Path: ...!weretis.net!feeder9.news.weretis.net!news.quux.org!eternal-september.org!feeder3.eternal-september.org!news.eternal-september.org!eternal-september.org!.POSTED!not-for-mail From: Moebius Newsgroups: sci.math Subject: Re: New way of dealing with complex numbers Date: Sun, 9 Mar 2025 00:31:24 +0100 Organization: A noiseless patient Spider Lines: 41 Message-ID: References: <8wfDWQnhQFqO9uBQ5h-nF8mcuFk@jntp> 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 00:31:24 +0100 (CET) Injection-Info: dont-email.me; posting-host="10511604d5b70f470a56010831842feb"; logging-data="403773"; mail-complaints-to="abuse@eternal-september.org"; posting-account="U2FsdGVkX1+xrLRN31XOlYwxP8RvoFCo" User-Agent: Mozilla Thunderbird Cancel-Lock: sha1:4aXK8Mo3jC0DO7mlVjLhAgBAgD0= In-Reply-To: Content-Language: de-DE Bytes: 2382 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*. :-) Trying to use crank strategies? .. .. ..