Re: How many different unit fractions are lessorequal than all unit fractions?

On 9/5/2024 4:44 PM, WM wrote:

On 05.09.2024 20:56, Jim Burns wrote:

On 9/5/2024 9:53 AM, WM wrote:

Le 04/09/2024 à 22:52, Jim Burns a écrit :

On 9/4/2024 3:10 PM, WM wrote:

Either
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
is wrong or Peano is wrong.

>
...or
natural numbers aren't what you think they are.

>
That is possible.

Insisting that ω-1 exists and that,
for b ≠ 0 and β < ω, β-1 exists
is
insisting that ω is finite.

I should have mentioned β+1 and well.order.
But, β+1 is β∪{β}, and,
ordinal.ness means well.ordered.ness.

No.

Peano refers to those β = {α:α<β} such that
⎛ {α:α<β} is well.ordered
⎜ ∀α: 0<α≦β ⇒ 0≦α-1<β  (stepped.down)
⎝ ∀α: 0≦α≦β ⇒ 0<α+1≦β+1 (stepped.up)
The Peano axioms are a different way of describing
the same numbers.
----
The condition for cisfinite induction
P(0) ∧ ∀β<ω:P(β)⇒P(β+1)
is the condition for no.first.counter.example
¬∃γ<ω:( ¬P(γ) ∧ (γ=0 ∨ P(γ-1)) )
Because they're ordinals,
no.first.counter.example requires no.counter.example.
No counter.example: induction.
The well.ordered stepped.down stepped.up numbers
are inductively.valid.
----
Consider {α:α<β}
Extend < from {α:α<β} to {α:α<β}∪{β} = {α:α<ᵦβ+1}
where {α:α<β} ᵉᵃᶜʰ<ᵦ β
0 = {α:α<0} = {} is well.ordered.
if {α:α<β} is well.ordered,
then {α:α<ᵦβ+1} is well.ordered.
0 = {α:α<0} = {} is stepped.down.
if {α:α<β} is stepped.down,
then {α:α<ᵦβ+1} is stepped.down.
0 = {α:α<0} = {} is stepped.up.
if {α:α<β} is stepped.up,
then {α:α<ᵦβ+1} is stepped.up.
By induction,
the inductively.valid numbers
are well.ordered stepped.down stepped.up.

Either
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
is wrong or Peano is wrong.

The most frugal explanation of your claim is that
you simply do not know what 'finite' means.

>
Finite means that
you can count from one end to the other.
Infinite means that
it is impossible to count from one end to the other.

It is impossible to count from ⅟1 down to
_anything_ below the visibleᵂᴹ unit.fractions,
  because
there is no step across
any ⅟ℕᵈᵉᶠ ᵉᵃᶜʰ<ᵉᵃᶜʰ ⅟ℕᵈᵃʳᵏ split,
  because
each step starting in ⅟ℕᵈᵉᶠ ends in ⅟ℕᵈᵉᶠ.
Analogously, ω is on the far side of
the cisfinite ᵉᵃᶜʰ<ᵉᵃᶜʰ transfinite split,
and there is no step from ω-1 to ω
tl;dr
ω-1 does not exist.
Any 'ω' for which 'ω-1' exists is
a fake 'ω' NOT on the far side of ℕᵈᵉᶠ ᵉᵃᶜʰ<ᵉᵃᶜʰ ℕᵈᵃʳᵏ

Do you believe that it needs a shift to state:
All different unit fractions are different.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
I can see no shift.

>
It needs a shift to conclude from
( for each ⅟j: there is ⅟k≠⅟j: ⅟k < ⅟j
that
( there is ⅟k: for each ⅟j≠⅟k: ⅟k < ⅟j
>
Have you evolved on that topic?

>
You are mistaken.
I do not conclude the latter from the former.
I conclude the latter from the fact that
NUF(0) = 0 and NUF(x>0) > 0
and never, at no x, NUF can increase by more than 1.
Try to understand that.

x > 0  ⇒
0 < ... < ⅟⌊4+⅟x⌋ < ⅟⌊3+⅟x⌋ < ⅟⌊2+⅟x⌋ < ⅟⌊1+⅟x⌋ < x
You're welcome.