The set Q of rational numbers is a field.
Proof
Q is a (commutative) ring with additive identity 01 (which we will write as 0 for short) and multiplicative identity 11 (which we will write as 1 for short): we check the axioms individually.
- + is commutative: ab+cd=ad+bcbd, which by commutativity of addition and multiplication in Z is cb+dadb=cd+ab
- 0 is an identity for +: have ab+0=ab+01=a×1+0×bb×1, which is ab because 1 is a multiplicative identity in Z and 0×n=0 for every integer n.
- Every rational has an additive inverse: ab has additive inverse −ab.
- + is associative: (a1b1+a2b2)+a3b3=a1b2+b1a2b1b2+a3b3=a1b2b3+b1a2b3+a3b1b2b1b2b3
which we can easily check is equal to a1b1+(a2b2+a3b3).
actually do this
- × is associative, trivially: (a1b1a2b2)a3b3=a1a2b1b2a3b3=a1a2a3b1b2b3=a1b1(a2a3b2b3)=a1b1(a2b2a3b3)
- × is commutative, again trivially: abcd=acbd=cadb=cdab
- 1 is an identity for ×: ab×1=ab×11=a×1b×1=ab by the fact that 1 is an identity for × in Z.
- + distributes over ×: ab(x1y1+x2y2)=abx1y2+x2y1y1y2=a(x1y2+x2y1)by1y2
while abx1y1+abx2y2=ax1by1+ax2by2=ax1by2+by1ax2b2y1y2=ax1y2+ay1x2by1y2
so we are done by distributivity of + over × in Z.
So far we have shown that Q is a ring; to show that it is a field, we need all nonzero fractions to have inverses under multiplication.
But if ab is not 0 (equivalently, a≠0), then ab has inverse ba, which does indeed exist since a≠0.
This completes the proof.