Re: Gaps... ;^)

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Sujet : Re: Gaps... ;^)
De : chris.m.thomasson.1 (at) *nospam* gmail.com (Chris M. Thomasson)
Groupes : sci.math
Date : 10. Sep 2024, 19:30:29
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <vbq385$33j9p$1@dont-email.me>
References : 1 2
User-Agent : Mozilla Thunderbird
On 9/9/2024 5:28 PM, Moebius wrote:
Am 10.09.2024 um 00:59 schrieb Chris M. Thomasson:
 
Between zero and any positive x there is a unit fraction small enough to fit in the ["]gap["].
 Right. This follows from the so called "Archimedean property" of the reals. From this property we get:
 For all x e IR, x > 0, there is an n e IN such that 1/n < x.
 See: https://en.wikipedia.org/wiki/Archimedean_property
 Of course, from this we get that there are infinitely many unit fractions smaller than x, say, 1/n, 1/(n + 1), 1/(n + 2), 1/(n + 3), ...
 We can even refer to such unit fraction "in terms of x":
 All of the following (infinitely many) unit fractions are smaller than x: 1/ceil(1/x + 1), 1/ceil(1/x + 2),  1/ceil(1/x + 3), ...
 
Between x and any y that is different than it (x), there will be a unit fraction to fit into the gap. infinitely many.... :^)
 Nope. There is no unit fraction (strictly) between, say, 1/2 and 1/1.
What about 1/4? Ahhhh! You mentioned the word _strictly_. Okay.
Humm... Well, if we play some "games" ;^), then 1/4 would sit in the center of the gap between 1/2 and 1/1 where:
p0 = 1/2
p1 = 1/1
dif = p1 - p0
unit_mid = dif / 2
So:
p0 = 1/2 = .5
p1 = 1/1 = 1
dif = p1 - p0 = .5
unit_mid = dif / 2 = .5 / 2 = .25 = 1/4
?
Or is this just crank bullshit that removes the strict factor?

In other words, there is no unit fraction u such that 1/2 < u < 1/1.
 
Say the gap is abs(x - y) where x and y can be real. If they are different (aka abs(x - y) does not equal zero), then there are infinitely many unit fractions that sit between them.
 Nope. See counter example above.
 
Any thoughts? Did I miss something? Thanks.
 Yes. It works for any (0, x) where x e IR, x > 0.
 But it does not work "in general" for (x, y) where x,y e IR, x,y > 0 and x < y (and hence abs(x - y) > 0).
 If you'd consider _rational numbers_ (or fractions) instead of unit fractions, your intuition would be right, though.
 
thanks.

Date Sujet#  Auteur
9 Sep 24 * Gaps... ;^)10Chris M. Thomasson
10 Sep 24 `* Re: Gaps... ;^)9Moebius
10 Sep 24  `* Re: Gaps... ;^)8Chris M. Thomasson
10 Sep 24   `* Re: Gaps... ;^)7Moebius
10 Sep 24    +* Re: Gaps... ;^)3Chris M. Thomasson
10 Sep 24    i`* Re: Gaps... ;^)2Moebius
10 Sep 24    i `- Re: Gaps... ;^)1Chris M. Thomasson
10 Sep 24    `* Re: Gaps... ;^)3Chris M. Thomasson
10 Sep 24     `* Re: Gaps... ;^)2Moebius
10 Sep 24      `- Re: Gaps... ;^)1Chris M. Thomasson

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