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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