TITLE: Circle group addition of fractional parts is well-defined
QUESTION [6 upvotes]: 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 .

REPLY [2 votes]: 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.