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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)?

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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 $a+b+c \ge 2$.

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If you know what a quotient group is then it is much easier to prove that $G$ is a group: the additive group of the integers $\mathbf Z$ is a subgroup of the additive group of the rational numbers $\mathbf Q$. Therefore, one has something as the quotient group $\mathbf Q/\mathbf Z$. One thinks of this group as the additive group of the rational numbers modulo $1$. Anyway, the map $$ f\colon G\rightarrow \mathbf Q/\mathbf Z, $$ defined by letting $f(x)$ be the left coset $x+\mathbf Z$, is a bijection, and satisfies $f(x *y)=f(x)+f(y)$. Since addition $+$ on $\mathbf Q/\mathbf Z$ is a group law, the corresponding operation $*$ on $G$ is one too.

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