Path: ...!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: Sat, 8 Mar 2025 23:55:20 +0100 Organization: A noiseless patient Spider Lines: 23 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: Sat, 08 Mar 2025 23:55:20 +0100 (CET) Injection-Info: dont-email.me; posting-host="bb9edeeeddeb8dcebd46aa6a250807e0"; logging-data="381656"; mail-complaints-to="abuse@eternal-september.org"; posting-account="U2FsdGVkX1+fU5cMJi4YTbq45vpSyMOD" User-Agent: Mozilla Thunderbird Cancel-Lock: sha1:foxUH7VKRze11PtQwE5E+lpRSEc= In-Reply-To: Content-Language: de-DE Bytes: 1872 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.