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

On 10/31/2024 4:18 PM, WM wrote:

On 31.10.2024 19:34, Jim Burns wrote:

On 10/31/2024 1:41 PM, WM wrote:

On 31.10.2024 13:22, Jim Burns wrote:

On 10/31/2024 4:13 AM, WM wrote:

Neither
  ∀n ∈ ℕ: 1/n - 1/(n+1) > 0
nor
  ∀ᴿx > 0: ∀n ∈ ℕ: ⅟⌈n+⅟x⌉ ∈ ⅟ℕ∩(0,x]
is wrong.

>
But the first formula predicts that
only single unit fractions
are existing on the real line.
How could NUF(x) grow from zero by more than 1?

>
Is
⎛  ∀ᴿx > 0:
⎜  ∀n ∈ ℕ:
⎝ ⅟⌈n+⅟x⌉ ∈ ⅟ℕ∩(0,x]
wrong?

>
One of two contradicting formulas must be dropped.

>
Which (inside.quantifiers) formula is
the last which you accept with all prior formulas?

Is _accepting_ too hard for you?
Are there _any (inside quantifiers) formulas at all_
which you accept with all prior formulas?
⎛  ∀ᴿx > 0:
⎜  ∀n ∈ ℕ:
⎜ x > 0
⎜ ⅟x > 0
⎜ n+⅟x ≥ ⅟x > 0
⎜ ⌈n+⅟x⌉ ≥ ⅟x > 0
⎜ ⌈n+⅟x⌉ ∈ ℕ
⎜ ⅟⌈n+⅟x⌉ ∈ ⅟ℕ
⎜ x⋅⌈n+⅟x⌉ ≥ x⋅⅟x > 0
⎜ x⋅⌈n+⅟x⌉⋅⅟⌈n+⅟x⌉ ≥ 1⋅⅟⌈n+⅟x⌉ > 0
⎜ x ≥ ⅟⌈n+⅟x⌉ > 0
⎜ ⅟⌈n+⅟x⌉ ∈ (0,x]  ∧  ⅟⌈n+⅟x⌉ ∈ ⅟ℕ
⎝ ⅟⌈n+⅟x⌉ ∈ ⅟ℕ∩(0,x]
⎛  ∀n ∈ ℕ⁺:
⎜ n > 0
⎜ n+1 > n > 0
⎜ ⅟n⋅(n+1)⋅⅟(n+1) > ⅟n⋅n⋅⅟(n+1)
⎜ ⅟n > ⅟(n+1)
⎜ ⅟n - ⅟(n+1) > ⅟(n+1) - ⅟(n+1)
⎝ ⅟n - ⅟(n+1) > 0

Which one requires that
NUF(x) can grow at an x ∈ ℝ by more than 1?

With a little more work, both
  ∀n ∈ ℕ: 1/n - 1/(n+1) > 0
and
  ∀ᴿx > 0: ∀n ∈ ℕ: ⅟⌈n+⅟x⌉ ∈ ⅟ℕ∩(0,x]
show that
NUF(x) grows at 0 to ℵ₀
Will you (WM) also be ignoring
the work which shows
NUF(x) grows at 0 to ℵ₀ ?

 Regards, WM