問題の解答
検索用コード（LaTeX）
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% 例題I2.1.8：命題の真偽と集合（One More）★
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次の命題の真偽を，集合の考えを用いて調べよ．
(1) $n$を自然数とする．$n$が6以下の正の偶数ならば，$n$は12の正の約数である．
(2) 実数$x$について，$0\leqq x\leqq 2$ならば，$|x|<2$
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% 例題I2.1.8の解答（One More）★
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(1) $P=\{2,4,6\},Q=\{1,2,3,4,6,12\}$とおくと，
\[
P\subset Q
\]
よって，命題は真である．
(2) $P=\{x\mid 0\leqq x\leqq 2\},Q=\{x\mid|x|<2\}$とおく．
$|x|<2$より，$-2<x<2$
$P,Q$を数直線上に表すと，右の図のようになる．
ここで，$x=2$のとき，$P$に属するが，$Q$には属さない．
したがって，$P\subset Q$は成り立たない．
よって，命題は偽である．
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% 問題I2.1.8：命題の真偽と集合（One More）★
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次の命題の真偽を，集合の考えを用いて調べよ．
(1) $n$を自然数とする．$n$が1桁の正の奇数ならば，$n$は15の正の約数である．
(2) 実数$x$について，$|x|<3$ならば，$x>-4$
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% 問題I2.1.8の解答（One More）★
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(1) $P=\{1,3,5,7,9\},Q=\{1,3,5,15\}$とおくと，$7\notin Q$であるから，$P\subset Q$は成り立たない．
よって，命題は偽である．
(2) $P=\{x\mid|x|<3\},Q=\{x\mid x>-4\}$とおく．
$|x|<3$より，$-3<x<3$
$P,Q$を数直線上に表すと，右の図のようになる．
したがって，$P\subset Q$が成り立つ．
よって，命題は真である．
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