Path: ...!news.nobody.at!eternal-september.org!feeder3.eternal-september.org!news.eternal-september.org!.POSTED!not-for-mail
From: Moebius <invalid@example.invalid>
Newsgroups: sci.math
Subject: Re: Does the number of nines increase?
Date: Tue, 16 Jul 2024 03:39:26 +0200
Organization: A noiseless patient Spider
Lines: 33
Message-ID: <v74j0e$ulge$4@dont-email.me>
References: <tJf9P9dALSN4l2XH5vdqPbXSA7o@jntp>
 <53214031-3ad1-48bd-8584-0720ec5b28dd@att.net>
 <bsFBaEx89RCdvkhqBwd1K4mh5ns@jntp>
 <d98d5c8a-041d-4ce6-b7c8-5a212a7bfa3c@att.net>
 <e3ZDe1OozyaPv_8HZy_kTDZtHJk@jntp> <v6spao$3diun$2@dont-email.me>
 <d6yZRpOl38J4dqE-n_qqzplqNmQ@jntp> <v6ul15$3ni5h$1@dont-email.me>
 <79JoZp5bHCH4hf4J9cxbLGeMvPE@jntp> <v70pd4$6n41$1@dont-email.me>
 <vv6K_9idTUEwX3W3ECn0Z9xK6Uk@jntp>
 <74edd85cf4bdd0aecaee742ef35763e9d9dc8741@i2pn2.org>
 <v745vl$skrg$1@dont-email.me> <v746qj$t0g9$1@dont-email.me>
 <v74d0v$u43b$1@dont-email.me> <v74dn3$u43c$1@dont-email.me>
 <v74ejo$tr01$1@dont-email.me> <v74gtv$ulge$1@dont-email.me>
 <v74ia4$uvo1$1@dont-email.me>
Reply-To: invalid@example.invalid
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8; format=flowed
Content-Transfer-Encoding: 7bit
Injection-Date: Tue, 16 Jul 2024 03:39:26 +0200 (CEST)
Injection-Info: dont-email.me; posting-host="f7b47e05a7b4f9518fb2a2693f1d498d";
	logging-data="1005070"; mail-complaints-to="abuse@eternal-september.org";	posting-account="U2FsdGVkX19931+06AHstreZhzciB2CO"
User-Agent: Mozilla Thunderbird
Cancel-Lock: sha1:YiM/WCDeIZ82JZo6CRXW2GLb810=
In-Reply-To: <v74ia4$uvo1$1@dont-email.me>
Content-Language: de-DE
Bytes: 2357

Am 16.07.2024 um 03:27 schrieb Chris M. Thomasson:

> Naturals:
> 
> a = any natural number
> b = a + 1 is the /successor/ of a?
> 
> ;^)

Hell, man, you are asking questions... :-)

Ok, starting with the Peano-axioms there is a functions called /the 
successor function/ s. Then THE successor of, say, n is s(n). :-)

Based on the Peano-axioms we usually define "addition" + ("recursively") 
the following way:

    n + 0 = n
    n + s(m) = s(n + m).

Moreover we (usually) have the definition:

    1 := s(0).
    "1 is the successor of 0".

Hence

n + 1 = n + s(0) = s(n + 0) = s(n).

So, yes, a + 1 is indeed the successor of a (for any a e IN).