proof with a contradictory result is
false. At least one side of this argument is flawed, and I THINK I know
which side it is, but until someone can show that the proof of 1
rational between ANY 2 irrationals is flawed, I think there is a
legitimate problem to work on.

Let x and y be irrationals with x < y, so that y-x > 0.
By the Archimedean property of the reals there in a natural n such that
n*(y-x) > 2.

Since y - x > 2/n, there must be some integer k with n*y > k > n*x,
or, equivalently, with y > k/n > x.

So that k/n is then a rational between x and y.

Phil


What IS the case, though it seems contradictory at first glance, is
(1) the number of irrationals between any two rationals is the same as
the number between any other two rationals, regardless of how far apart
they are, and
(2) the number of rationals between any two irrationals is the same as
the number between any other two irrationals, regardless of how far
apart they are.

Note, for example, that if a and b are rational, with a < b, then
x -> (b-a)x + a is an order preserving bijection of the reals to the
reals carrying rationals to rationals and irrationals to irrationals,
regardless of the distance between a and b.



Like in a picket fence, there

are as many spaces between pickets as there are pickets. Alright,
one less, unless the fence is a closed curve, or infinitely long.

This would take some careful explaning to sort out, and I'm not
the guy to do it, because I'm unfamiliar with that particular
proof, though I've seen it mentioned before.

--
hz
.

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