Sujet : Re: How many different unit fractions are lessorequal than all unit fractions? (infinitary)
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : sci.mathDate : 02. Nov 2024, 21:30:53
Autres entêtes
Organisation : i2pn2 (i2pn.org)
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On 11/2/24 1:56 PM, WM wrote:
On 02.11.2024 17:02, Richard Damon wrote:
On 11/2/24 6:23 AM, WM wrote:
If NUF(x) has grown to ℵ₀ at x₀, then ℵ₀ unit fractions must be between 0 and x₀. Hence at least ℵ₀ points with ℵ₀ intervals of uncountably many points must be between 0 and x₀. That cannot happen at x₀ = 0.
>
Right, NUF(x)is 0 at x - 0, and Aleph_0 at any x > 0, since as you said, for any positive finite x, there are Aleph_0 unit fractions below it.
No, neither did I say so nor is it correct. Only for x > x₀ are ℵ₀ smaller unit fractions existing.
Regards, WM
So, for what finite number x > 0 is there less than that?
There isn't one, as we CAN PROVE that for all x > 0, there exist and infinite number of unit fractions less that x, we just need to find largest, which is the m = ceil(1/x) and then we can use 1/x, 1/(x+1),
1/(m+2), 1/(m+3), 1/(m+4), ... 1/(m+n), ... for all Aleph_0 values of possible n.
The fact that the Naturals are closed over addition, so given m and n Natrual Numbers, m+n is a Natural Number, we have Aleph_0 values of n, all different, and Aleph_0 values of m+n, all dfferent, and 1/(m+n) will be less than the initial x.
Thus, for all finite x > 0, NUF(x) = Alpeh_0.
Anything else just proves your mathematics can't handle the complete set of Natural Numbers.