Re: abundance of irrationals!)
From: Jesse F. Hughes (jesse_at_phiwumbda.org)
Date: 02/07/05
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Date: Mon, 07 Feb 2005 15:39:56 +0100
mueckenh@rz.fh-augsburg.de writes:
> Jesse F. Hughes wrote:
>> mueckenh@rz.fh-augsburg.de writes:
>>
>> >> So bloody what?
>> >>
>> >> What is the contradiction?
>> >
>> > The contradiction is that my proof is valid for each n, i.e. for
> every
>> > sequence of terminating rationals. You should be able to see that,
> in
>> > principle.
>>
>> Again, I ask so what?
>>
>> That doesn't prove any contradiction at all.
>>
>> Look, a contradiction is a statement of the form P & ~P. Now, have
>> another go at it.
>
> P: There is always a terminating rational between two irrationals.
As usual, you're incoherent.
There are three distinct interpretations of P.
(1) For all irrational x and y such that x < y, there is a
"terminating" rational z such that x < z < y.
(2) For all sets X c R and all irrational x, y in X such that x < y,
there is a terminating rational z such that x < z < y.
(3) For all sets X c R such that Q n X = Dn for some n, for all
irrational x, y in X such that x < y, there is a z in Dn such that
x < z < y.
Claim (1) is clearly true, but you haven't proved its negation. You
have proved that for all n, there are irrational x and y such that
there is no terminating rational z of length n such that x < z < y.
So what?
You have proved the negation of claims (2) and (3) but nobody has ever
argued that they are true. Except maybe you.
You are incompetent to teach mathematics.
-- "So, at this time, I'd like to assure you that I am not interested in making sure mathematicians worldwide get fired."--JSH Apr 28, 2003 "I'll have prosecutors knocking on your doors. I have no problem with any number of mathematicians spending time in jail."--JSH Jun 10, 2003
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