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 Proof Ackermann function extended to reals cannot be commutative/associative JmsNxn Long Time Fellow Posts: 456 Threads: 84 Joined: Dec 2010 09/08/2011, 01:37 AM (This post was last modified: 09/08/2011, 08:07 PM by JmsNxn.) Well, the proof is really simple, but it works; Lets assume we have an operator $\otimes_q$ where $0\le q \le1$, and $\otimes_q$ is the super operator of $\otimes_{q-1}$, furthermore, $\otimes_{0} = +$ and $\otimes_{1} = \cdot$ Start off by making our only assumption that $\otimes_q$ and $\otimes_{q-1}$ are commutative and associative. start off with the basic formula: $a_1\,\otimes_{q-1}\,a_2\,\otimes_{q-1}\,a_3\,...\,\otimes_{q-1}\, a_n\,=\,a\,\otimes_{q}\,n$ Now, since $\otimes_{q-1}$ is commutative and associative, we can rearrange them in the following manner if $m + k = n$: so that: $(a\,\otimes_{q}\,m)\,\otimes_{q-1}\,(a\,\otimes_{q}\,k) = \,a\,\otimes_{q}\,n$ therefore, for any a,b,c: $(a\,\otimes_{q}\,b)\,\otimes_{q-1}\,(a\,\otimes_{q}\,c) = a \,\otimes_{q}\,(b+c)$ given this law, if we set a = S(q) or the identity for operator $\,\otimes_{q}\,$ we instantly see that $b\,\otimes_{q-1}\,c\,=\,b\,+\,c$ the only assumption we made was that $\otimes_{q-1}$ and $\otimes_{q}$ be commutative and associative. I think maybe this proof is inadequate at proving it cannot be commutative, but I think it'd be on shaky ground to say they are commutative. But forsure, not associative. Edit: The proof to make it non-commutative is as follows. Since the ackermann function is defined as: $\vartheta(a, b, \sigma) = a \, \otimes_\sigma\,b$ where the only law it must obey is: $a \,\otimes_{\sigma - 1}\,(a\,\otimes_{\sigma}\,b) = a \, \otimes_\sigma\,(b+1)$ If we want $\vartheta$ to be analytic over $\sigma$ (which we do), we cannot have $\otimes_{\sigma}$ being commutative over any strip. because if perhaps we say: all operators including and below multiplication are commutative. This would mean: $\vartheta(a, b, \sigma) = \vartheta(b, a, \sigma)\,\,\R(\sigma)\le 1$ but if two functions are analytic and they equal each other over a strip then they must be the same function therefore: $\vartheta(a, b, \sigma) = \vartheta(b, a, \sigma)$, for all $\sigma$, but this is clearly untrue because exponentiation is not commutative. Therefore $\sigma$ is only commutative at addition (0) and multiplication (1). I guess our rational operators are going to have to behave like exponentiation, I'm really curious about an analytic and integral calculus attack at this problem. Maybe dynamics ain't the right field. I think logarithmic semi operators are as close as it'll get. Maybe there's a more natural equation that may have some aesthetic properties in terms of relations to trigonometric functions, or other established functions with maybe some fancy constants involved. MphLee Fellow Posts: 172 Threads: 15 Joined: May 2013 06/15/2013, 08:02 PM I'm a bit confused You said that the only assumptions were the commutativity of two operations, but you did not use the assumption of the the commutativity of the operation $\otimes_q$ in the proof. (in fact it is not important imo): you asumptions are: 1-commutativity and associativity of the operator $\otimes_{q-1}$. 2-the existence of a unique right identity element (that is the left id. element too if it is commutative) of the operation $\otimes_q$. And these assuptions make you proof valid to show that the only operation $\otimes_{q-1}$ with these properties is the addition. MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ « Next Oldest | Next Newest »

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