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

Virgil wrote:

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.

Sorry, I meant that there must be a sequence of irrationals -- without worrying about their being "next to" each other -- with no rationals between them. If a rational number exists between ANY two irrationals, meaning we cannot under ANY circumstances have several irrationals "in a row," with no rationals between ANY of the irrationals in question, then there cannot even be more irrationals than irrationals, let alone infinitely more.


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.

Well, I'm not saying that proofs don't exist that rationals exist between ANY two irrationals, I'm just pointing out that this REALLY IS incompatible with the proofs that there are infinitely many more irrationals. If you have 10 of x for every 1 of y, there simply is no way to arrange these x's and y's such that for ANY two x's, there is a y. Now, if we want to say that we are limited by potential infinity when mentally examining a structure with actually infinitely many elements, I would certainly agree that we cannot mentally examine the elements individually, but in the limit, where actual infinity either CAN be used, or at least predicted, there cannot be a y between ANY two x's.


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.

I don't have a problem with that, any more than I have a problem with the Achilles saying that we can never leave the room, but I want to point out that this proof uses the assumptions of potential infinity, not actual infinity. Not that there's anything wrong with that, but it should be remembered, especially when combining these results with the results from other proofs that use actual, not potential, infinity.


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.

Oh, it could easily be a proof, although it is almost certain that either (1) one of the proofs is wrong, or (2) they use different premises, such as, one uses potential infinity, while the other uses actual infinity. This would allow them to both be valid and correct, but applicable to different realms, a bit like Euclidean and Riemannian geometry.


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.

Okay, without reading the rest of your proof yet, and not saying this is bad, but you do know that the Archimedean postulate is a way of (1) eliminating infinitesimals, by (2) again placing us under the limitations of potential infinity.

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.

Without question, under potential infinity, two irrationals are a FINITE distance apart -- since infinitesimals don't exist under either potential infinity or the Archimedean postulate (the Archimedean postulate basically DEFINES a "finite region," in which any two reals have a ratio that is "finite but unbounded," i.e., limited by potential infinity) -- and we can ALWAYS find a rational between two real numbers that have a finite difference between them. Just don't be surprised if the proofs that the irrationals are infinitely more numerous than the rationals use actual, rather than potential, infinity. Cantor's diagonal proof, for example, uses aleph-0 rationals, infinitely many rationals, to produce infinitely many more irrationals.

By the way, did you ever ask yourself how INFINITELY many reals can exist on the line segment [0,1] (probably proved using actual infinity) with a FINITE distance between each and every one of them (probably proved using potential infinity)? And here I thought that the sum of infinitely many finite distances always equaled an infinite distance; shows you how dumb I am! Or for that matter, why modern mathematicians believe that they can take the results of proofs that ASSUME potential infinity, and combine them with results from proofs that ASSUME actual infinity, and NOT create a contradictory mess? Just thought I would ask ...

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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