The rationals form a field

by Patrick Stevens Jul 1 2016 updated Jul 6 2016

The set $~$\mathbb{Q}$~$ of rational numbers is a field.


$~$\mathbb{Q}$~$ is a (commutative) ring with additive identity $~$\frac{0}{1}$~$ (which we will write as $~$0$~$ for short) and multiplicative identity $~$\frac{1}{1}$~$ (which we will write as $~$1$~$ for short): we check the axioms individually.

So far we have shown that $~$\mathbb{Q}$~$ is a ring; to show that it is a field, we need all nonzero fractions to have inverses under multiplication. But if $~$\frac{a}{b}$~$ is not $~$0$~$ (equivalently, $~$a \not = 0$~$), then $~$\frac{a}{b}$~$ has inverse $~$\frac{b}{a}$~$, which does indeed exist since $~$a \not = 0$~$.

This completes the proof.