Sujet : Re: Replacement of Cardinality
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.logic sci.mathDate : 10. Aug 2024, 18:28:11
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <75e2ce0e-7df8-4266-968b-9c58e4140b03@att.net>
References : 1 2 3 4 5 6 7 8 9 10 11
User-Agent : Mozilla Thunderbird
On 8/10/2024 11:54 AM, WM wrote:
Le 09/08/2024 à 02:32, Jim Burns a écrit :
Surely, a lawyer wouldn't think that
"Boom! Here's the conclusion"
is an _argument_ ?
>
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
<place for argument>
Therefore there can only be a single first unit fraction.
No one has said there are two first unit.fractions.
What forbids zero first unit.fractions?
What causes an exception: nₓ ∈ ℕ without ⅟(nₓ+1) ?
What else do you need?
That which you've not yet done,
show the opposite of this argument:
⎛ ∀n ∈ ℕ: ⅟n - ⅟(n+1) > 0
⎜
⎜ ∀n ∈ ℕ: ∃k ∈ ℕ: ⅟k = ⅟(n+1) ∧ ⅟n > ⅟k
⎜
⎜ ∀n ∈ ℕ: ∃k ∈ ℕ: ⅟n > ⅟k
⎜
⎜ ⅟n is the first unit fraction ⇔
⎜ ¬∃k ∈ ℕ: ⅟n > ⅟k
⎜
⎜ ∀n ∈ ℕ: ¬(⅟n is the first unit fraction)
⎜
⎜ ¬∃n ∈ ℕ: ⅟n is the first unit fraction
⎜
⎝ There is no first unit fraction.
…