
Re: Can addition be defined in terms of multiplication?
Posted:
Aug 18, 2013 11:07 AM


Jim Burns <burns.87@osu.edu> writes:
> On 8/18/2013 8:02 AM, Ben Bacarisse wrote: >> William Elliot <marsh@panix.com> writes: >> >>> 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. >> >> 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. >> >> <snip> >> > > Does "first order theory" mean no recursive definitions or > no recursion at all?
No, but it means a weaker induction axiom than the one usually used in a secondorder theory. At this point you bump up hard against my knowledge boundary and I can't really say much more about it.
I should pass on the rest. Someone Who Knows will come along clear it up, I hope.
<snip>  Ben.

