Sujet : Re: How many different unit fractions are lessorequal than all unit fractions?
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.mathDate : 23. Sep 2024, 19:53:36
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <7dffb42d-c049-4553-a96a-c20cf8ad5caa@att.net>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
User-Agent : Mozilla Thunderbird
On 9/23/2024 8:56 AM, WM wrote:
On 23.09.2024 00:53, Richard Damon wrote:
Count down from infinity please,
what is the actual first number you use,
(not w-1, that isn't a number).
>
The problem is it doesn't exist,
because the numbers don't have a bound
on that end.
>
The problem is not non-existence.
Some things which we can describe
would require gibberish, impossibilities
if they existed.
Because we do not,
we _can not_ use gibberish,
we conclude that
gibberish.requirers do not exist.
You (WM) can declare that gibberish.requirers exist
until you're blue in the face;
gibberish.requirers will persist in
requiring gibberish.
We _can't_ accept gibberish.
There is _nothing there to accept_
Proof:
Chosse the greates existing number m.
Consider Boolos's ST
⎛ ∃{}
⎜ ∀x∀y∃z=x∪{y}
⎝ {set}.extensionality
∀m∃z=m∪{m}
¬∃m¬∃z=m∪{m}
Along with the claim ∃m¬∃z=m∪{m}
that a greatest number exists
we have
¬∃m¬∃z=m∪{m} ∧ ∃m¬∃z=m∪{m}
which is gibberish.
Then double it to get 2m.
Which is gibberish in the context of
m being the largest number.
Why didn't you choose 2m originally?
Because largest.number m is gibberish.
Obviously it exists.
largest m and m+1 2m and mᵐ
are all gibberish.
Talking about them
as though they make sense
does not change gibberish into sense.