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

Proof

$~$\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.

• $~$+$~$ is commutative: $~$\frac{a}{b} + \frac{c}{d} = \frac{ad+bc}{bd}$~$, which by commutativity of addition and multiplication in $~$\mathbb{Z}$~$ is $~$\frac{cb+da}{db} = \frac{c}{d} + \frac{a}{b}$~$
• $~$0$~$ is an identity for $~$+$~$: have $~$\frac{a}{b}+0 = \frac{a}{b} + \frac{0}{1} = \frac{a \times 1 + 0 \times b}{b \times 1}$~$, which is $~$\frac{a}{b}$~$ because $~$1$~$ is a multiplicative identity in $~$\mathbb{Z}$~$ and $~$0 \times n = 0$~$ for every integer $~$n$~$.
• Every rational has an additive inverse: $~$\frac{a}{b}$~$ has additive inverse $~$\frac{-a}{b}$~$.
• $~$+$~$ is associative: $$~$\left(\frac{a_1}{b_1}+\frac{a_2}{b_2}\right)+\frac{a_3}{b_3} = \frac{a_1 b_2 + b_1 a_2}{b_1 b_2} + \frac{a_3}{b_3} = \frac{a_1 b_2 b_3 + b_1 a_2 b_3 + a_3 b_1 b_2}{b_1 b_2 b_3}$~$$ which we can easily check is equal to $~$\frac{a_1}{b_1}+\left(\frac{a_2}{b_2}+\frac{a_3}{b_3}\right)$~$. [todo: actually do this]
• $~$\times$~$ is associative, trivially: $$~$\left(\frac{a_1}{b_1} \frac{a_2}{b_2}\right) \frac{a_3}{b_3} = \frac{a_1 a_2}{b_1 b_2} \frac{a_3}{b_3} = \frac{a_1 a_2 a_3}{b_1 b_2 b_3} = \frac{a_1}{b_1} \left(\frac{a_2 a_3}{b_2 b_3}\right) = \frac{a_1}{b_1} \left(\frac{a_2}{b_2} \frac{a_3}{b_3}\right)$~$$
• $~$\times$~$ is commutative, again trivially: $$~$\frac{a}{b} \frac{c}{d} = \frac{ac}{bd} = \frac{ca}{db} = \frac{c}{d} \frac{a}{b}$~$$
• $~$1$~$ is an identity for $~$\times$~$: $$~$\frac{a}{b} \times 1 = \frac{a}{b} \times \frac{1}{1} = \frac{a \times 1}{b \times 1} = \frac{a}{b}$~$$ by the fact that $~$1$~$ is an identity for $~$\times$~$ in $~$\mathbb{Z}$~$.
• $~$+$~$ distributes over $~$\times$~$: $$~$\frac{a}{b} \left(\frac{x_1}{y_1}+\frac{x_2}{y_2}\right) = \frac{a}{b} \frac{x_1 y_2 + x_2 y_1}{y_1 y_2} = \frac{a \left(x_1 y_2 + x_2 y_1\right)}{b y_1 y_2}$~$$ while $$~$\frac{a}{b} \frac{x_1}{y_1} + \frac{a}{b} \frac{x_2}{y_2} = \frac{a x_1}{b y_1} + \frac{a x_2}{b y_2} = \frac{a x_1 b y_2 + b y_1 a x_2}{b^2 y_1 y_2} = \frac{a x_1 y_2 + a y_1 x_2}{b y_1 y_2}$~$$ so we are done by distributivity of $~$+$~$ over $~$\times$~$ in $~$\mathbb{Z}$~$.

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.