
Re: Can addition be defined in terms of multiplication?
Posted:
Aug 18, 2013 9:57 AM


On Sun, 18 Aug 2013 03:09:00 0700, William Elliot <marsh@panix.com> wrote:
>On Sun, 18 Aug 2013, Peter Percival wrote: >> > >> > > > > Can addition be defined in terms of multiplication? I.e., >> > > > > is there a formula in the language of arithmetic >> > > > > x + y = z <> ... >> > > > > >> > > > > such that in '...' any of the symbols of arithmetic >> > > > > except + may occur? >> > > > > >> > > > > The symbols of arithmetic (for the purpose of this question) are >> > > > > either >> > > > > individual variables, (classical) logical constants including =, >> > > > > S, +, *, and punctuation marks; >> > > > > or the above with < as an additional binary predicate symbol. >> > > > >> > > > How about >> > > > x + y = z <> 2^x * 2^y = 2^z >> > > > >> > > > where 2^x is just an abbreviation for the function 2pwr: N > N, >> > > > defined by >> > > > 2pwr(0) = 1 >> > > > 2pwr( Sx ) = 2 * 2pwr( x ) >> > > That goes beyond what I defined as the language of arithmetic. >> > >> > It does not. It quite definable with Peano's axioms >> > which may be presumed to be what you intend because >> > of the inclusion of S in the symbols of arithematic. >> >> 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.
Yes. Now give a defintion of 2^n _in the language of arithmetic_.
> >> > If you want it for the reals, >> >> Which I don't. >> >> > then 2^x, x in is >> > definable with <= and a whole lot of logical overhand.

