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Topic: ? 533 Proof
Replies: 8   Last Post: Aug 4, 2014 1:24 PM

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Ben Bacarisse

Posts: 1,294
Registered: 7/4/07
Re: ? 533 Proof
Posted: Aug 3, 2014 9:10 PM
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mueckenh@rz.fh-augsburg.de writes:

> On Sunday, 3 August 2014 16:51:30 UTC+2, Ben Bacarisse wrote:
>

>> > Sorry, you seem to have overlooked this sentence: Assuming this kind
>> > of infinity (finished infinity), you get from set theory the result
>> > that all rationals can be enumerated. My proof shows that this is
>> > wrong for every n and even in the limit.

>>
>> No, not overlooked. It's not provable (at least you never show a
>> proof).
>>

> I have proven that all naturals do not index all rationals because all
> naturals leave infinitely many rationals not indexed. In the eyes of
> all mathematicians with no matheological indoctrination this is a
> proof.


You are confused. When I asked you to write that out "properly" you
replied with the statement of the theorem in 533 -- something that I
accept as a valid theorem. Does that mean that I now understand and
accept what you mean by the words "all naturals do not index all
rationals"?

Meanwhile, there are lots of bijections between N and Q+, none of which
you have any objection to. Where's the conflict? Whatever you mean by
the deliberately tricksy wording "all naturals do not index all
rationals" it certainly is not that there are no bijections between N
and Q+.

<snip>
> Your objection is unfounded as well and therefore is of little weight
> in mathematics.


My objection was that you never show a proof of any contradiction. That
is founded on the total absence of any proof of a contradiction. The
lack of a proof is usually considered to be an objection that carries
some weight in mathematics, but I see in another post that you have
wisely decided to redefine mathematics, presumably because the usual
desire to see proofs is problematic for you.

>> We all know that there are enumerations of the rationals (in set theory,
>> not WMaths -- even you know that),

>
> You all are taking the first few numbers for "all". That is rather
> naive.


No, I am talking about a bijection. A bijection that has N as its
domain and Q+ as it's image. No member of Q+ is missing from the image,
and every member of N is in the domain. Logic lets me flip these
statements around: that every q is the image of some n means that no
rational is missing (and so on). You can do this too with potentially
infinite sets. In fact you do when you say that f(x) = x + 1 is a
bijection between Z and Z.

<snip>
>> that your
>> theorem 533 will do as a statement of not(exists enumeration of Q+).

>
> Would you believe that 60 out of 60 understand?


No, not if you are referring to your students. At least one seemed very
confused.

<snip>
--
Ben.



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