Path: ...!news.misty.com!weretis.net!feeder9.news.weretis.net!news.nk.ca!rocksolid2!i2pn2.org!.POSTED!not-for-mail From: Richard Damon Newsgroups: sci.math Subject: Re: Replacement of Cardinality Date: Mon, 2 Sep 2024 13:13:46 -0400 Organization: i2pn2 (i2pn.org) Message-ID: References: MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 7bit Injection-Date: Mon, 2 Sep 2024 17:13:47 -0000 (UTC) Injection-Info: i2pn2.org; logging-data="602294"; mail-complaints-to="usenet@i2pn2.org"; posting-account="diqKR1lalukngNWEqoq9/uFtbkm5U+w3w6FQ0yesrXg"; User-Agent: Mozilla Thunderbird X-Spam-Checker-Version: SpamAssassin 4.0.0 In-Reply-To: Content-Language: en-US Bytes: 2741 Lines: 28 On 9/2/24 12:56 PM, WM wrote: > On 01.09.2024 21:09, Chris M. Thomasson wrote: >> On 8/31/2024 8:27 PM, Moebius wrote: > >>> In general, for all x e IR, x > 0: NUF(x) = aleph_0. > > Don't get confused by that nonsense. Everybody knows that unit fractions > are different from each other. Therefore they cannot be counted at the > same x, let alone at less than all positive x, i.e., at zero. >>> >> How does that fit with WM who thinks there is a smallest unit fraction >> to start counting from? > > I do not think that but I prove that by the simple fact that not more > than one unit fraction can be lessequal than all unit fractions. > > Regards, WM > Except that it starts with the incorrect assumption that such a unit fraction exists. Since for *ANY* unit fraction, (including your claimed one) there always exist at least one more (and in fact an infinite number of them) that is actually smaller than it, there can not be a "smallest" one. Since for any finite x > 0 you choose, there exists an n = floor(1/x) and the values of 1/(n+1), 1/(n+2), 1/(n+3) ... 1/(n+k) ... that are smaller than x, we see that there can not be a smallest unit fraction.