Prove whether $([0,1),*)$ is group or not, where $x*y=\begin{cases}x+y &: x+y<1\\ x+y-1 &:x+y \ge 1.\end{cases}$ Problem. Given a nonempty set $G=\lbrace x \in \mathbb{R} | 0 \leq x < 1\rbrace$. Let * be a binary operation on $G$ that defined below.
(i) $x*y = x+y,$ if $x+y<1$.
(ii) $x*y=x+y-1,$ if $x+y \geq 1$.
Is $G$ a group?
Here my solution so far. Please correct it clearly.
(i) If $x+y<1$, then $x*y=x+y$.
We will show that $*$ is assosiative. Let $a,b,c \in G$. We'll show that $(a*b)*c=a*(b*c)$. Now,
$(a*b)*c=(a+b)*c = (a+b)+c=a+(b+c)=a*(b+c)=a*(b*c)$ Hence, $*$ is assosiative.
Now, $0 \in G$. Let $a \in G$, then
$0*a=0+a=a=a+0=a*0$. Hence, $0$ is an identity element.
Now, let $a,m \in G$ and $m$ be an inverse of $a$. Then,
$m*a=m+a=0 \Rightarrow m=-a$. But, $-a \notin G$. So, $a$ has no inverse.
Hence, $(G,*)$ is not group.
(ii) If $x+y \geq 1$, then $x*y=x+y-1$.
It's easy to prove that $*$ assosiative. Now, let $a,i \in G$ and $i$ be the inverse of $a$. Then, $i*a=i+a-1=a=a+i-1=a*i$. Its follows that $i=1$. But, $1 \notin G$. So, there's no an identity element of $a$.
It's follows that there's no inverse for $a$.
Hence, $(G*)$ is not group.
 A: First, it's worth verifying that * is well-defined: because $0 \leq x, y  < 1$, either 
$0 \leq x+y < 1$ 
or $1 \leq x+y < 2$ which means $0 \leq x+y-1 < 1$
so either way $0 \leq x*y < 1$ ensuring $x*y \in G$.
Existence of an identity and inverses are both straightforward: 
[identity] If $x \in G$ then $0 \leq 0+x = x <1$ so $0 * x = x*0 = x$. 
[inverse] If $x \in G$ then either $x=0$ and $0*0=0$ or... 
$$0 < x <1$$
$$0 < 1-x <1$$
$$x+(1-x) = x+1-x = 1$$
$$x*(1-x) = x+(1-x)-1 =0$$
So every element has an inverse.
Finally for associativity, note that $x*y = x+y+n$ for some $n \in \lbrace 0,-1 \rbrace$.  Using the associativity (and commutativity) of addition:$$(a*b)*c = (a+b+n_1)*c = a+b+n_1+c+n_2 = a+b+c+(n_1+n_2) \space [A]$$
$$a*(b*c) = a*(b+c+n_3) = a+(b+c+n_3)+n_4 = a+b+c+(n_3+n_4) \space [B]$$
where each $n_i$ is an integer.
Because of what I set out at the top of this answer, we know that both [A] and [B] are in G, so $$0 \leq a+b+c+(n_1+n_2) < 1$$ and $$0 \leq a+b+c+(n_3+n_4) < 1.$$  Writing the latter as $$-1 < -a-b-c-(n_3+n_4) \leq 0$$ and adding $$-1 < (n_1+n_2) - (n_3+n_4) <1.$$ But as $(n_1+n_2) - (n_3+n_4)$ is an integer its only possible value is $0$.  This means $$(n_1+n_2) = (n_3+n_4)$$ so $$ a+b+c+(n_1+n_2) = a+b+c+(n_3+n_4)$$ so $$(a*b)*c = a*(b*c)$$ as required.
In conclusion, $(G,*)$ is a group - the mistake in your solution was to consider separately the cases of $x+y$ being less than or greater than 1, as if there would be an identity $e_1$ satisfying $e_1 + y < 1$ and a different identity $e_2$ satisfying $e_2+y \geq 1$.  That's bound to fail because there is only one identity, namely $0$. 
A: Hint. Define $\{x\} := x - \lfloor x\rfloor$, $x\in\mathbb R$, where $\lfloor \cdot\rfloor$ is the floor function. Prove that $\{x+y\} = \{x\}*\{y\}$. Observe that for $x\in [0,1)$ we have $\{x\} = x$.


*

*$(x*y)*z = (\{x\}*\{y\})*\{z\} = \{x+y\}*\{z\} = \{(x+y)+z\}=\ldots =x*(y*z),$

*$x*0 = \{x\}*\{0\} = \{x+0\} = \{x\} = x$, $0*x=\{0\}*\{x\}=\ldots = x$,

*If $x = 0$, then $0*0 = 0$. If $x\neq 0$, then $0 = \{0\} = \{x +(-x)\} = \{x\} * \{-x\} = \{x\}*\{1-x\} = x*(1-x)$,

*$x*y = \{x\}*\{y\} = \{x+y\} = \{y+x\} =  \{y\}*\{x\} = y*x.$
A: Hint. Consider $\mathbb{R}$ under addition (clearly a group), and then quotient out by the integers to get $\mathbb{R/Z}$. Again, this is a group, and indeed is the group you are looking at here. So focus on this quotient group and work out what the operation should be, and prove that it is the operation you have here.
