help with a distributivity law - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: help with a distributivity law (/showthread.php?tid=699) help with a distributivity law - JmsNxn - 09/18/2011 Well I've been working on trying to generalize a set of operators that are commutative and associative, and everything works fairly fine until I come to the distributivity law across addition and f-addition. The concepts are very simple and the algebra is very simple, but the contradiction just keeps me bewildered. Any help would be greatly appreciated. We start off by defining f-multiplication: $a\,\otimes_f\,b = f(f^{-1}(a)\,+\,f^{-1}(b))$ right off the bat we see $\otimes_f$ is commutative and associative. We can create it's super-operator, called f-exponentiation: $a^{\otimes_f b} = f(b\cdot f^{-1}(a))$ which is distributive across f-multiplication: $(a\,\otimes_f\,b)^{\otimes_f c} = (a^{\otimes_f\,b})\,\otimes_f\,(b^{\otimes_f\, c})$ The trouble comes when we try to develop f-addition, which is defined such that f-multiplication is it's super operator. For this definition we first need to define f-division, which is the inverse of f-multiplication: $a\,\oslash_f\,b= a\,\otimes_f\,(b^{\otimes_f\, -1}) = f(f^{-1}(a)\, -\, f^{-1}(b))$ Therefore f-addition is given by the usual super function/abel function equation: $a\,\oplus_f\,b = b\, \otimes_f\,((a\,\oslash_f\,b) + 1)$ furthermore we also know this extends more generally to: $a\,\oplus_f\,(b\,\otimes_f\, k)= b \, \otimes_f \, ((a\,\oslash_f\,b)\,+\,k)$ That is, if we give the condition that $f(0)=1$ which I do. and here's where we get our contradiction. if we let $u = a\,\oslash_f\,b$ we instantly see: $b \, \otimes_f \, (u\,+\,k) = (b\,\otimes_f\,u)\,\oplus_f\,(b\,\otimes_f\,k)$ and now if we let $b = f(0)$ which is the identity of $\otimes_f$, we get: $f(0) \, \otimes_f \, (u\,+\,k) = (f(0)\,\otimes_f\,u)\,\oplus_f\,(f(0)\,\otimes_f\,k)$ $u \,+\, k = u\,\oplus_f\,k$ This is obviously false so I'm wondering, can we simply not have the distributivity law? And if so, why? What exact step am I doing that is inconsistent? furthermore if we have the distributive law, we also get another distributive law thanks to the commutative and associative nature of $\otimes_f$ $a\,\otimes_f\,(b\,\oplus_f\,c) = (a\,\otimes_f\,b) + (a\,\oplus_f\,c)$ so it's like f-multiplication converts f-addition and addition back and forth. Which is consistent when we set $f(x) = b^x$ and f-multiplication becomes multiplication. but otherwise it becomes oddly inconsistent. again, anyhelp would be greatly appreciated. RE: help with a distributivity law - tommy1729 - 09/21/2011 hmm although you made a typo , your derivation seems correct this time. well to avoid " loss of information " we always need an ordinary addition somewhere - possibly hidden - you have assumed an identity of f-multiplication. i think the identity causes the problem. if you could avoid exp type solutions and identities you might have some luck ... however i think that would require non-complex numbers ... RE: help with a distributivity law - tommy1729 - 09/21/2011 as a sidenote to the above : notice that if a function has no identity , this often results in the equivalent of a value that cannot be attained. if f(z) is entire and has values that cannot be attained ; it is of type exp(entire(z)) + Constant. hence we tend to have solutions of type exp^[a](exp^[b](x) + exp^[b](y)). but more can be said. you might wanna read this old but good thread containing an intresting proof by bo : http://math.eretrandre.org/tetrationforum/showthread.php?tid=125 regards tommy1729 RE: help with a distributivity law - JmsNxn - 09/22/2011 Yes I saw the little mistake I made where I assumed $f(0) = 1$. The reason is because I've only really been observing functions which meet that requirement. I think I'm not doing anything inconsistent but instead we have to create the law, which is not dissimilar to division by zero: $f(0)\,\otimes_f\,(a + b) = a + b \neq\, a\, \oplus_f\,b$ or that the distributive law fails when f-multiplied by the identity. I looked at that other thread too, very interesting. I had of hunch bo's proof but it's nice to see it proved. but furthermore, this gives some very strange laws for multiplication: $v \,\otimes_f\,(k\cdot a) = v\,\otimes_f\,k\,\otimes_f\,a = k\,\otimes_f\,(v \cdot a)$ which means for exponentiation: $v \,\otimes_f\,(a^k) = v \, \otimes_f\,(a^{k-1})\,\otimes_f\,a$ which means f-multiplying a number to the power of another number we convert exponentiation to f-exponentiation: $v \,\otimes_f\,(a^k) = v\,\otimes_f\,(a^{\otimes_f\,k})$ which again is very very inconsistent. I must be doing something incorrect. I think we cannot give the distribution law, but that's not enough for me. I'd really like to know why. I'm absolutely puzzled.