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

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 Jim Burns Posts: 1,200 Registered: 12/6/04
Re: Can addition be defined in terms of multiplication?
Posted: Aug 18, 2013 9:04 AM
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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 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.

>
> Stepping out of my comfort zone here, but I think the point is that
> allowing recursive definitions makes the theory second-order, and raises
> the question of why one would not simply define + directly that way too.
>
> Broadly speaking, you can either have a second-order theory in which +
> and * and so on are not in the signature of the language (but are
> defined recursively) or you can have a first-order 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 first-order theory where recursive
> definitions are not permitted.
>
> <snip>
>

Does "first order theory" mean no recursive definitions or
no recursion at all?

I have grown used to gliding past the logical implications of
definitions, assuming (possibly incorrectly) that there were
no *logical* implications, that they were solely a matter of
convenience for the writer and reader -- although a very
important matter of enormous convenience in practice,
making the otherwise incomprehensible comprehensible.
How wrong am I?

It seems to me that one is very limited in what one can do
without recursion. It might well be possible to prove that
Goldbach's Conjecture or the similar Nam's Conjecture is
unprovable in such a system (though one may have to use
recursion to prove that).

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

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