The rationals form a field

Written by Patrick Stevens, et al. last updated
Requires: Algebraic field

The set of rational numbers is a field.

Proof

is a (commutative) ring with additive identity (which we will write as for short) and multiplicative identity (which we will write as for short): we check the axioms individually.

  • is commutative: , which by commutativity of addition and multiplication in is
  • is an identity for : have , which is because is a multiplicative identity in and for every integer .
  • Every rational has an additive inverse: has additive inverse .
  • is associative: which we can easily check is equal to .
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  • is associative, trivially:
  • is commutative, again trivially:
  • is an identity for : by the fact that is an identity for in .
  • distributes over : while so we are done by distributivity of over in .

So far we have shown that is a ring; to show that it is a field, we need all nonzero fractions to have inverses under multiplication. But if is not (equivalently, ), then has inverse , which does indeed exist since .

This completes the proof.