Re: Replacement of Cardinality

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.