整数剰余類と合同式法、合同、和、差、積

学習環境

親切な代数学演習新装2版―整数・群・環・体 (加藤明史(著)、現代数学社)の第Ⅰ部(整数)、第3章(剰余類と合同式)の問8の解答を求めてみる。

以下法はm。

\begin{array}{l} a - b \equiv 0 \\ c - d \equiv 0 \end{array}

なので

\begin{array}{l} a - b = m k_{0} \\ c - d = m k_{1} \end{array}

よって

\begin{array}{l} (a - b) + (c - d) \\ = m k_{0} + m k_{1} \\ = m (k_{0} + k_{1}) \end{array}

ゆえに、

\begin{array}{l} (a - b) + (c - d) \equiv 0 \\ (a + c) - (b + d) \equiv 0 \\ a + c \equiv b + d \end{array}

\begin{array}{l} (a - b) - (c - d) \\ = m k_{0} - m k_{1} \\ = m (k_{0} - k_{1}) \end{array}

よって、

\begin{array}{l} (a - b) - (c - d) \equiv 0 \\ a - b \equiv c - d \end{array}

\begin{array}{l} (a - b) c + (c - d) b = m k_{0} c + m k_{1} b \\ a c - b c + b c - b d = m (k_{0} c + k_{1} b) \\ a c - b d = m k_{2} \end{array}

よって、

\begin{array}{l} a c - b d \equiv 0 \\ a c \equiv b d \end{array}