# The rationals form a field

https://arbital.com/p/rationals_are_a_field

by Patrick Stevens Jul 1 2016 updated Jul 6 2016

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.