Proof
The first three parts of the proof suffice to establish that Q(R) is a ring, the last part that it is a field.
(Q(R),+,0) is an additive group.
(Q(R),*,1) is a commutative monoid.
Distributivity.
Each nonzero element of Q(R) has a multiplicative inverse.
R, with
n, n', n'' 
0.
Then, by associativity of + on R,
(t/n + t'/n') + t''/n''
| = |
(tn' + t'n)/nn' + t''/n''
| = |
(tn'n''+ t'nn''+ t''nn')/nn'n''
| = |
t/n + (t'n''+ t''n')/n'n''
| = |
t/n + (t'/n' + t''/n'').
|
We have shown that + is associative on Q(R).
We leave the (easier) verifications that
0/n is the zero element, that -t/n is the inverse of
t/n and that + is commutative to the reader.
R, with
n, n', n'' 0.
Then, by associativity of * on R,
(t/n * t'/n') * t''/n''
| = |
(t * t')/nn'
| = |
(tt't'')/nn'n''
| = |
t/n * (t't'')/n'n''
| = |
t/n * (t'/n' * t''/n'').
|
We have shown that * is associative on Q(R).
We leave the (easier) verifications that * is commutative and
that 1/n is the unit element to the reader.
R, with
n, n', n'' 0.
Then, by distributivity of R,
t/n * ( t'/n' + t''/n'')
| = |
t/n * (t'n'' +
t''n')/n'n''
| = |
(tt'n''+ tt''n')/nn'n''
| = |
(tt'nn''+ tt''nn')/n2n'n''
| = |
tt'/nn' + tt''/nn''
| = |
tt'/nn' + tt''/nn''
| = |
t/n * t'/n' + t/n * t''/n''.
|
We have shown left distributivity.
In view of commutativity of *,
there is no need to prove right distributivity.
This establishes
that a/r is
invertible in Q(R) with inverse
r/a.