問題の解答
検索用コード（LaTeX）
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題I2.1.1：集合の表し方（One More）★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
(1) $A=\{x\mid x\text{は15以下の素数}\}$とする．次の$\boxed{\text{}}$の中に，$\in$または$\notin$のいずれか適するものを書き入れよ．
(i) $2\boxed{\text{}}A$
(ii) $10\boxed{\text{}}A$
(iii) $13\boxed{\text{}}A$
(2) 次の集合を要素を書き並べて表せ．
(i) 12の正の約数全体の集合
(ii) $\{x\mid-3\leqq x<4,x\text{は整数}\}$
(3) 次の2つの集合$A,B$の間に成り立つ包含関係をいえ．
(i) $A=\{2n-1\mid 0\leqq n\leqq 5,n\text{は整数}\},B=\{6n-5\mid 1\leqq n\leqq 2,n\text{は整数}\}$
(ii) $A=\{2n+1\mid n=1,2\},B=\{x\mid(x-3)(x-5)=0,x\text{は整数}\}$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題I2.1.1の解答（One More）★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
(1)
(i) $2$は$15$以下の素数であるから，$2\in A$
(ii) $10$は$15$以下の素数ではないから，$10\notin A$
(iii) $13$は$15$以下の素数であるから，$13\in A$
(2)
(i) $\{1,2,3,4,6,12\}$
(ii) $\{-3,-2,-1,0,1,2,3\}$
(3)
(i)
\begin{align*}
A&=\{2\cdot 0-1,2\cdot 1-1,2\cdot 2-1,2\cdot 3-1,2\cdot 4-1,2\cdot 5-1\}\\
&=\{-1,1,3,5,7,9\},\\
B&=\{6\cdot 1-5,6\cdot 2-5\}=\{1,7\}
\end{align*}
よって，$B\subset A$
(ii) $A=\{2\cdot 1+1,2\cdot 2+1\}=\{3,5\}$
また，$(x-3)(x-5)=0$を解くと，$x=3,5$
したがって，$B=\{3,5\}$
よって，$A=B$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題I2.1.1：集合の表し方（One More）★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
(1) $A=\{x\mid x\text{は20以下の正の奇数}\}$とする．次の$\boxed{\text{}}$の中に，$\in$または$\notin$のいずれか適するものを書き入れよ．
(i) $7\boxed{\text{}}A$
(ii) $12\boxed{\text{}}A$
(2) 次の集合を要素を書き並べて表せ．
(i) 16の正の約数全体の集合
(ii) $\{x\mid-5\leqq x<3,x\text{は整数}\}$
(3) 次の2つの集合$A,B$の間に成り立つ包含関係をいえ．
(i) $A=\{4n+1\mid 0\leqq n\leqq 1,n\text{は整数}\},B=\{2n-1\mid-1\leqq n\leqq 3,n\text{は整数}\}$
(ii) $A=\{2n+1\mid n=0,1\},B=\{x\mid(x-1)(x-3)=0,x\text{は整数}\}$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題I2.1.1の解答（One More）★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
(1)
(i) $7$は$20$以下の正の奇数であるから，$7\in A$
(ii) $12$は$20$以下の正の奇数ではないから，$12\notin A$
(2)
(i) $\{1,2,4,8,16\}$
(ii) $\{-5,-4,-3,-2,-1,0,1,2\}$
(3)
(i)
\begin{align*}
A&=\{4\cdot 0+1,4\cdot 1+1\}=\{1,5\},\\
B&=\{2\cdot(-1)-1,2\cdot 0-1,2\cdot 1-1,2\cdot 2-1,2\cdot 3-1\}\\
&=\{-3,-1,1,3,5\}
\end{align*}
よって，$A\subset B$
(ii) $A=\{2\cdot 0+1,2\cdot 1+1\}=\{1,3\}$
また，$(x-1)(x-3)=0$を解くと，$x=1,3$
したがって，$B=\{1,3\}$
よって，$A=B$
あわせて読みたい
【数学I】2章：集合と命題（基本事項）
検索用コード（LaTeX） 本文・解答側注 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % 基本事項I2.1.1：集合（One More） %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%...
あわせて読みたい
【数学I】2章：集合と命題（節末問題・章末問題）
節末I2.1.1〜I2.1.6の解答 節末I2.1.1節末I2.1.2節末I2.1.3節末I2.1.4節末I2.1.5節末I2.1.6 リンク（関連例題） https://onemath.net/onemorei-reidai2-1-3 https://one...