【数学I】2章：集合と命題（節末問題・章末問題）

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% 節末問題I2.1.1：（One More）★
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$P=\{a,b,c\}$の部分集合をすべて求めよ．

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% 節末問題I2.1.1の解答（One More）★
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\[
\varnothing,\{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\},\{a,b,c\}
\]

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% 節末問題I2.1.2：（One More）★★
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実数全体を全体集合とし，その2つの部分集合を$A=\{x\mid|x-1|<\sqrt{6}\},B=\{x\mid-a\leqq x\leqq a\}$とするとき，次の問いに答えよ．ただし，$a$は正の定数とする．

(1) $A\supset B$となる$a$の値の範囲を求めよ．

(2) $A\cup B$に属する整数の個数が9個となる$a$の値の範囲を求めよ．

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% 節末問題I2.1.2の解答（One More）★★
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(1) 不等式$|x-1|<\sqrt{6}$を解くと

$-\sqrt{6}<x-1<\sqrt{6}$より，$1-\sqrt{6}<x<1+\sqrt{6}$

したがって，$A=\{x\mid 1-\sqrt{6}<x<1+\sqrt{6}\}$であるから，右の数直線より，$A\supset B$となるのは

\[
1-\sqrt{6}<-a
\]

よって，$a>0$より，$0<a<-1+\sqrt{6}$

(2) $A$に属する整数は，$-1,0,1,2,3$の5個であるから，$A\cup B$に属する整数が，$-4,-3,-2,-1,0,1,2,3,4$の9個になればよい．

よって，

\[
4\leqq a<5
\]

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% 節末問題I2.1.3：（One More）★★★
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次の命題の真偽を調べよ．また，真のときにはその証明をし，偽のときは具体的な反例を挙げよ．ただし，$x,y$は実数とし，$\sqrt{2},\sqrt{5}$は無理数であることを用いてもよい．

(1) $x$が無理数，$y$が有理数ならば，$x+y$は無理数である．

(2) $x^2-x$が有理数ならば，$x$は有理数である．

(3) $x,y$がともに無理数ならば，$x+y,x^2+y^2$のうち少なくとも一方は無理数である．

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% 節末問題I2.1.3の解答（One More）★★★
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(1) $x$が無理数かつ$y$が有理数ならば，$x+y$が有理数であると仮定すると，

\[
x=(x+y)-y
\]

となる．しかし，これは左辺が無理数，右辺が有理数となり，$x$が無理数であることに矛盾する．

したがって，$x$が無理数かつ$y$が有理数ならば，$x+y$は無理数である．

よって，命題は真である．

(2) $x^2-x=1$とすると，$x^2-x-1=0$

これを解いて，$x=\frac{1\pm\sqrt{5}}{2}$

$\sqrt{5}$は無理数であるから，$x$は無理数である．

よって，命題は偽である．反例は，$x^2-x=1$

(3) $x=\sqrt{2},y=-\sqrt{2}$のとき，$x,y$はともに無理数であるが，$x+y=0,x^2+y^2=4$であるから，$x+y,x^2+y^2$はどちらも無理数ではない．

よって，命題は偽である．反例は，$x=\sqrt{2},y=-\sqrt{2}$

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% 節末問題I2.1.4：（One More）★★★
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次の$\boxed{\text{}}$に最も適するものを，(i)〜(iv)から選べ．ただし，$a,b,c$は実数とする．

(1) $a=b$は，$ac=bc$であるための$\boxed{\text{}}$

(2) $a=b=c$は，$a^2+b^2+c^2-ab-bc-ca=0$であるための$\boxed{\text{}}$

(3) $a^2>b^2$は，$a>b$であるための$\boxed{\text{}}$

(i) 必要条件であるが十分条件ではない

(ii) 十分条件であるが必要条件ではない

(iii) 必要十分条件である

(iv) 必要条件でも十分条件でもない

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% 節末問題I2.1.4の解答（One More）★★★
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(1) 「$a=b\Longrightarrow ac=bc$」は，真である．

「$ac=bc\Longrightarrow a=b$」は，偽である．反例は，$a=1,b=2,c=0$

よって，(ii)十分条件であるが必要条件ではない

(2)

\begin{align*}
a^2+b^2+c^2-ab-bc-ca=0&\Longleftrightarrow(a-b)^2+(b-c)^2+(c-a)^2=0\\
&\Longleftrightarrow a=b=c
\end{align*}

よって，(iii)必要十分条件である

(3) 「$a^2>b^2\Longrightarrow a>b$」は，偽である．反例は，$a=-1,b=0$

「$a>b\Longrightarrow a^2>b^2$」は，偽である．反例は，$a=0,b=-1$

よって，(iv)必要条件でも十分条件でもない

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% 節末問題I2.1.5：（One More）★★
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次の命題が成り立つことを対偶を用いて証明せよ．

