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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:

  1. associativity of addition and multiplication
  2. commutativity of addition and multiplication
  3. distributivity of multiplication over addition
  4. existence of identity elements for addition and multiplication
  5. existence of additive inverse
  6. existence of mulitiplicative inverse where a cannot be 0
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    $\begingroup$ Try finding a map that converts $\#$ and $\ast$ into ordinary addition and multiplication. $\endgroup$ – Daniel Fischer Jun 5 '14 at 20:38
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    $\begingroup$ another hint along Daniel Fischer's lines: $a\# b = 1-(1-a)(1-b)$. $\endgroup$ – Dustan Levenstein Jun 9 '14 at 20:32
  • $\begingroup$ It seems to me condition, or rather definition, $(2)$ is nested. The operation $*$ appears on the left hand side as well as in the exponent in the right hand side. Continued exponential(similar to continued fractions)? $\endgroup$ – ReverseFlow Jun 14 '14 at 4:26
  • $\begingroup$ @Genomeme According to answers below it seems in $(2)$ it is the standard multiplication in the exponent $\endgroup$ – Arun Kumar Jun 16 '14 at 16:18
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+50
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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.

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  • $\begingroup$ so by mapping it this way, it is not neccesary to run through all the properties? Since it is understood that the mapping already has all the properties? $\endgroup$ – cele Jun 11 '14 at 16:23
  • $\begingroup$ This is a general principle called, transfer of structure, you needn't verify all the properties, but it is a good exercise to see using the definition how properties of $\mathbb{R}$ are transferred to $F$. $\endgroup$ – Omran Kouba Jun 11 '14 at 16:28

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