【数学I】例題2.1.6：集合の包含関係・相等の証明（One More）★★★★

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% 例題I2.1.6：集合の包含関係・相等の証明（One More）★★★★
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$\mathbb{Z}$を整数全体の集合とするとき，次のことを証明せよ．

(1) $A=\{6x+1\mid x\in\mathbb{Z}\},B=\{3x+1\mid x\in\mathbb{Z}\}$であるとき，$A\subset B$かつ$A\neq B$

(2) $A=\{4x-3y\mid x\in\mathbb{Z},y\in\mathbb{Z}\},B=\{5x+2y\mid x\in\mathbb{Z},y\in\mathbb{Z}\}$であるとき，$A=B$

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% 例題I2.1.6の解答（One More）★★★★
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(1) $\alpha\in A$とすると，$\alpha=6x+1$ $(x\in\mathbb{Z})$と表すことができる．

このとき，$\alpha=3\times 2x+1$と表されるので，$2x=y$とおくと，$\alpha=3y+1(y\in\mathbb{Z})$

したがって，$\alpha\in B$

よって，$\alpha\in A$ならば，$\alpha\in B$が成り立つから，$A\subset B$

また，$4\in B$であるが$4\notin A$であることから，$A\neq B\blacksquare$

(2)

(i) $\alpha\in A$とすると，$\alpha=4x-3y(x\in\mathbb{Z},y\in\mathbb{Z})$と表すことができる．

$4=5\times 0+2\times 2,3=5\times 1+2\times(-1)$より，

\[
\alpha=(5\times 0+2\times 2)x-\{5\times 1+2\times(-1)\}y=5\times(-y)+2(2x+y)
\]

$x\in\mathbb{Z},y\in\mathbb{Z}$より，$-y\in\mathbb{Z},2x+y\in\mathbb{Z}$であるから，$\alpha\in B$

したがって，$A\subset B$が成り立つ．

(ii) $\beta\in B$とすると，$\beta=5x+2y(x\in\mathbb{Z},y\in\mathbb{Z})$と表すことができる．

$5=4\times 2+3\times(-1),2=4\times(-1)+3\times 2$より，

\[
\beta=\{4\times 2+3\times(-1)\}x+\{4\times(-1)+3\times 2\}y=4(2x-y)-3(x-2y)
\]

$x\in\mathbb{Z},y\in\mathbb{Z}$より，$2x-y\in\mathbb{Z},x-2y\in\mathbb{Z}$であるから，$\beta\in A$

したがって，$B\subset A$が成り立つ．

よって，(i)，(ii)より，$A\subset B$かつ$B\subset A$であるから，$A=B$が成り立つ．$\blacksquare$

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% 問題I2.1.6：集合の包含関係・相等の証明（One More）★★★★
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$\mathbb{Z}$を整数全体の集合とするとき，次のことを証明せよ．

(1) $A=\{6x+5\mid x\in\mathbb{Z}\},B=\{3x-1\mid x\in\mathbb{Z}\}$であるとき，$A\subset B$

(2) $A=\{5x+2y\mid x\in\mathbb{Z},y\in\mathbb{Z}\},B=\{4x+3y\mid x\in\mathbb{Z},y\in\mathbb{Z}\}$であるとき，$A=B$

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(1) $\alpha\in A$とすると，$\alpha=6x+5$ $(x\in\mathbb{Z})$と表すことができる．

このとき，$\alpha=6(x+1)-1=3\cdot 2(x+1)-1$

$2(x+1)=y$とおくと，$\alpha=3y-1(y\in\mathbb{Z})$

したがって，$\alpha\in B$

よって，$\alpha\in A$ならば$\alpha\in B$が成り立つから，$A\subset B\blacksquare$

(2)

(i) $\alpha\in A$とすると，$\alpha=5x+2y(x\in\mathbb{Z},y\in\mathbb{Z})$と表すことができる．

$5=4\times 2+3\times(-1),2=4\times(-1)+3\times 2$より，

\[
\alpha=\{4\times 2+3\times(-1)\}x+\{4\times(-1)+3\times 2\}y=4(2x-y)+3(-x+2y)
\]

$x\in\mathbb{Z},y\in\mathbb{Z}$より，$2x-y\in\mathbb{Z},-x+2y\in\mathbb{Z}$であるから，$\alpha\in B$

したがって，$A\subset B$が成り立つ．

(ii) $\beta\in B$とすると，$\beta=4x+3y(x\in\mathbb{Z},y\in\mathbb{Z})$と表すことができる．

$4=5\times 0+2\times 2,3=5\times 1+2\times(-1)$より，

\[
\beta=\{5\times 0+2\times 2\}x+\{5\times 1+2\times(-1)\}y=5y+2(2x-y)
\]

$x\in\mathbb{Z},y\in\mathbb{Z}$より，$2x-y\in\mathbb{Z}$であるから，$\beta\in A$

したがって，$B\subset A$が成り立つ．よって，(i)，(ii)より，$A\subset B$かつ$B\subset A$であるから，$A=B$が成り立つ．$\blacksquare$

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% 例題I2.1.6の側注：集合の包含関係・相等の証明（One More）★★★★
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% 側注1
$4x-3y$を$5\times\text{(整数)}+2\times\text{(整数)}\cdots\text{（ア）}$の形で表すために，$4$と$3$を（ア）の形で表す．なお，$3=5\times(-1)+2\times 4$などとしてもよい．

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$5x+2y$を$4\times\text{(整数)}-3\times\text{(整数)}\cdots\text{（イ）}$の形で表すために，$5$と$2$を（イ）の形で表す．

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$\alpha\in B$を示すために，$3\times\text{(整数)}+1$の形にする．

% 側注4
$B$の要素であるが，$A$の要素ではないものを1つ挙げることで，$A\neq B$が示せる．

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% 問題I2.1.6の側注：集合の包含関係・相等の証明（One More）★★★★
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% 側注1
$\alpha\in B$を示すために，$6x+5$を$3\times$(整数)$-1$の形で表す．

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% 例題I2.1.6の考え方・補足など：集合の包含関係・相等の証明（One More）★★★★
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% 考え方
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(1) $A$のすべての要素が$B$の要素であること，すなわち，$\alpha\in A$ならば，$\alpha\in B$が成り立つとき，$A$を$B$の部分集合という．よって，「$\alpha\in A$ならば，$\alpha\in B$」を示せばよい．また，$A\neq B$であることを示すためには，$B$の要素であるが$A$の要素ではないものを1つ挙げればよい．

(2) $A=B$（2つの集合の相等）を示すためには，$A\subset B$かつ$B\subset A$を示せばよい．なお，$A\subset B$の証明は，任意の整数$x,y$に対して，「$4x-3y=5\times\text{(整数)}+2\times\text{(整数)}$」で表せることを示す（$B\subset A$も同様の手法で示すことができる）．

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$A\subset B$の証明は，$x\in A$ならば，$x\in B$を示す．

$A=B$の証明は，$A\subset B$かつ$B\subset A$を示す．

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% 注意
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$4\times 1-3\times 1=1$のようになることから，$4\times n-3\times n=n$であることがわかる．つまり，$x=y=n$のとき，$4x-3y=n$となり，$A$は整数全体となる（$B$も同様に$x=n,y=-2n$のときを考えれば，整数全体となることがわかる）．