【数学I】例題2.1.11：「すべて」「ある」の否定（One More）★★★

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% 例題I2.1.11：「すべて」「ある」の否定（One More）★★★
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次の命題の否定を述べよ．また，もとの命題とその否定の真偽を答えよ．

(1) すべての実数$x$について，$x^2>0$

(2) ある自然数$n$について，$n^2=2n$

(3) 任意の実数$x,y$について，$x^2-2xy+y^2>0$

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% 例題I2.1.11の解答（One More）★★★
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(1) 否定は，「ある実数$x$について，$x^2\leqq 0$」

これは，$x=0$のとき，$x^2=0$

よって，真である．

また，もとの命題は$x=0$のとき，$x^2=0$であるから，$x^2>0$は成り立たない．

よって，偽である．

(2) 否定は，「すべての自然数$n$について，$n^2\neq 2n$」

これは，$n=2$のとき，$n^2=2n$であるから，$n^2\neq 2n$は成り立たない．

よって，偽である．

また，もとの命題は$n=2$のとき，$n^2=2n$であるから，真である．

(3) 否定は，「ある実数$x,y$について，$x^2-2xy+y^2\leqq 0$」

これは，$x=y=0$のとき，$x^2-2xy+y^2=0$

よって，真である．

また，もとの命題は$x=y=1$とすると，$x^2-2xy+y^2=1^2-2\cdot 1\cdot 1+1^2=0$

よって，偽である．

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% 問題I2.1.11：「すべて」「ある」の否定（One More）★★★
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次の命題の否定を述べよ．また，もとの命題とその否定の真偽を答えよ．

(1) ある整数$k$について，$k^2=3k$

(2) 任意の実数$x,y$について，$(x+y)^2>x^2+y^2$

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% 問題I2.1.11の解答（One More）★★★
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(1) 否定は，「すべての整数$k$について，$k^2\neq 3k$」

これは，$k=3$のとき，$k^2=3k$であるから，$k^2\neq 3k$は成り立たない．

よって，否定は偽である．

また，もとの命題は$k=3$のとき，$k^2=3k$であるから，真である．

(2) 否定は，「ある実数$x,y$について，$(x+y)^2\leqq x^2+y^2$」

これは，$x=0,y=0$のとき，$(x+y)^2=x^2+y^2=0$

よって，否定は真である．

また，もとの命題は，$x=1,y=-1$とすると，$(x+y)^2=(1-1)^2=0$であり，$x^2+y^2=2$である．

よって，偽である．

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% 例題I2.1.11の側注：「すべて」「ある」の否定（One More）★★★
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% 側注1
「すべて」を「ある」に入れ替える．

% 側注2
\begin{align*}
&x^2-2xy+y^2=0\\
&\Longleftrightarrow(x-y)^2=0\\
&\Longleftrightarrow x=y
\end{align*}

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% 問題I2.1.11の側注：「すべて」「ある」の否定（One More）★★★
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% 側注1
$k=0$も反例である．

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% 例題I2.1.11の考え方・補足など：「すべて」「ある」の否定（One More）★★★
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% 考え方
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命題の否定において「すべて」と「ある」を含むとき，「すべて」と「ある」を入れ替えて，もとの命題の結論を否定すればよい．一般に，$p$が真のときその否定$\overline{p}$は偽であり，$p$が偽のときその否定$\overline{p}$は真であることが成り立つ．また，解答では否定の真偽の理由を記しているが，もとの命題の真偽がわかっている場合は，必ずしも記す必要はない（例えば，「もとの命題が真なので，否定は偽である」などと否定の理由を省略してよい．なお，否定の真偽がわかっている場合も同様である）．

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命題の否定は，「すべて」と「ある」に注意する．

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「すべての$x$」を，「任意の$x$」ということもある（「常に」を用いて表されることもある）．また，「ある$x$」を，「適当な$x$」や「少なくとも1つの$x$」ということもある．