Re: abundance of irrationals

mueckenh_at_rz.fh-augsburg.de
Date: 01/30/05 

Date: 30 Jan 2005 08:51:16 -0800

briggs@encompasserve.org wrote:

> > Then we come to the next lesson: Try to imagine the set of all
whole
> > numbers n and the set of all irrational numbers formed by n + pi.
In
> > the interval (0, oo) we can obtain an alternating sequence
(alternating
> > between rational and irratiuonal values).
>
> Fair enough. We're all with you so far.
>
> 1, 1.1415..., 2, 2.1415..., 3, 3.1415...
>
> Yes, this sequence alternates.
>
> And this sequence includes neither all rationals nor all irrationals.
>
> > Let's continue with the next lesson: Try to imagine the set of all
> > rational numbers D1 which, in decimal representation, have only one
> > digit behind the point. Imagine set of these numbers D1 and the set
of
> > all irrational numbers formed by D1 + pi. In the interval (0, oo)
we
> > can obtain an alternating sequence (alternating between rational
and
> > irratiuonal values).
>
> Sure.
>
> 1, 1.04159..., 1.1, 1.14159..., 1.2, 1.24159...
>
> Again, we get an alternating sequence
> And again we get neither all rationals nor all irrationals.
>
> > Continue with rationals D2 with 2 digits behind the point.
>
> Yes, the extension to 2 digits is obvious.
>
> > Come to rationals Dn with n digits behind the point.
>
> Yes, the extension to an arbitrary finite number of digits is
> also obvious. We can take this as far as we like and
> preserve the three invariant features:
>
> o The sequence alternates
> o The sequence does not contain all rationals
> o The sequence does not contain all irrationals.

I did not say I did.
>
> > Remain there. Because any pair of irrational numbers is separated
by a
> > terminating rational number Dn (i.e., a rational with finite
decimal
> > sequence),
>
> Yes. Between each pair of irrational numbers on the list there is
> a terminating rational on the list.

Wonderful.
>
> > we need not pass into any infinities, but use only an
> > argument, Cantor used at his time: "Continue including all natural
> > (thus finite) numbers n." The set of all Dn and the set of all Dn +
pi
> > taken together, form an alternating sequence in (0, oo).
>
> Here you go astray. Although every Dn U Dn + pi forms an alternating
> sequence, it need not be the case that the infinite union of all
> ( Dn U Dn + pi ) as n ranges over N forms an alternating sequence.

There is no infinite union! Because all numbers Dm with m < n are
automatically included in the set of all numbers Dn. (You see: 0.20 is
also present in 0.200, which is a member of D3.)
>
> And, in point of fact, it is not the case.

There is no union. And, as you admitted, for any n my arguing holds.
>
> > Introducing
> > another irrational, like e, leads to two irrationals which are not
> > separated by a terminating rational. This is a contradiction.
>
> Let's continue the lesson.
>
> 1, 1.14159..., 1.71828..., 2, 2.14159..., 2.71828..., 3, ...
>
> A rational, 2 irrationals, a rational, 2 irrationals.
>
> Yet surely you would not assert that there is no terminating rational
> between 1.14159... and 1.71828...
>
> Indeed, if we go to the next iteration, we find
>
> 1, 1.01828..., 1.04159..., 1.1, 1.11828..., 1.14159..., 1.2,
1.21828...
>
> And right there, plain as day is a rational number 1.2 which falls
> between 1.71828 and 2.14159
>
> You can't look at the absence of intervening numbers at step n
> to prove that there are no intervening numbers at step n+1.

But at step n+1 you obtain two adjacent irrationals again! This fact
remains as long as the natural numbers reach. There is no n for which
it would become false. But we know from basic mathematics that there is
always a terminating rational between two irrationals!
>
> Of course, you will object that 1.71828 and 2.14159 are not
> adjacent irrationals in the limiting sequence.
>
> The problem is that there are _NO_ adjacent irrationals in the
> limiting sequence. The limiting sequence is densely ordered.
> The intermediate sequences are not.

There is no limiting number in Cantor's list. (You would not be able to
determine the position of the diagonal digit in that case.) And there
is no limiting sequence in my proof. n IS ALWAYS A NATURAL NUMBER!!! IT
IS THE SAME ARGUING AS WITH CANTOR'S LIST.

Set theory can be destroyed only by its own weapons, because the
obvious truth that infinity is a nonsense per se will not be accepted
by many mathematicians.
>
>
> In general, you cannot look at invariants that hold at every finite
> step in an infinite progression and properly conclude that they hold
> in the limit as well.
>
> Every leading substring of pi is rational.
> pi itself is not.

So what? What has that to do with my arguing? Just another paradox of
actual infinity.
>
> Every Dn is an alternating sequence.
> Doo is not.
>
> Every irrational in every Dn has an immediate successor
> No irrational in Doo has an immediate successor.
>
> Every rational in every Dn has an immediate successor.
> No rational in Doo has an immediate successor.

There is no Doo because it would not be a terminating rational. I use
terminating rationals only. So your argument does not apply.

But thank you for your constructive and unpolemic response.
Regards, WM

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