I would like to define a function $\oplus$ on $\mathbb Q$ ($\oplus:\mathbb Q^2\mapsto \mathbb Q$), such that (for all $a,b,c$):
- (commutativity) $a\oplus b=b\oplus a$
- (associativity) $(a\oplus b)+c=a\oplus (b\oplus c)$
- (positivity) $a\oplus b>a$ with $>$ the usual order on $\mathbb Q$
- (growing) $b>c \Rightarrow a\oplus b>a\oplus c$
Is it possible (For now, I'm sure it is) and how can I give an example of such a function ?
I found a simple solution. The idea came from this question (and the answer from Did).
Let $f(x)=\left\{\begin{array}{cc}\frac{1}{1-x}&\mbox{if } x\le0\\ x+1&\mbox{if }x\ge 0\\\end{array}\right.$
$f$ is a bijection from $\mathbb Q$ to $\mathbb Q^+$
Let $a\oplus b=f^{-1}\left(f(a)+f(b)\right)$.
Commutativity and associativity are obvious. The other properties come from the fact that $f$ is an increasing function.