Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

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

 Messages: [ Previous | Next ]
 Graham Cooper Posts: 4,495 Registered: 5/20/10
Re: Can addition be defined in terms of multiplication?
Posted: Aug 18, 2013 7:13 AM

On Sunday, August 18, 2013 3:17:34 AM UTC-7, Peter Percival wrote:
> William Elliot 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.
>
> >>>>>
>
>
> >>>>> 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.
>
>
>
> And the magic formulae
>
>
>
> 2^x = ...
>
>
>
> 2^x = y <-> ...
>
>
>

First you need to define a power sequence..

t(t(t(1))))

is 2^3

--------------

pot [t[t[t 1]]] ?

****** QUERY ******
1 pot
21 t
221 t
2221 t
2222 1
*******************

1 X (22:1) = t
2 X (22:21) = t
3 X (22:22) = 1
pot [ t X ]
TAIL 1
pot X
pot [ t t 1 ]

****** QUERY ******
1 pot
21 t
221 t
222 1
*******************

4 X (22:1) = t
5 X (22:2) = 1
pot [ t X ]
TAIL 1
pot X
pot [ t 1 ]

****** QUERY ******
1 pot
21 t
22 1
*******************

6 X (:) = 1
pot [ t X ]
TAIL 1
pot X
pot [ 1 ]

****** QUERY ******
1 pot
2 1
*******************

pot 1
MATCH
TRUE 1
MATCH
TRUE 1
MATCH
TRUE 1
MATCH

as opposed to 'addition based' arithmetic.

http://phpprolog.com/demo/pp7.png
http://phpprolog.com/demo/pp9.png
http://phpprolog.com/demo/pp1.png

Herc
--

www.phpPROLOG.com

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