Sujet : Re: Replacement of Cardinality
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.logic sci.mathDate : 15. Aug 2024, 23:51:01
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <0aae40c7-1092-4b12-99dc-290aa5a94021@att.net>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13
User-Agent : Mozilla Thunderbird
On 8/15/2024 3:37 PM, Moebius wrote:
Am 15.08.2024 um 21:28 schrieb Jim Burns:
Defining ⅟𝔊 to be the smallest unit fraction
isn't a claim that ⅟𝔊 exists.
>
You need an existence proof [...]
BEFORE stating a/the proper definition.
Proving that mimsy borogoves exist
before defining mimsy borogoves
seems like a pretty steep hill to climb.
That's a hill which I doubt we need to climb.
'⅟𝔊' can be used to derive contradictions,
>
No, it can't.
Half or more of my proofs to WM say
"Assume otherwise... However... Contradiction."
But maybe I'm crazy.
There's an old, old proof that √2 is irrational.
If that's wrong, everyone but you is crazy.
√2 is irrational.
⎛ Assume otherwise.
⎜ Assume p₃,q₃ ∈ ℕ₁: p₃⋅p₃ = 2⋅q₃⋅q₃
⎜
⎜ p₃ ∈ {p ∈ ℕ₁: ∃q ∈ ℕ₁: p⋅p = 2⋅q⋅q}
⎜ p₂ = min.{p ∈ ℕ₁: ∃q ∈ ℕ₁: p⋅p = 2⋅q⋅q}
⎜ ∃q₂ ∈ ℕ₁: p₂⋅p₂ = 2⋅q₂⋅q₂
⎜ ¬∃p₁ < p₂: ∃q₁ ∈ ℕ₁: p₁⋅p₁ = 2⋅q₁⋅q₁
⎜
⎜ However,
⎜ p₂⋅p₂ = 2⋅q₂⋅q₂
⎜ 2 is prime.
⎜ 2|p₂ or 2|p₂
⎜ p₂ = 2⋅p₁
⎜ 2⋅p₁⋅2⋅p₁ = 2⋅q₂⋅q₂
⎜ 2⋅p₁⋅p₁ = q₂⋅q₂
⎜ 2|q₂ or 2|q₂
⎜ q₂ = 2⋅q₁
⎜ 2⋅p₁⋅p₁ = 2⋅q₁⋅2⋅q₁
⎜ p₁⋅p₁ = 2⋅q₁⋅q₁ and p₁ < p₂
⎝ Contradiction.
The proof uses properties such as
well.order and unique.prime.factorization
which p₂ and q₂ would have if they existed
to show that they don't exist.
Doing that isn't a problem.