Re: Bruijn babbles bull***

Horand.Gassmann@xxxxxxxxxxxxxx wrote:

On Feb 4, 1:37 pm, Han de Bruijn <Han.deBru...@xxxxxxxxxxxxxx> wrote:

Yeah, you're right. But that's not what I mean. What I mean is to find
_new_ decimals in the expansion of e.g. Pi every time again. And again,
because it's never ending. So there is _a_ (finite) decimal expansion
available, despite of the fact that many of us imagine it as the first
portion of THE infinite decimal expansion. I would say that the latter
does not exist.

I don't get that. Can you explain (one more time)? I believe you said
that you accept the existence of N, _the_ set of integers. Is this
correct?

No. I've always said that THE set of ALL integers does not exist.

In particular, if you and I apply the axioms (0 \in N; if n
\in N, then also s(n) \in N), do you agree that we end up with the
same model for the integers?

I do not accept the ZFC axiom of infinity. And I'm not able to apply it.
To be more specific: i.m.o N cannot be defined as a (completed infinite)
set, therefore I do not comprehend what "\in" means in this context.

Do you accept _the_ set of prime numbers?

No.

Do you accept the infinite expansion 0.3333... as an equivalent to the
ratio 1/3?

Yes. (Because the infinite expansion can be understood as a limit)

Would you agree that the finite approximation 0.3333...3 (n threes) is
_not_ equal to 1/3, no matter how large n?

No. Any such finite approximation can be equal to 1/3, depending on the
error allowed in equality. (Absolute rigour is a phantom)

Do you accept the existence of Champernowne's number
0.123456789101112131415161718192021...?

I see it for the first time in my life, so I have to think about it.
But yes, it seems to be well defined. So I'm reluctant to accept it.

Would you agree that I can produce for you any digit you care to name
in the decimal expansion of this number?

Looking up the Wikipedia: yes.

Do you agree that pi exists?

Yes.

Would you agree that I can produce for you any digit you care to name
in the decimal expansion of this number, given enough time and
resources?

Yes. But "any" is not "all".

Han de Bruijn

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