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text: 'The set $\\mathbb{Q}$ of [4zq rational numbers] is a [481 field].\n\n# Proof\n\n$\\mathbb{Q}$ is a ([3jb commutative]) [3gq 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.\n\n- $+$ is commutative: $\\frac{a}{b} + \\frac{c}{d} = \\frac{ad+bc}{bd}$, which by commutativity of addition and multiplication in $\\mathbb{Z}$ is $\\frac{cb+da}{db} = \\frac{c}{d} + \\frac{a}{b}$\n- $0$ 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 $n$.\n- Every rational has an additive inverse: $\\frac{a}{b}$ has additive inverse $\\frac{-a}{b}$.\n- $+$ is [3h4 associative]: $$\\left(\\frac{a_1}{b_1}+\\frac{a_2}{b_2}\\right)+\\frac{a_3}{b_3} = \\frac{a_1 b_2 + b_1 a_2}{b_1 b_2} + \\frac{a_3}{b_3} = \\frac{a_1 b_2 b_3 + b_1 a_2 b_3 + a_3 b_1 b_2}{b_1 b_2 b_3}$$\nwhich 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]\n- $\\times$ is associative, trivially: $$\\left(\\frac{a_1}{b_1} \\frac{a_2}{b_2}\\right) \\frac{a_3}{b_3} = \\frac{a_1 a_2}{b_1 b_2} \\frac{a_3}{b_3} = \\frac{a_1 a_2 a_3}{b_1 b_2 b_3} = \\frac{a_1}{b_1} \\left(\\frac{a_2 a_3}{b_2 b_3}\\right) = \\frac{a_1}{b_1} \\left(\\frac{a_2}{b_2} \\frac{a_3}{b_3}\\right)$$\n- $\\times$ is commutative, again trivially: $$\\frac{a}{b} \\frac{c}{d} = \\frac{ac}{bd} = \\frac{ca}{db} = \\frac{c}{d} \\frac{a}{b}$$\n- $1$ 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 $\\mathbb{Z}$.\n- $+$ distributes over $\\times$: $$\\frac{a}{b} \\left(\\frac{x_1}{y_1}+\\frac{x_2}{y_2}\\right) = \\frac{a}{b} \\frac{x_1 y_2 + x_2 y_1}{y_1 y_2} = \\frac{a \\left(x_1 y_2 + x_2 y_1\\right)}{b y_1 y_2}$$\nwhile $$\\frac{a}{b} \\frac{x_1}{y_1} + \\frac{a}{b} \\frac{x_2}{y_2} = \\frac{a x_1}{b y_1} + \\frac{a x_2}{b y_2} = \\frac{a x_1 b y_2 + b y_1 a x_2}{b^2 y_1 y_2} = \\frac{a x_1 y_2 + a y_1 x_2}{b y_1 y_2}$$\nso we are done by distributivity of $+$ over $\\times$ in $\\mathbb{Z}$.\n\nSo 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.\nBut 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 $a \\not = 0$.\n\nThis completes the proof.',
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