Solutions To Mathematics Textbooks/Proofs and Fundamentals/Chapter 5

Section 5.1

Remember the definitions:
Definition 5.3.1 Let $A$ and $B$ be non-empty sets, let $R$ be a relation from $A$ to $B$, and let $x\in A$. The relation class of $x$ with respect to $R$, denoted $R[x]$, is the set defined by $R[x]=\{y\in B|xRy\}$.
Definition 2.2.1 Let $a$ and $b$ be integers. The number $a$ divides the number $b$ if there is some integer $q$ such that $aq=b$. If $a$ divides $b$, we write $a|b$, and we say that $a$ is a factor of $b$, and that $b$ is divisible by $a$.

5.1.1

1. Let $aSb\iff a=|b|$, for all $a,b\in \mathbb{Z}$. Then
   $S[3]=\{b\in \mathbb{Z}:3Sb\}$, but by the definition of the relation $S$ is $S[3]=\{b\in \mathbb{Z}:3=|b|\}$ and the only elements that satisfy this property are $3$ and $-3$, since $3=|3|=|-3|$ and therefore $S[3]=\{-3,3\}$. Analogously, we have to:
   $S[-3]=\{b\in \mathbb{Z}:-3Sb\}=\{b\in \mathbb{Z}:-3=|b|\}=\mathrm{\varnothing }$.
   $S[6]=\{b\in \mathbb{Z}:6Sb\}=\{b\in \mathbb{Z}:6=|b|\}=\{-6,6\}$.
2. Let $aDb\iff a|b$, for all $a,b\in \mathbb{Z}$. Then
   $D[3]=\{b\in \mathbb{Z}:3Db\}=\{b\in \mathbb{Z}:\mathrm{\exists }k\in \mathbb{Z}\text{ and }b=3k\}=\{...,-6,-3,-1,0,1,3,6,...\}$.
   $D[-3]=\{b\in \mathbb{Z}:-3Db\}=\{b\in \mathbb{Z}:\mathrm{\exists }k\in \mathbb{Z}\text{ and }b=-3k\}=\{...,-6,-3,-1,0,1,3,6,...\}$.
   $D[6]=\{b\in \mathbb{Z}:6Db\}=\{b\in \mathbb{Z}:\mathrm{\exists }k\in \mathbb{Z}\text{ and }b=6k\}=\{...,-12,-6,0,6,12,...\}$.
3. Let $aTb\iff b|a$, for all $a,b\in \mathbb{Z}$. Then
   $T[3]=\{b\in \mathbb{Z}:3Tb\}=\{b\in \mathbb{Z}:b|3\}=\{b\in \mathbb{Z}:\mathrm{\exists }k\in \mathbb{Z}\text{ and }3=bk\}=\{-3,-1,1,3\}$.
   $T[-3]=\{b\in \mathbb{Z}:-3Tb\}=\{b\in \mathbb{Z}:b|-3\}=\{b\in \mathbb{Z}:\mathrm{\exists }k\in \mathbb{Z}\text{ and }-3=bk\}=\{-3,-1,1,3\}$.
   $T[6]=\{b\in \mathbb{Z}:6Tb\}=\{b\in \mathbb{Z}:b|6\}=\{b\in \mathbb{Z}:\mathrm{\exists }k\in \mathbb{Z}\text{ and }6=bk\}=\{-6,-3,2,-1,1,2,3,6\}$.
4. Let $aQb\iff a+b=7$, for all $a,b\in \mathbb{Z}$. Then
   $Q[3]=\{b\in \mathbb{Z}:3Qb\}=\{b\in \mathbb{Z}:3+b=7\}=\{4\}$.
   $Q[-3]=\{b\in \mathbb{Z}:-3Qb\}=\{b\in \mathbb{Z}:-3+b=7\}=\{10\}$.
   $Q[6]=\{b\in \mathbb{Z}:6Qb\}=\{b\in \mathbb{Z}:6+b=7\}=\{1\}$.

5.1.2

1. Let $S$ be the relation defined by $(x,y)S(z,w)\iff y=3w\text{ for all }(x,y),(z,w)\in {ℝ}^2$.
   $S[(0,0)]=\{(z,w)\in {ℝ}^2:(0,0)S(z,w)\}=\{(z,0)\}$. Because $y=3w\Rightarrow 0=3w$, and therefore $w=0$. The geometric description of the relation class are: the $x$-axis.
   $S[(3,4)]=\{(z,w)\in {ℝ}^2:(3,4)S(z,w)\}=\{(z,\frac{4}{3})\}$. Because $y=3w\Rightarrow 4=3w$, and therefore $w=\frac{4}{3}$. The geometric description of the relation class are: the the line whose equation is $y=\frac{4}{3}$.
2. Let $T$ be the relation defined by $(x,y)T(z,w)\iff x^2+3y^2=7z^2+w^2\text{, for all }(x,y),(z,w)\in {ℝ}^2$.
   $T[(0,0)]=\{(z,w)\in {ℝ}^2:(0,0)T(z,w)\}=(0,0).Because{0}^2+3\cdot 0^2=7z^2+w^2\Rightarrow Z=0andw=o$.
   $T[(3,4)]=\{(z,w)\in {ℝ}^2:(3,4)T(z,w)\}=\{(z,\sqrt{57-7z^2})\}$. Because $3^2+3\cdot 4^2=7z^2+w^2\Rightarrow w=\sqrt{57-7z^2}$. The geometric description of the relation class are the graph of $y=\sqrt{57-7z^2}$.
3. Let $Z$ be the relation defined by $(x,y)T(z,w)\iff x=z\vee y=w\text{, for all }(x,y),(z,w)\in {ℝ}^2$.
   $Z[(0,0)]=\{(z,w)\in {ℝ}^2:(0,0)Z(z,w)\}=\{(0,w)\vee (z,0)\}$.
   $Z[(3,4)]=\{(z,w)\in {ℝ}^2:(3,4)Z(z,w)\}=\{(3,w)\vee (z,4)\}$.

5.1.3

Let $A=\{1,2,3\}$. Each of the following subsets of $A\times A$ defines a relation on $A$. Is each relation reflexive, symmetric and/or transitive?

1. $M=\{(3,3),(2,2),(1,2),(2,1)\}$. is symmetric only
2. $N=\{(1,1),(2,2),(3,31),(1,2)\}$. is reflexive only