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 \(\frac{a_1}{b_1}+\left(\frac{a_2}{b_2}+\frac{a_3}{b_3}\right)\). 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.