Re: how

Le 26/05/2024 à 23:21, Jim Burns a écrit :

On 5/26/2024 3:15 PM, WM wrote:

Le 25/05/2024 à 19:23, Jim Burns a écrit :

On 5/23/2024 8:10 AM, WM wrote:

 

There is no unit fraction smaller than all x > 0, [A]

>
Also true:
There is no x > 0 smaller than all unit fractions.  [B]

 

Note that
points on the real axis are fixed and
can be subdivided into two sets, namely
the set of unit fractions and
the set of positive non-unit fractions.

 Agreed.
 

If A is true, then there is
a positive non-unit fraction smaller than
all unit fractions.

 No.
 | Assume that claim is correct.
| Assume that x > 0 is smaller than
| all unit.fractions.
|
| However,
| ⅟⌊(1+⅟x)⌋ is a unit.fraction < x
| Contradiction.

Not possible with dark number x.

 That claim is incorrect.

for visible numbers.

 

If B is true, then there is
a unit fraction smaller than
all positive non-unit fraction.

 No.
 | Assume that claim is correct.
| Assume that ⅟n is smaller than
| all positive non.unit.fractions.
|
| However,
| ⅟(n+π) is a non.unit.fraction < ⅟n
| Contradiction.

Not possible with dark number n.

 That claim is incorrect.

for visible numbers.

There is no third alternative.

 Consider ⅟⌊(1+⅟x)⌋ and ⅟(n+π)

I consider that points on the real axis are fixed and can be subdivided into two sets.

The decision is that not all subsets
of unit.fractions or of non.unit.fractions
have two ends.

The real axis and all point sets in it have an end at or before zero.
Regards, WM