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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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 Rock Brentwood Posts: 129 Registered: 6/18/10
Re: Can addition be defined in terms of multiplication?
Posted: Aug 22, 2013 6:59 PM

On Friday, August 16, 2013 5:41:00 PM UTC-5, Rotwang wrote:
> I guess, no, but it is but a guess. I think the answer is no. Note that in any ring, x.0 = x.0 + (x.x - x.x) = x.(0 + x) - x.x = x.x - x.x = 0 So if M with ordinary multiplication could be extended to a ring, there would be an element of M which multiplied by every other element to give itself. No such element exists (since if e.g. x.4 = x then x = 0).

This answer is generic to rings, and may not adequaetly adress the particular features of the natural number system. Similarly, the answer based on the decitability result is not necessarily a valid deduction; for one isn't asking whether the theory of arithmetic (i.e. a free semiring PLUS the essentially 2nd order "axiom of induction" (which is what makes arithmetic undecitable) can be reconstructed from the multiplication operation, but only whether the structure of the free semiring.

Ignore the 0 for the following and just treat the system as {1,2,3,...}.

The multiplication operator yields a structure that is captured by the prime number decomposition result; namely, a countably cartesian product of free monoids, restricted so that all elements have "finite support". Thus, if the associated factor monoids are M_2, M_3, M_5, M_7, ... then a generic element is a = (a_2, a_3, a_5, a_7, ...) corresponding to the number 2^{a_2} 3^{a_3} 5^{a_5} 7^{a_7} .... The "finite support" property is that all but a finite number of components of the sequence a are equal to the respective monoid identity 0.

So, the question on hand is to define an addition operator that is distributive with respect to the multiplication (and satisfies the other properties of an Abelian semigroup: commutativity and associativity).

A "non-standard" addition operator is then one that is NOT generated from the S: x |-> x + 1 operator.

If there's an undecidability result to be found anywhere (and if this is even relevant here), then it may simply by that the actual formula for S(x) -- expressed in terms of the factor monoids and their exponents -- may be non-recursive. But this need not exclude the possibility that the addition operator is uniquely given by the distributivity property and the other properties that define a semiring.

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