# 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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