The set $Q$ of rational numbers is a field.

Proof

$Q$ 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 $Z$ 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 $Z$ 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: $$(\frac{a_1}{b_1}+\frac{a_2}{b_2})+\frac{a_3}{b_3}=\frac{a_1{b}_2+b_1{a}_2}{b_1{b}_2}+\frac{a_3}{b_3}=\frac{a_1{b}_2{b}_3+b_1{a}_2{b}_3+a_3{b}_1{b}_2}{b_1{b}_2{b}_3}$$ which we can easily check is equal to $\frac{a_1}{b_1}+(\frac{a_2}{b_2}+\frac{a_3}{b_3})$.

  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 $Z$.

• $+$ distributes over $\times$: $$\frac{a}{b}(\frac{x_1}{y_1}+\frac{x_2}{y_2})=\frac{a}{b}\frac{x_1{y}_2+x_2{y}_1}{y_1{y}_2}=\frac{a(x_1{y}_2+x_2{y}_1)}{b{y}_1{y}_2}$$ 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 $Z$.

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 $\frac{a}{b}$ is not $0$ (equivalently, $a\ne 0$), then $\frac{a}{b}$ has inverse $\frac{b}{a}$, which does indeed exist since $a\ne 0$.

This completes the proof.