Circle group addition of fractional parts is well-defined Let $G=\{x \in \mathbb{R}\mid 0 \leq x < 1 \}$ and for $x,y \in G$ let $x\star y$ be the fractional part of $x+y$ (i.e $x\star y=x+y-\lfloor x+y \rfloor$ where $\lfloor a \rfloor $ is the greatest integer less than or equal to a). Then, how do I prove that $\star$ is a well defined binary operation on $G$ and that $G$ is an abelian group under $\star$?
Thank you.I have just started group theory.My progress on this is minimal and next to 0.
Edited .
 A: There is no issue of "well-definedness": the elements of $G$ are real numbers, and for every real number $r$, $r-\lfloor r\rfloor$ is a real number that lies in $[0,1)$; thus, $\star$ is a function from $\mathbb{R}\times\mathbb{R}$ to $[0,1)$, and hence by restriction $\star$ is a function from $G\times G$ to $[0,1)\subseteq G$.
(Note: The next paragraph was written when the definition of $G$ was that $G$ included $1$; this has since been changed.)
However, as given, $G$ is not a group under the operation: note that if $(G,\cdot)$ is a group, then in particular for every $g\in G$ there exist $x,y\in G$ with $x\cdot y = g$. However, there are no elements of your $G$ which satisfy $x\star y = 1$, even though $1\in G$.
If, however, you change $G$ to be $G=\{x\in\mathbb{R}\mid 0\leq x\lt 1\}$, then the set is indeed a group (It is, in fact, isomorphic to $\mathbb{R}/\mathbb{Z}$). Note that $0$ is a two-sided identity, since if $x\in G$ then $\lfloor x\rfloor = 0$; and that if $x\neq 0$ is in $G$, then $1-x\in G$ and $x\star(1-x) = 0$. And trivially, since addition of reals is commutative, $x\star y = y\star x$. 
So the only thing that needs to be proven is associativity. 
The key is to note that for all real numbers $r$ and all integers $n$,
$$\lfloor r-n\rfloor = \lfloor r\rfloor - n.$$ 
So, if $x,y,z\in [0,1)$, then:
$$\begin{align*}
(x\star y)\star z &= \Bigl( x+y - \lfloor x+y\rfloor\Bigr)\star z\\
&= \Bigl( x+y-\lfloor x+y\rfloor + z\Bigr) - \lfloor x+y - \lfloor x+y\rfloor + z\rfloor\\
&= x+y+z - \lfloor x+y\rfloor -\Bigl( \lfloor x+y+z\rfloor - \lfloor x+y\rfloor\Bigr)\\
&= x+y+z - \lfloor x+y+z\rfloor;\\
x\star(y\star z) &= x\star\Bigl( y+z - \lfloor y+z\rfloor\Bigr)\\
&= x + y + z - \lfloor y+z\rfloor - \lfloor x+y+z-\lfloor y+z\rfloor\rfloor\\
&= x+y+z - \lfloor y+z\rfloor - \bigl( \lfloor x+y+z\rfloor - \lfloor y+z\rfloor\bigr)\\
&= x+y+z - \lfloor x+y+z\rfloor\\
&= (x\star y)\star z.
\end{align*}$$
You can also go the longer route and consider the possibilities of $x+y\lt 1$, $y+z\lt 1$, $x+y+z\lt 1$; or $x+y\lt 1$, $1\leq y+z\lt 2$ and $1\leq x+y+z\lt 2$;  $x+y\geq 1$, $y+z\geq 1$, and $1\leq x+y+z\lt 2$; and $x+y\geq 1$, $y+z\geq 1$, and $2\leq x+y+z\lt 3$. But the observation above is much easier.
A: Hint $ $ The only difficult part is proving associativity. Denote the fractional part of a real by $\, f(r) = r - \lfloor r\rfloor = r\bmod 1.\,$ Note that $\,x\star y = f(x+y).\,$ CLAIM: $\ f(x+f(y)) = f(x+y)\,$
Proof: let $\,f(y) = r,\ y = r + n,\ n\in \mathbb Z.\,$ The claim is equivalent to $\,f(x+r) = f(x+r+n)\,$
which is true: adding an integer $\,n\,$ doesn't alter the fractional part. Hence applying claim twice
$$ x\star(y\star z)\ =\ f(x+f(y+z))\ =\ f(x+y+z)\ =\ f(f(x+y)+z)\ =\ (x\star y)\star z\qquad $$
Note $ $ The same proof yields associativity of integer addition $\!\bmod m,\,$ with $\,f(k) = k\bmod m.\,$ The essence of the matter will be much clearer after you learn about quotient groups. Then you will see that this group is simply the real circle group $\,\Bbb R/\Bbb Z = \Bbb R\bmod\Bbb Z = $ reals modulo $1\,.\,$ That the group laws are preserved in every modular image becomes a triviality from this general standpoint.
A: I'm expanding Bill Dubuque's note:
Forget about fractional parts for the moment. Call two real numbers $x$, $y$ equivalent if $y-x\in{\mathbb Z}$. Denote the equivalence class of $x\in{\mathbb R}$ by $[x]$ and the set of all equivalence classes by $F$. The definition
$$[x]+[y]\ :=\ [x+y]$$
defines addition uniquely on $F$ (check this!), $[0]$ is the neutral element, and the addition inherits from ${\mathbb R}$ the familiar properties: commutativity, associativity and existence of (additive) inverses. Therefore $F$ is an abelian group.
Now each equivalence class $[x]\in F$ contains exactly one element of your set $G:=[0,1[\ \subset{\mathbb R}$, namely the (well defined!) number $x-\lfloor x\rfloor$. It then follows that $G$ is a complete set of representatives for $F$. So instead of dealing with classes you can just add their representatives, and that's what has been set up in your problem.
A: If in doubt about what well defined means, read the last sentence of page 1 of the textbook.
Proof:
Let a,b,c∈G.
When 0≤a+b<1, then ⌊a+b⌋=0 (see  “Identity” section for proof that -0=0). When 1≤a+b<2, then ⌊a+b⌋=1. Thus G is closed under ⋆.
The mapping a⋆b ∶ G×G → G is defined for all a,b∈G because G is closed under ⋆.
Define a binary relation ~ on G as follows:
a ~ b if and only if a⋆b=a+b-⌊a+b⌋ (i.e., (a,b)∈G×G)