$x,y$がともに正の数のとき，$x^2+y^2\geqq 4$ならば，$x\geqq\sqrt{2}$または$y\geqq\sqrt{2}$である．

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% 節末問題I2.1.5の解答（One More）★★
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もとの命題の対偶「$x,y$がともに正の数のとき，$x<\sqrt{2}$かつ$y<\sqrt{2}$ならば，$x^2+y^2<4$」を証明する．

$x$は正の数であり，$x<\sqrt{2}$より，$x^2<2\cdots(\mathrm{i})$

$y$は正の数であり，$y<\sqrt{2}$より，$y^2<2\cdots(\mathrm{ii})$

したがって，(i)と(ii)の辺々を足し合わせて，$x^2+y^2<4$

ゆえに，対偶は真である．

よって，対偶が証明されたから，もとの命題も成り立つ．$\blacksquare$

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% 節末問題I2.1.6：（One More）★★★★
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整数$a,b$を係数とする2次方程式$x^2+ax+b=0$が有理数の解$r$をもつならば，$r$は整数であることを証明せよ．

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% 節末問題I2.1.6の解答（One More）★★★★
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$x^2+ax+b=0$が有理数の解$r$をもつから，

$r=\frac{m}{n}(m$と$n$は1以外に公約数をもたない整数，$n\neq 0$)

と表される．

このとき，$(\frac{m}{n})^2+a\cdot\frac{m}{n}+b=0$

すなわち，$m^2+amn+bn^2=0$

したがって，$m^2=-n(am+bn)\cdots(\mathrm{i})$

$n\neq\pm 1$と仮定すると，$n$はある素数$p$を約数にもつ．

このとき，(i)より$m^2$は素数$p$を約数にもつ．

$m^2$が素数$p$を約数にもてば，$m$も素数$p$を約数にもつ．

これは，$m$と$n$が1以外に公約数をもたないことに矛盾する．

したがって，$n=\pm 1$

よって，$r$は整数である．$\blacksquare$

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% 章末問題I2.1：（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．

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

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% 章末問題I2.1の解答（One More）★★★
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(i) $a\in A$とすると，$a=5x+2y(x,y\in\mathbb{Z})$と表すことができる．

$5x+2y$は整数であるから，$a\in\mathbb{Z}$

すなわち，$a\in A$ならば，$a\in\mathbb{Z}$であるから，$A\subset\mathbb{Z}$

(ii) $a\in\mathbb{Z}$とすると，$a=5a+2(-2a)$であり，$a,-2a$はともに整数であるから，

\[
5a+2(-2a)\in A
\]

すなわち，$a\in\mathbb{Z}$ならば，$a\in A$であるから，$\mathbb{Z}\subset A$

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

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% 章末問題I2.2：（One More）★★★
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次の命題の真偽を調べよ．また，真のときにはその証明をし，偽のときには具体的な反例を挙げよ．ただし，$a,b$を自然数とする．

(1) $a$が奇数かつ$b$が奇数ならば，$a^2+b^2$が偶数

(2) $a^2+b^2$が偶数ならば，$a$が偶数かつ$b$が偶数

(3) $a^2+b^2$が奇数ならば，$a$が奇数または$b$が奇数

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% 章末問題I2.2の解答（One More）★★★
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(1) $a,b$は奇数であるから，$a=2m+1,b=2n+1(m,n$は整数)とおくと，

\begin{align*}
a^2+b^2&=(2m+1)^2+(2n+1)^2\\
&=(4m^2+4m+1)+(4n^2+4n+1)\\
&=2(2m^2+2m+2n^2+2n+1)
\end{align*}

したがって，$2m^2+2m+2n^2+2n+1$は整数であるから，$a^2+b^2$は偶数である．

よって，命題は真である．

(2) $a=3,b=3$のとき，$a^2+b^2=18$となり，$a^2+b^2$は偶数であるが，$a,b$はともに奇数である．

よって，命題は偽である．反例は，$a=3,b=3$

(3) もとの命題の対偶「$a$が偶数かつ$b$が偶数ならば，$a^2+b^2$が偶数」が正しいことを証明する．

$a,b$は偶数であるから，$a=2m,b=2n(m,n$は整数)とおくと，

\[
a^2+b^2=(2m)^2+(2n)^2=4m^2+4n^2=2(2m^2+2n^2)
\]

$2m^2+2n^2$は整数であるから，$a^2+b^2$は偶数である．

よって，対偶が証明されたので，もとの命題は真である．

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% 章末問題I2.3：（One More）★★
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次の命題を証明せよ．ただし，$m,n$は正の整数，$m>n$とする．

$\frac{m+n}{m-n}$が既約分数ならば，$\frac{n}{m}$は既約分数である．

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% 章末問題I2.3の解答（One More）★★
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もとの命題の対偶「$m,n$が共通の素因数をもつならば，$m+n,m-n$は共通の素因数をもつ」を証明する．

$m,n$に共通の素因数を$k$として，$m=ka,n=kb$（$a,b$は整数）とおくと，

\[
m+n=k(a+b),m-n=k(a-b)
\]

