
Re: Can addition be defined in terms of multiplication?
Posted:
Aug 19, 2013 7:23 AM


Ben Bacarisse <ben.usenet@bsb.me.uk> writes:
> William Elliot <marsh@panix.com> writes: > >> On Sun, 18 Aug 2013, Peter Percival wrote: ... >>> Then I think the onus is on you to produced definitions in one or both of >>> these forms: >>> x + y = ... >>> x + y = z <> ... >>> >>> where the only nonlogical symbols (baring punctuation) in the ... are from >>> this set: {*,S,0} or this set: {*,S,0,<}. I wouldn't be surprised if + can be >>> defined (in the way requested) from {*,S,0} or {*,S,0,<} but I would like >>> either to see it spelt out, or to be given a reference. >> >> As Jim Burns said >> z = x + y iff 2^z = 2^x * 2^y >> >> where 2^n is defined by induction 2^0 = 1, 2^1 = 1 and 2^(n+1) = 2*2^n >> all of which can be done with Peano's axioms. > > Stepping out of my comfort zone here, but I think the point is that > allowing recursive definitions makes the theory secondorder, and raises > the question of why one would not simply define + directly that way too. > > Broadly speaking, you can either have a secondorder theory in which + > and * and so on are not in the signature of the language (but are > defined recursively) or you can have a firstorder theory where + and * > and so on are added to the signature, with axioms used to induce the > usual meaning. > > I suspect Peter is talking about a firstorder theory where recursive > definitions are not permitted.
I do too; it can be done, but it is not easy.
See Goedel on defining exponentiation from plus and times via the Chinese remainder theorem.
 Alan Smaill

