Let G = {x in Q : 0≤x<1}. Define the operation on G:
a•b = a+b if 0≤a+b<1, a+b-1 if a+b≥1
Prove that (G,*) is an Abelian group.
Attempt: (commutativity was easy). For associativity I got a+b≥1 as the last condition for (a•b)•c but b+c≥1 for a•(b•c). How do I fix this?
For inverses (-a) and a±1 are all out of bound so what else can be used as the inverse (the identity is 0)?
Best Answer
If $a$ is to have an inverse $b$, then either $a+b < 1$ and $a+b = 0$ but that makes $b$ negative (except for $a=0$); OR, $a+b \ge 1$ and $a+b-1 = 0$. This tells you what the inverse $b$ could be - namely, $1-a$. Exception, the inverse of $0$ is itself, not $1$ (which is not in $G$).
For associativity, show that both the LHS and the RHS can be written as $a+b+c$ if $a+b+c <1$, as $a+b+c-1$ if $1 \le a+b+c < 2$ and as $a+b+c-2$ if $a+b+c \ge 2$.