【数学I】例題2.1.16：背理法を用いた証明2（One More）★★★

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% 例題I2.1.16：背理法を用いた証明2（One More）★★★
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$a,b$を有理数とするとき，次の問いに答えよ．ただし，$\sqrt{2}$が無理数であることを用いてもよい．

(1) $a+b\sqrt{2}=0$ならば，$a=0$かつ$b=0$であることを証明せよ．

(2) $a(3+\sqrt{2})+b(4-\sqrt{2})=6+2\sqrt{2}$を満たす$a,b$の値を求めよ．

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% 例題I2.1.16の解答（One More）★★★
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(1) $b\neq 0$と仮定する．

$a+b\sqrt{2}=0$より，$\sqrt{2}=-\frac{a}{b}$

ここで，$a,b$は有理数より，$-\frac{a}{b}$も有理数となるが，これは$\sqrt{2}$が無理数であることに矛盾する．

したがって，$b=0$である．

これを$a+b\sqrt{2}=0$に代入すると，$a=0$

よって，$a,b$が有理数のとき，$a+b\sqrt{2}=0$ならば，$a=0$かつ$b=0$である．$\blacksquare$

(2) $a(3+\sqrt{2})+b(4-\sqrt{2})=6+2\sqrt{2}$を整理すると，

\[
(3a+4b-6)+(a-b-2)\sqrt{2}=0
\]

$a,b$が有理数より，$3a+4b-6,a-b-2$は有理数となる．したがって，(1)より，

\[
\begin{cases}
3a+4b-6=0\\
a-b-2=0
\end{cases}
\]

よって，これを解いて，$a=2,b=0$

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% 問題I2.1.16：背理法を用いた証明2（One More）★★★
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$a,b$を有理数とするとき，次の問いに答えよ．ただし，$\sqrt{3}$が無理数であることを用いてもよい．

(1) $a+b\sqrt{3}=0$ならば，$a=0$かつ$b=0$であることを証明せよ．

(2) $a(2+\sqrt{3})+b(5-\sqrt{3})=13+3\sqrt{3}$を満たす$a,b$の値を求めよ．

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% 問題I2.1.16の解答（One More）★★★
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(1) $b\neq 0$と仮定する．

$a+b\sqrt{3}=0$より，$\sqrt{3}=-\frac{a}{b}$

ここで，$a,b$は有理数より，$-\frac{a}{b}$も有理数であるが，これは$\sqrt{3}$が無理数であることに矛盾する．

したがって，$b=0$である．

これを$a+b\sqrt{3}=0$に代入すると，$a=0$

よって，$a,b$が有理数のとき，$a+b\sqrt{3}=0$ならば，$a=0$かつ$b=0$である．$\blacksquare$

(2) $a(2+\sqrt{3})+b(5-\sqrt{3})=13+3\sqrt{3}$を整理すると，

\[
(2a+5b-13)+(a-b-3)\sqrt{3}=0
\]

$a,b$が有理数より，$2a+5b-13,a-b-3$は有理数である．したがって，(1)より，

\[
\begin{cases}
2a+5b-13=0\\
a-b-3=0
\end{cases}
\]

よって，これを解いて，$a=4,b=1$

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% 例題I2.1.16の側注：背理法を用いた証明2（One More）★★★
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% 側注1
$b=0$であることのみを導いている．次に，$b=0$を用いて$a=0$を導く．

% 側注2
$\sqrt{2}$について整理する．

% 側注3
$3a+4b-6,a-b-2$がともに有理数であることを記すように注意すること．

% 側注4
連立方程式を解く．

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% 問題I2.1.16の側注：背理法を用いた証明2（One More）★★★
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% 側注1
$b=0$であることのみを導いている．

% 側注2
$\sqrt{3}$について整理する．

% 側注3
$2a+5b-13,a-b-3$がともに有理数であることを記すように注意すること．

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% 例題I2.1.16の考え方・補足など：背理法を用いた証明2（One More）★★★
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% 考え方
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直接証明するのは難しいので，背理法を用いる．(1)の結論は，「$a=0$かつ$b=0$」なので，その否定である「$a\neq 0$または$b\neq 0$」を仮定してもよいが，ここでは$b\neq 0$と仮定するとうまくいく．結論が「$p$かつ$q$」という場合は，一方の否定のみを仮定して矛盾を導いた方が考えやすい．

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$a,b,c,d$が有理数，$\sqrt{A}$が無理数のとき，次のことが成り立つ．

\[
a+b\sqrt{A}=c+d\sqrt{A}\text{ならば，}a=c,b=d
\]

とくに，$a+b\sqrt{A}=0$ならば，$a=b=0$