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.
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.