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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 .

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    If the issue here about "well-defined" is what I think it is (that $x\star y\in G$ when $x,y\in G$), then this is more generally refered to as "closure" rather than "well-definedness". The issue of "well-defined function" in group theory usually refers to something else. See some of the comments [here](http://math.stackexchange.com/questions/60781/symmetric-groups-on-sets-with-the-same-number-of-elements-are-isomorphic/60788#60788)2011-09-29

4 Answers 4

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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    =a+(-0)    =a+(-1)0 (Proposition 7.1(4))    =a+0 (Proposition 7.1(1))    =a (group axiom (ii)) 

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.

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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.

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HINT $\ $ The only difficult part is proving associativity. Denote the fractional part of a real by $\rm\: f(r) = r - \lfloor r\rfloor = r\ mod\ 1\:.\:$ Note that $\rm\:x\star y = f(x+y)\:.\:$ We CLAIM $\rm\ f(x+f(y)) = f(x+y)\:$
Proof: let $\rm\:f(y) = r,\ y = r + n,\ n\in \mathbb Z\:.\:$ Then the claim is equivalent to $\rm\:f(x+r) = f(x+r+n)\:$
which is true: adding an integer $\rm\:n\:$ doesn't alter the fractional part. Hence applying claim twice

$\rm\ x\star(y\star z)\ =\ f(x+f(y+z))\ =\ f(x+y+z)\ =\ f(f(x+y)+z)\ =\ (x\star y)\star z $

NOTE $\ $ The same proof yields associativity of integer addition $\rm\:mod\ m\:,\:$ with $\rm\:f(k) = k\ mod\ m\:.\:$ 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" $\rm\:\mathbb R\ mod\ \mathbb Z\:$ of reals modulo $1\:.\:$ That the group laws are preserved in every modular image becomes a triviality from this general standpoint.

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    Thank you.I will make it a point to return to this after I learn quotient groups.2011-09-29
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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.