数学 – 代数学 - 整数 - 同値関係、合同式(剰余(mod))

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代数系入門 (松坂和夫(著)、岩波書店)の第1章(整数)、6(同値関係、合同式)、問題1.を取り組んでみる。

問題1.

\begin{array}{l} a - b = m l \\ a - (b + m k) \\ = (a - b) - m k \\ = m l - m k \\ = m (l - k) \\ a \equiv b + m k (\mod 5) \end{array}

問題2

\begin{array}{l} a c - b c = m l \\ (a - b) c = m l \\ (c, m) = 1 \\ a - b = m k \\ a \equiv b (\mod m) \end{array}

問題3

\begin{array}{l} a k - b k \\ = (a - b) k \\ = m l k \\ = l m k \\ a k \equiv b k (\mod m k) \end{array}

問題4

\begin{array}{l} a - b = m k \\ a_{1} d - b_{1} d = m_{1} d k \\ a_{1} - b_{1} = m_{1} k \\ a_{1} \equiv b_{1} (\mod m_{1}) \end{array}

問題5

\begin{array}{l} a - b = m k = m' d k = m' k d \\ a \equiv b (\mod d) \end{array}

問題6

\begin{array}{l} a \equiv b (\mod m m'_{1}), a \equiv b (\mod m m'_{k}) \\ a \equiv b (\mod m) \end{array}

問題7

\begin{array}{l} {(x + y)}^{p} - (x^{p} + y^{p}) \\ = \sum_{r = 0}^{p} (\begin{matrix} p \\ r \end{matrix}) x^{n - r} y^{r} - (x^{p} + y^{p}) \\ = x^{p} + y^{p} + \sum_{r = 1}^{p - 1} (\begin{matrix} p \\ r \end{matrix}) x^{n - r} y^{r} - (x^{p} + y^{p}) \\ = \sum_{r = 1}^{p - 1} (\begin{matrix} p \\ r \end{matrix}) x^{n - r} y^{r} \\ = p k \\ {(x + y)}^{p} \equiv x^{p} + y^{p} (\mod p) \end{array}

問題8

\begin{array}{l} {(x_{1} + \cdot \cdot \cdot + x_{n} + x_{n + 1})}^{p} \\ = {((x_{1} + \cdot \cdot \cdot + x_{n}) + x_{n + 1})}^{p} \\ \equiv {(x_{1} + \cdot \cdot \cdot + x_{n})}^{p} + x_{n + 1}^{p} \\ \equiv x_{1}^{p} + \cdot \cdot \cdot + x_{n}^{p} + x_{n + 1}^{p} (\mod p) \end{array}

(帰納法)

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