On 4/6/2024 3:49 PM, WM wrote:
Le 06/04/2024 à 17:49, Jim Burns a écrit :
On 4/6/2024 9:44 AM, WM wrote:
But we can use the ordinal axis
as Cantor has described it
0, 1, 2, 3, ...,
ω, ω + 1, ..., ω + k, ...,
ω + ω (= ω2), ω2 + 1, ..
and multiply 0, 1, 2, 3, ..., ω, by 2.
>
Two times anything on the first row
is something on the first row.
>
That is contradicted by
...your anti.arithmetism.
tl;dr
The successor operation is closed in
the natural numbers.
Natural.number.addition is closed in
the natural numbers.
Natural.number.multiplication is closed in
the natural numbers.
Those claims are theorems.
In order to be theorems,
it is important to know the meaning of
successor, natural number, addition and multiplication.
If you want to see the proofs, ask.
Consider the successor.operation ⁺¹ which is
non.0, 1.to.1, and closed in the successor.havers.
i⁺¹≠0 ∧ ¬∃h≠i:h⁺¹=i⁺¹ ∧ ∃k=i⁺¹⁺¹ ∧ 1=0⁺¹
For some successor.havers k
a set ⟦0,k⦆ exists such that
predecessor k⁻¹ of k and
predecessor i⁻¹ of each i≠0 in ⟦0,k⦆ and
0 are in ⟦0,k⦆
∀i ∈ ⟦0,k⦆: i⁺¹ ∈ ⦅0,k⟧
The natural numbers ℕ are
0 and successor.havers with ⟦0,k⦆ like that
k ∈ ℕ ⟺
k=0 ∨ ∃⟦0,k⦆: ∀i ∈ ⟦0,k⦆: i⁺¹ ∈ ⦅0,k⟧
That successor operation ⁺¹ is closed in ℕ
If
∃⟦0,k⦆: ∀i ∈ ⟦0,k⦆: i⁺¹ ∈ ⦅0,k⟧
then
∃⟦0,k⁺¹⦆: ∀i ∈ ⟦0,k⁺¹⦆: i⁺¹ ∈ ⦅0,k⁺¹⟧
Thus,
if k ∈ ℕ
then k⁺¹ ∈ ℕ
Addition in ℕ is defined so that
k+m=n ⟺
a sequence fₖ₊ₘ₌ₙ: ⟦0,m⟧ ⟶ 𝔸 exists of addition.facts
starting with ⟨k+0=k⟩, ending with ⟨k+m=n⟩
and ⟨k+i=j⟩ ⇔ ⟨k+i⁺¹=j⁺¹⟩
∀k,m,n ∈ ℕ:
k+m=n ⟺
∃fₖ₊ₘ₌ₙ: ⟦0,m⟧ ⟶ 𝔸:
fₖ₊ₘ₌ₙ(0)=⟨k+0=k⟩ ∧ fₖ₊ₘ₌ₙ(m)=⟨k+m=n⟩ ∧
∀i ∈ ⟦0,m⦆:
fₖ₊ₘ₌ₙ(i)=⟨k+i=j⟩ ⇔ fₖ₊ₘ₌ₙ(i⁺¹)=⟨k+i⁺¹=j⁺¹⟩
Addition in ℕ is closed in ℕ
If
∀i ∈ ⟦0,k⦆: i⁺¹ ∈ ⦅0,k⟧ and
∀i ∈ ⟦0,m⦆: i⁺¹ ∈ ⦅0,m⟧
then
∀i ∈ ⟦0,k+m⦆: i⁺¹ ∈ ⦅0,k+m⟧
Thus,
if k,m ∈ ℕ
then k+m ∈ ℕ
Multiplication in ℕ is defined so that
k⋅m=n ⟺
a sequence fₖ.ₘ₌ₙ: ⟦0,m⟧ ⟶ 𝕄 exists of multiplication.facts
starting with ⟨k⋅0=0⟩, ending with ⟨k⋅m=n⟩
and ⟨k⋅i=j⟩ ⇔ ⟨k⋅i⁺¹=j+k⟩
∀k,m,n ∈ ℕ:
k⋅m=n ⟺
∃fₖ.ₘ₌ₙ: ⟦0,m⟧ ⟶ 𝕄:
fₖ.ₘ₌ₙ(0)=⟨k⋅0=0⟩ ∧ fₖ.ₘ₌ₙ(m)=⟨k⋅m=n⟩ ∧
∀i ∈ ⟦0,m⦆:
fₖ.ₘ₌ₙ(i)=⟨k+i=j⟩ ⇔ fₖ.ₘ₌ₙ(i⁺¹)=⟨k+i⁺¹=j+k⟩
Multiplication in ℕ is closed in ℕ
If
∀i ∈ ⟦0,k⦆: i⁺¹ ∈ ⦅0,k⟧ and
∀i ∈ ⟦0,m⦆: i⁺¹ ∈ ⦅0,m⟧
then
∀i ∈ ⟦0,k⋅m⦆: i⁺¹ ∈ ⦅0,k⋅m⟧
Thus,
if k,m ∈ ℕ
then k⋅m ∈ ℕ
That is contradicted by this argument:
Two times all numbers of 1, 2, 3, .., ω
will double the distance between ℕ and ω
to that between 2ℕ and 2ω.
No, it won't.
V
Re: V