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 .
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.
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.
Generally ring $\rm\color{#c00}{identities}$ (e.g. associative, commutative, distributive laws) that are true in $\,\Bbb Z\,$ are preserved in $\,\Bbb Z_m = \Bbb Z\bmod m\,$ simply because the map $\,h\,:\, a\mapsto a_m\,$ is a surjective (onto) ring homomorphism. Similarly for other quotient structures such as the above circle group. Follow the link for details.
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.
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.
