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: efji Newsgroups: sci.math Subject: Re: New way of dealing with complex numbers Date: Sun, 9 Mar 2025 00:26:21 +0100 Organization: A noiseless patient Spider Lines: 33 Message-ID: References: <8wfDWQnhQFqO9uBQ5h-nF8mcuFk@jntp> MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit Injection-Date: Sun, 09 Mar 2025 00:26:21 +0100 (CET) Injection-Info: dont-email.me; posting-host="46f014286ea6fe11ade7226b58030362"; logging-data="369024"; mail-complaints-to="abuse@eternal-september.org"; posting-account="U2FsdGVkX1976ZMW6NsKa31WxLX27mKg" User-Agent: Mozilla Thunderbird Cancel-Lock: sha1:PHPCY5rUl9BHuLUc0bZXw6yYOT0= Content-Language: fr, en-US In-Reply-To: Bytes: 2078 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) -- F.J.