$S := \{ (a, b) \in \mathbb{Z}^{2}\,|\,b \neq 0\}$
$S$ に同値関係$\sim$を, $(a, b)\sim(c, d) \underset{def.}{\Leftrightarrow} ad - bc = 0$ で定める.
$\mathbb{Q} := S/\sim$
$(a, b)$ の同値類のことを,
$\begin{eqnarray*} \dfrac{a}{b} \nonumber \nonumber \end{eqnarray*}$ と書く.
$\begin{eqnarray*} \dfrac {a}{b} + \dfrac {c}{d} & := & \dfrac {ad + bc}{bd} \nonumber \\ \dfrac {a}{b} \times \dfrac {c}{d} & := & \dfrac {ac}{bd} \nonumber \end{eqnarray*}$
と定める.
well-defined を言う.
$\begin{eqnarray*} \dfrac{a}{b} = \dfrac{a^{'}}{b^{'}} \nonumber \\ \dfrac{c}{d} = \dfrac{c^{'}}{d^{'}} \nonumber \end{eqnarray*}$
とすると,
$\begin{eqnarray*} {a}{b^{'}} - {a^{'}}{b} = 0 \nonumber \\ {c}{d^{'}} - {c^{'}}{d} = 0 \nonumber \end{eqnarray*}$
$\begin{eqnarray*} \dfrac{ad+bc}{bd} = \dfrac{a^{'}d^{'}+b^{'}c^{'}}{b^{'}d^{'}} \nonumber \end{eqnarray*}$
$\begin{eqnarray*} (ad + bc)b^{'}d^{'} - bd(a^{'}d^{'} + b^{'}c^{'}) & = & adb^{'}d^{'} + bcb^{'}d^{'} - bda^{'}d^{'} - bdb^{'}c^{'} \nonumber \\ & = & dd^{'}(ab^{'} - a^{'}b) + bb^{'}(cd^{'} - c^{'}d) \nonumber \\ & = & 0 \nonumber \end{eqnarray*}$
$\begin{eqnarray*} \dfrac{ac}{bd} = \dfrac{a^{'}c^{'}}{b^{'}d^{'}} \nonumber \end{eqnarray*}$
$\begin{eqnarray*} {ac}{b^{'}d^{'}} - {a^{'}c^{'}}{bd} & = & {ab^{'}}{cd^{'}} - {a^{'}b}{c^{'}d} \nonumber \\ & = & {ab^{'}}{c^{'}d} - {ab^{'}}{c^{'}d} \nonumber \\ & = & 0 \nonumber \end{eqnarray*}$
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