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Topic: function arity > 2
Replies: 7   Last Post: Jan 17, 2013 1:37 AM

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Butch Malahide

Posts: 894
Registered: 6/29/05
Re: function arity > 2
Posted: Jan 16, 2013 1:51 AM
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On Jan 15, 11:44 pm, JohnF <j...@please.see.sig.for.email.com> wrote:
> As in universal algebra, where introductory discussions
> typically suggest arity>2 not often used. Are there any
> functions f:N^3-->N (domain integers) that can't be
> decomposed into some g,h:N^2-->N, where f(i,j,k)=g(i,h(j,k))?
> If so (i.e., if arity>2 needed), got an example? If not,
> got a proof? And, if not for integers, is there any domain D
> where f:D^3-->D can't be decomposed like that (example or
> proof again appreciated)?


Hmm. Your equation "f(i,j,k) = g(i,h(j,k))" seems kind of special, and
it's not clear to me that it covers all ways of expressing a ternary
operation in terms of binary operations. Anyway:

If D is an infinite domain, assuming the axiom of choice, there is an
injection h:DxD-->D, i.e., a pairing function. Given any function
f:D^3-->D, we can then define g(i,h(j,k)) = f(i,j,k).

If D = {0,1}, the function f:D^3-->D such that f(0,j,k) = j, f(1,j,k)
= k, can not be expressed in your form f(i,j,k) = g(i,h(j,k)).

If D is an n-element set, 2 < n < infinity, we can generalize the
example for n = 2, or we can just use a counting argument: the number
of ternary operations on D is n^(n^3), whereas the number of ordered
pairs of binary operations is only n^(2n^2).



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