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Path: ...!eternal-september.org!feeder3.eternal-september.org!news.eternal-september.org!eternal-september.org!.POSTED!not-for-mail From: Moebius <invalid@example.invalid> Newsgroups: sci.math Subject: Re: New way of dealing with complex numbers Date: Sun, 9 Mar 2025 00:47:25 +0100 Organization: A noiseless patient Spider Lines: 57 Message-ID: <vqikud$cgtp$1@dont-email.me> References: <kRgli3QEdimCvJ9569p9c9pq7Kc@jntp> <vqemhv$2gck$1@news.muc.de> <h8RR2Nzw97n_q0rv1uxcQeGImmk@jntp> <vqfm6n$3ndao$2@dont-email.me> <pPxKmW5wI10Jy2mOG_Y_84dfZXg@jntp> <vqhanm$4cd3$1@dont-email.me> <8wfDWQnhQFqO9uBQ5h-nF8mcuFk@jntp> <vqhgtr$5je4$1@dont-email.me> <vqihdj$bkmo$2@dont-email.me> <vqihso$bkmo$3@dont-email.me> <vqijmt$b8c0$2@dont-email.me> <vqik0c$ca9t$1@dont-email.me> <vqiknp$b8c0$3@dont-email.me> Reply-To: invalid@example.invalid MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit Injection-Date: Sun, 09 Mar 2025 00:47:27 +0100 (CET) Injection-Info: dont-email.me; posting-host="10511604d5b70f470a56010831842feb"; logging-data="410553"; mail-complaints-to="abuse@eternal-september.org"; posting-account="U2FsdGVkX1/Gt+JksXAGU2HCpkf2bWpU" User-Agent: Mozilla Thunderbird Cancel-Lock: sha1:j4Une150DFXm/d/VROHN6kZ+BQI= Content-Language: de-DE In-Reply-To: <vqiknp$b8c0$3@dont-email.me> Bytes: 3125 Am 09.03.2025 um 00:43 schrieb efji: > Le 09/03/2025 à 00:31, Moebius a écrit : >> Am 09.03.2025 um 00:26 schrieb efji: >>> Le 08/03/2025 à 23:55, Moebius a écrit : >>>> Am 08.03.2025 um 23:47 schrieb Moebius: >>>>> Am 08.03.2025 um 14:32 schrieb efji: >>>>>> Le 08/03/2025 à 14:18, Richard Hachel a écrit : >>>>> >>>>>> Associativity is MANDATORY to be able to write something like i^4 >>>>>> = i*i*i*i. >>>>>> >>>>>> For a non associative operator, i^4 means NOTHING. >>>>> >>>>> Oh, i^(n+1) just might mean (i^n) * i (with n e IN). >>>>> >>>>> [And i^0 = 1.] >>>>> >>>>> Then: i^4 = ((i*i)*i)*i. >>>>> >>>>> [Hint: recursive definition: >>>>> x^0 = 1 >>>>> x^(n+1) = x^n * x (for all n e IN)] >>>> >>>> x^0 = 1 >>>> x^(n+1) = (x^n) * x (for all n e IN)] >>>> >>>> ... if you like. >>> >>> I don't like. >>> What if * is not commutative ? >>> >>> (x^n) * x =/= x * (x^n) >> >> Might be the case, yes. So what? :-P >> >> But -hint- you talked about *associativity*, not about >> *commutativity*. :-) > > I just pointed out the fact that the notation x^n is never used in the > case of non associative operators because it is ambiguous without > further definition. Think about the vector product in R^3 for example, > which is not associative, and not commutative too. Nobody would write > x^3 for (x \wedge x)\wedge x. > > In the case of Hachel's delirium, the product is obviously associative, > thus i^2 = -1 and i^4 = -1 makes no sense. > > And of course, even with your recursive definition, it makes no sense. > >> >> Trying to use crank strategies? > > fighting fire with fire :) :-P