+- Tetration Forum (https://tetrationforum.org)
+-- Forum: Tetration and Related Topics (https://tetrationforum.org/forumdisplay.php?fid=1)
+--- Forum: Hyperoperations and Related Studies (https://tetrationforum.org/forumdisplay.php?fid=11)
+--- 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.