# The rationals form a field

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.

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