Re: abundance of irrationals

briggs_at_encompasserve.org
Date: 01/28/05 

Date: 28 Jan 2005 09:43:05 -0600

In article <1106917840.695486.191830@z14g2000cwz.googlegroups.com>, mueckenh@rz.fh-augsburg.de writes:
>
> Jesse F. Hughes wrote:
>
>>
>> What does "actual existence" mean? Do some elements in a model
>> "really have actual existence" and others not?
>
> In set theory it is pretended that all numbers do exist, in particular
> that it is usefull to "count" them by cardinal numbers. It would be
> nonsense, if they didn't exist.
>>
>> > But if you are unable to grasp it at one, try the following model.
>> > Imagine all the rational numbers. Add one irrationale number
>> > like pi to this ordered (but not well ordered) set. Can you imagine
>> > this?
>>
>> I guess it depends. Am I also supposed to be pretending that
>> irrational numbers don't exist?
>>
>> Or is that a different game of make-believe?
>
> That is a different game. But the present one leads to the same
> conclusion.
>>
>> If I am "allowed" to know that pi exists, then of course I can
>> imagine the set Q u {pi}. So what?
>
> 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.

> 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.

> 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.

And, in point of fact, it is not the case.

> 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.

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.

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.

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.

        John Briggs

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