I think I may need a refresher in logs here. The question is:
F=$\{a \in R \vert a<1\} 1<t \in R$
(1)$a\#b= a+b-ab$ for all $a,b \in$ F
(2)$a*b=1-t^{log_t(1-a) * log_t(1-b)}$ for all $a,b \in F$ where # and * are just function notation.
Is this a field?
Well, starting with (1), I stated that a=1/2 and b=2/3, then
$1/2\#2/3= 1/2 + 2/3 - (1/2)(2/3)= 5/6$ which is less than 1
(2) then is $(1/2)*(2/3)=1-t^{log_t(1-a) log_t(1-b)}$
Here is where I'm lost on how to work the rest. I'm pretty sure its going to turn out to be less than 1 since we have 1-something bigger than 1. After, this I believe I then have to use the 6 conditions that determine a field. They are:
- associativity of addition and multiplication
- commutativity of addition and multiplication
- distributivity of multiplication over addition
- existence of identity elements for addition and multiplication
- existence of additive inverse
- existence of mulitiplicative inverse where a cannot be 0
Consider the bijective mapping $$ \psi:F\to\mathbb{R},\psi(x)=\log_t(1-x) $$ with $\psi^{-1}(x)=1-t^x$. Then $$\eqalign{ a\#b&=\psi^{-1}(\psi(a)+\psi(b))\cr a*b&=\psi^{-1}(\psi(a)\cdot\psi(b))} $$ This proves that $\psi$ transfers the field structure of $(\mathbb{R},+,0)$ to $(F,\#,*)$. So, $(F,\#,*)$ is a field isomorphic to the field of real numbers.