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.