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.