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Topic: Can addition be defined in terms of multiplication?
Replies: 58   Last Post: Aug 23, 2013 3:56 PM

 Messages: [ Previous | Next ]
 David C. Ullrich Posts: 3,555 Registered: 12/13/04
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 non-logical 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.

Date Subject Author
8/16/13 Peter Percival
8/16/13 William Elliot
8/16/13 Peter Percival
8/16/13 David C. Ullrich
8/16/13 namducnguyen
8/17/13 Peter Percival
8/17/13 namducnguyen
8/17/13 fom
8/23/13 tommy1729_
8/16/13 Peter Percival
8/16/13 Robin Chapman
8/16/13 Helmut Richter
8/16/13 Rotwang
8/16/13 Virgil
8/22/13 Rock Brentwood
8/16/13 Shmuel (Seymour J.) Metz
8/17/13 Helmut Richter
8/16/13 Jim Burns
8/16/13 fom
8/17/13 Robin Chapman
8/17/13 fom
8/17/13 Peter Percival
8/17/13 fom
8/17/13 Peter Percival
8/17/13 Peter Percival
8/18/13 William Elliot
8/18/13 Peter Percival
8/18/13 William Elliot
8/18/13 Peter Percival
8/18/13 Graham Cooper
8/18/13 David C. Ullrich
8/18/13 David C. Ullrich
8/17/13 Graham Cooper
8/18/13 David Bernier
8/18/13 Ben Bacarisse
8/18/13 Peter Percival
8/18/13 Jim Burns
8/18/13 fom
8/18/13 Ben Bacarisse
8/18/13 Graham Cooper
8/18/13 Graham Cooper
8/18/13 Graham Cooper
8/18/13 Graham Cooper
8/19/13 Graham Cooper
8/19/13 Alan Smaill
8/19/13 fom
8/19/13 Alan Smaill
8/20/13 Alan Smaill
8/20/13 Peter Percival
8/20/13 Graham Cooper
8/20/13 Graham Cooper
8/22/13 David Libert
8/22/13 Peter Percival
8/20/13 fom