# Tetration Forum

Full Version: help with a distributivity law
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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.
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 ...
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/tetrationforu...hp?tid=125

regards

tommy1729
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.