Liste des Groupes | Revenir à s math |
Am 13.08.2024 um 19:02 schrieb Jim Burns:Of course you are right. 100% right. Any marginally decent high schoolOn 8/13/2024 10:21 AM, WM wrote:Well, no, this just isn't a proof, sorry about that, Jim.[There is] a real [number] x with NUF(x) = 1.>
INVNUF(1) > ⅟ ⌊⅟INVNUF(1) +1⌋ > ⅟ ⌊⅟INVNUF(1) +2⌋
>
NUF(INVNUF(1)) > 1
Contradiction.
(Hint: The term "INVNUF(1)" is not defined. Hence you may not even use it in a proof by contradiction. Actually, you didn't state an assumption in your "proof".)
Proof by contradiction:
Assume that there is an x e IR such that NUF(x) = 1. Let x0 e IR such that NUF(x0) = 1. This means that there is exactly one unit fraction u such that u < x0. Let's call this unit fraction u0. Then (by definition) there is a (actually exactly one) natural number n such that u0 = 1/n. Let n0 e IN such that u0 = 1/n0. But then (again by definition) 1/(n0 + 1) is an unit fraction which is smaller than u0 and hence smaller than x0. Hence NUF(x0) > 1. Contradiction!
(Of course, it's clear that I'm using the same "proof idea" as you used in your attempt of a proof.)
Les messages affichés proviennent d'usenet.