【数学I】例題1.2.9：整数部分と小数部分（One More）★★★

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% 例題I1.2.9：整数部分と小数部分（One More）★★★
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$\frac{1}{3-\sqrt{5}}$の整数部分を$a$，小数部分を$b$とする．

(1) $a,b$の値を求めよ．

(2) $a^2+ab$の値を求めよ．

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(1) $\frac{1}{3-\sqrt{5}}=\frac{3+\sqrt{5}}{(3-\sqrt{5})(3+\sqrt{5})}=\frac{3+\sqrt{5}}{4}$

$2<\sqrt{5}<3$であるから，$5<3+\sqrt{5}<6$

したがって，$\frac{5}{4}<\frac{3+\sqrt{5}}{4}<\frac{3}{2}$

ゆえに，$a=1$

よって，

\[
b=\frac{3+\sqrt{5}}{4}-a=\frac{3+\sqrt{5}}{4}-1=\frac{\sqrt{5}-1}{4}
\]

(2) (1)より，

\[
a^2+ab=a(a+b)=1\cdot\frac{1}{3-\sqrt{5}}=\frac{3+\sqrt{5}}{4}
\]

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% 問題I1.2.9：整数部分と小数部分（One More）★★★
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$\frac{3}{4-\sqrt{7}}$の整数部分を$a$，小数部分を$b$とする．

(1) $a,b$の値を求めよ．

(2) $a+\frac{1}{b}$の値を求めよ．

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(1) $\frac{3}{4-\sqrt{7}}=\frac{3(4+\sqrt{7})}{(4-\sqrt{7})(4+\sqrt{7})}=\frac{4+\sqrt{7}}{3}$

$2<\sqrt{7}<3$であるから，$6<4+\sqrt{7}<7$

したがって，$\frac{6}{3}<\frac{4+\sqrt{7}}{3}<\frac{7}{3}$

ゆえに，$a=2$

よって，

\[
b=\frac{4+\sqrt{7}}{3}-a=\frac{4+\sqrt{7}}{3}-2=\frac{\sqrt{7}-2}{3}
\]

(2) (1)より，

\[
a+\frac{1}{b}=2+1\div(\frac{\sqrt{7}-2}{3})=2+\frac{3}{\sqrt{7}-2}=2+\frac{3(\sqrt{7}+2)}{(\sqrt{7}-2)(\sqrt{7}+2)}={4+\sqrt{7}}
\]

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% 例題I1.2.9の側注：整数部分と小数部分（One More）★★★
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% 側注1
分母を有理化する．

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$2^2<5<3^2$より，$2<\sqrt{5}<3$

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$\frac{5}{4}=1.25,\frac{3}{2}=1.5$

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小数部分は，

\[
(\text{もとの数})-(\text{整数部分})
\]

で求められる．

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$a+b$はもとの数$\frac{1}{3-\sqrt{5}}$であることを用いる．

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分母を有理化する．

% 側注2
$2^2<7<3^2$より，$2<\sqrt{7}<3$

% 側注3
$\frac{6}{3}=2,\frac{7}{3}=2.333\ldots$

% 側注4
小数部分は，

\[
(\text{もとの数})-(\text{整数部分})
\]

で求められる．

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$(\text{もとの数})=(\text{整数部分})+(\text{小数部分})$に分けられるので，小数部分は，$(\text{小数部分})=(\text{もとの数})-(\text{整数部分})$と考えることができる．例えば$\sqrt{5}=2.236\ldots$は，右のように$\sqrt{5}=2+0.236\ldots$と分けられるので，$0.236\ldots=\sqrt{5}-2$と考えることができる．

\begin{align*}
\sqrt{5}&=2.236\ldots\\
&=\underbrace{2}_{\text{整数部分}}+\underbrace{0.236\ldots}_{\text{小数部分}}
\end{align*}

なお，$\sqrt{5}$の整数部分は，$n^2\leqq 5<(n+1)^2$となる整数$n$を見つけることで求められる．

例：$\sqrt{5}$は，$2^2<5<3^2$より$2<\sqrt{5}<3$であるから，$\sqrt{5}=2.\ldots$となる．よって，整数部分は$2$である．

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$(\text{小数部分})=(\text{もとの数})-(\text{整数部分})$

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実数$x$の整数部分は，$n\leqq x<n+1$を満たす整数$n$である．

例：$4\leqq x<5$のとき，$x$の整数部分は$4$，小数部分は$x-4$