これらは，共通の素因数$k$をもつ．

よって，対偶が証明されたから，もとの命題も成り立つ．$\blacksquare$

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% 章末問題I2.4：（One More）★★★
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$\sqrt{a}+\sqrt{b}$が有理数ならば，$\sqrt{a},\sqrt{b}$はともに有理数であることを証明せよ．ただし，$a,b$を正の有理数とする．

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% 章末問題I2.4の解答（One More）★★★
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$\sqrt{a}+\sqrt{b}=r$（$r$は正の有理数）とおくと，$\sqrt{b}=r-\sqrt{a}\cdots(\mathrm{i})$

(i) の両辺を2乗すると，$b=(r-\sqrt{a})^2$

したがって，$b=r^2-2r\sqrt{a}+a$

ゆえに，$2r\sqrt{a}=r^2+a-b$

$r$は正の有理数であるから，両辺を$2r(\neq 0$)で割って，

\[
\sqrt{a}=\frac{r^2+a-b}{2r}
\]

ここで，$a,b,r$は正の有理数であるから，$\frac{r^2+a-b}{2r}$は有理数であり，$\sqrt{a}$も有理数となる．

また，$\sqrt{a}$が有理数のとき，(i)より，$\sqrt{b}$も有理数である．

よって，$\sqrt{a}+\sqrt{b}$が有理数ならば，$\sqrt{a},\sqrt{b}$はともに有理数である．$\blacksquare$

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% 章末問題I2.5：（One More）★★
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三角形の内角で，$60^\circ$以上のものが少なくとも1つ存在することを証明せよ．

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% 章末問題I2.5の解答（One More）★★
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$\triangle\mathrm{ABC}$において，すべての内角が$60^{\circ}$未満であると仮定すると，

\[
\angle\mathrm{A}<60^{\circ},\angle\mathrm{B}<60^{\circ},\angle\mathrm{C}<60^{\circ}
\]

これらの角度を足し合わせると，

\[
\angle\mathrm{A}+\angle\mathrm{B}+\angle\mathrm{C}<180^{\circ}
\]

これは，三角形の内角の和が$180^{\circ}$であることに矛盾する．

よって，三角形の内角で$60^{\circ}$以上のものが少なくとも$1$つ存在する．$\blacksquare$

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% 節末問題I2.1.1の側注：（One More）★
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% 側注1
$\varnothing$や$P$も部分集合である．また，$\{\varnothing\}$は空集合を1つの要素にもつ集合となり，$\varnothing$とは意味が異なるので波括弧をつけないように注意すること．

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% 節末問題I2.1.2の側注：（One More）★★
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% 側注1
$a>0$を忘れないように注意すること．

% 側注2
等号を含むか否かに注意すること．

% 側注3
$|x|<a$（$a>0$）の解は，$-a<x<a$

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% 節末問題I2.1.3の側注：（One More）★★★
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% 側注1
\begin{align*}
&\text{(有理数})-(\text{有理数})\\
&=(\text{有理数})
\end{align*}

% 側注2
$ax^2+bx+c=0$の解は，
\[
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\]

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% 節末問題I2.1.4の側注：（One More）★★★
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% 側注1
$a^2+b^2+c^2-ab-bc-ca=0,a^2+b^2+c^2+ab+bc+ca=0$はそれぞれ，$(a-b)^2+(b-c)^2+(c-a)^2=0,(a+b)^2+(b+c)^2+(c+a)^2=0$と式変形することができる．展開して，同値の関係が成り立つことを確かめるとよい．

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% 節末問題I2.1.5の側注：（One More）★★
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% 側注1
「$p\Longrightarrow q$」の対偶は，「$\overline{q}\Longrightarrow\overline{p}$」

% 側注2
$0<x<\sqrt{2}$より，$0<x^2<2$

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% 節末問題I2.1.6の側注：（One More）★★★★
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% 側注1
有理数の解$r$は正であるとは限らないことに注意すること．

% 側注2
$m,n$が互いに素であるとき，$r$が整数であることを示すには，$n=\pm 1$を示せばよい．そこで，$n\neq\pm 1$と仮定する．

% 側注3
$m$と$n$が素数$p$を公約数にもつことになる．これは，公約数をもたないことに矛盾する．

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% 章末問題I2.1の側注：（One More）★★★
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% 側注1
すべての整数は$5\times(\text{整数})+2\times(\text{整数})$で表すことができることを示す．

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% 章末問題I2.2の側注：（One More）★★★
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% 側注1
$2m^2+2m+2n^2+2n+1$が整数であることを記すように注意すること．

% 側注2
もとの命題のかわりに，対偶を用いて証明する．

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% 章末問題I2.3の側注：（One More）★★
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% 側注1
もとの命題のかわりに，対偶を用いて証明する．

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% 章末問題I2.4の側注：（One More）★★★
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% 側注1
有理数の和，差，積，商（0で割る場合を除く）は有理数である．

% 側注2
$r$と$\sqrt{a}$が有理数であるから，$\sqrt{b}=r-\sqrt{a}$も有理数である．

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% 章末問題I2.5の側注：（One More）★★
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% 側注1
背理法を用いて証明する．