Reflexive:
(a,a)∈G×G because a⋆a is defined for all a∈G.

Symmetric:
(a,b)∈G×G⇒a⋆b=a+b-⌊a+b⌋
            =b+a-⌊b+a⌋ (ring axiom (i))
            =b⋆a
         ⇒(b,a)∈G×G

Transitive:
 (a,b),(b,c)∈G×G
⇒a⋆b=a+b+(-⌊a+b⌋)
    =b+a+(-⌊b+a⌋) (ring axiom (i))
    =b+(a+(-⌊b+a⌋)) (Proposition 1.1(5))

  and

  b⋆c=b+c+(-⌊b+c⌋)
     =b+(c+(-⌊b+c⌋)) (Proposition 1.1(5))
⇒a⋆b+(-(a+(-⌊b+a⌋)))=b+(a+(-⌊b+a⌋))+(-(a+(-⌊b+a⌋)))
                    =b+0 (group axiom (ii)) 

 and

 b⋆c+(-(c+(-⌊b+c⌋)))=b+(c+(-⌊b+c⌋))+(-(c+(-⌊b+c⌋)))
                    =b+0 (group axiom (ii))
⇒b⋆c+(-(c-⌊b+c⌋))=a⋆b+(-(a+(-⌊b+a⌋)))
⇒c⋆b+(-(c-⌊b+c⌋))=a⋆b+(-(a+(-⌊b+a⌋))) (⋆ is symmetric)
⇒c=a
⇒2c-⌊a+c⌋=a+c-⌊a+c⌋
         =a⋆c
⇒(a,c)∈G×G

Therefore since ⋆ is reflexive, symmetric, and transitive, ⋆ is a binary equivalence relation. By Proposition 2(1), the set of equivalence classes of ⋆ form a partition of G×G. This shows that ⋆ is well defined.
Since a⋆b is defined for all a,b∈G, ⋆ is well defined, and G is closed under ⋆, thus ⋆ is a well defined binary operation.
Associativity:
(a⋆b)⋆c=(a+b-⌊a+b⌋)+c-⌊(a+b-⌊a+b⌋)+c⌋
       =a+(b+(-⌊a+b⌋)+c)-⌊a+(b+(-⌊a+b⌋)+c)⌋ (group axiom (i))
       =a+(b+c+(-⌊a+b⌋))-⌊a+(b+c+(-⌊a+b⌋))⌋ (ring axiom (i))
       =a+(b+c-⌊a+b⌋)-⌊a+(b+c-⌊a+b⌋)⌋
       =a+(b+c-⌊c+b⌋)-⌊a+(b+c-⌊c+b⌋)⌋ (a=c by transitivity of ⋆ because a⋆b and b⋆c)
       =a+(b+c-⌊b+c⌋)-⌊a+(b+c-⌊b+c⌋)⌋  (ring axiom (i))
       =a⋆(b⋆c)

Identity:
0⋆a=a⋆0 (see “Symmetric”)
   =a+0-⌊a+0⌋
   =a-⌊a⌋ (group axiom (ii))
   =a-0 or =a-1
   =a+(-0) or =a-1
   =a+(-1)0 (Proposition 7.1(4)) or =a-1
 ⇒a=a-1 (Proposition 7.1(1))                              (1)

Thus, subracting a from both sides of (1) yields 0=-1. Substituting 0=-1 into equation (1) gives that a=a and therefore 0⋆a=a⋆0=a as desired.
Inverses:
Let d∈G-{0} and note that 1-d∈G-{0} for all d∈G-{0}. The “Identity” section shows that the inverse of 0 is 0.
(1-d)⋆d=d⋆(1-d) (see “Symmetric”)
       =d+(1-d)-⌊d+(1-d)⌋
       =d+(1+(-d))+(-⌊d+(1+(-d))⌋)
       =d+((-d)+1)+(-⌊d+((-d)+1)⌋) (ring axiom (i))
       =d+(-d)+1+(-⌊d+(-d)+1⌋)  (Proposition 1.1(5))       
       =0+1+(-⌊0+1⌋) (group axiom (iii))
       =1+(-⌊1⌋) (group axiom (ii))
       =1+(-1) 
       =0 (group axiom (iii))

Abelian:
See “Symmetric.”
Hence (G,⋆) is an abelian group.
