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: $\frac{a}{b} + \frac{c}{d} = \frac{a d + b c}{b d}$ , which by commutativity of addition and multiplication in $Z$ is $\frac{c b + d a}{d b} = \frac{c}{d} + \frac{a}{b}$

$0$ 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 $\frac{a_{1}}{b_{1}} + (\frac{a_{2}}{b_{2}} + \frac{a_{3}}{b_{3}})$ .

$\times$ is associative, trivially: $(\frac{a_{1}}{b_{1}} \frac{a_{2}}{b_{2}}) \frac{a_{3}}{b_{3}} = \frac{a_{1} a_{2}}{b_{1} b_{2}} \frac{a_{3}}{b_{3}} = \frac{a_{1} a_{2} a_{3}}{b_{1} b_{2} b_{3}} = \frac{a_{1}}{b_{1}} (\frac{a_{2} a_{3}}{b_{2} b_{3}}) = \frac{a_{1}}{b_{1}} (\frac{a_{2}}{b_{2}} \frac{a_{3}}{b_{3}})$

$\times$ is commutative, again trivially: $\frac{a}{b} \frac{c}{d} = \frac{a c}{b d} = \frac{c a}{d b} = \frac{c}{d} \frac{a}{b}$

$1$ 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, $a \neq 0$ ), then $\frac{a}{b}$ has inverse $\frac{b}{a}$ , which does indeed exist since $a \neq 0$ .

This completes the proof.

Requires: Algebraic field

The set of rational numbers is a field.

Proof

$+$ is commutative: $\frac{a}{b} + \frac{c}{d} = \frac{a d + b c}{b d}$ , which by commutativity of addition and multiplication in $Z$ is $\frac{c b + d a}{d b} = \frac{c}{d} + \frac{a}{b}$

$0$ 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 $\frac{a_{1}}{b_{1}} + (\frac{a_{2}}{b_{2}} + \frac{a_{3}}{b_{3}})$ .

$\times$ is associative, trivially: $(\frac{a_{1}}{b_{1}} \frac{a_{2}}{b_{2}}) \frac{a_{3}}{b_{3}} = \frac{a_{1} a_{2}}{b_{1} b_{2}} \frac{a_{3}}{b_{3}} = \frac{a_{1} a_{2} a_{3}}{b_{1} b_{2} b_{3}} = \frac{a_{1}}{b_{1}} (\frac{a_{2} a_{3}}{b_{2} b_{3}}) = \frac{a_{1}}{b_{1}} (\frac{a_{2}}{b_{2}} \frac{a_{3}}{b_{3}})$

$\times$ is commutative, again trivially: $\frac{a}{b} \frac{c}{d} = \frac{a c}{b d} = \frac{c a}{d b} = \frac{c}{d} \frac{a}{b}$

$1$ 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 $Z$ .

This completes the proof.

LESSWRONG
LW

The rationals form a field

Proof

LESSWRONG
LW

The rationals form a field

Proof