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 $$R[x] = \{y\in B | x R y\}$$. 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 $$a S b \Leftrightarrow a = |b| $$, for all  $$a,b\in \mathbb{Z}$$. Then $$S[3]=\{ b\in \mathbb{Z}: 3 S b \}$$, 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: $$S[-3]=\{ b\in \mathbb{Z}: -3 S b \}=\{b\in \mathbb{Z}: -3=|b| \} = \emptyset $$. $$S[6]=\{ b\in \mathbb{Z}: 6 S b \}=\{b\in \mathbb{Z}: 6=|b| \} = \{-6,6\} $$.
 * 2) Let $$a D b \Leftrightarrow a|b $$, for all  $$a,b\in \mathbb{Z}$$. Then $$D[3]=\{ b\in \mathbb{Z}: 3 D b \} = \{b\in \mathbb{Z}: \exists k\in \mathbb{Z} \text{ and } b=3k\}=\{...,-6,-3,-1,0,1,3,6,...\}$$. $$D[-3]=\{ b\in \mathbb{Z}: -3 D b \} = \{b\in \mathbb{Z}: \exists k\in \mathbb{Z} \text{ and } b=-3k\}=\{...,-6,-3,-1,0,1,3,6,...\}$$. $$D[6]=\{ b\in \mathbb{Z}: 6 D b \} = \{b\in \mathbb{Z}: \exists k\in \mathbb{Z} \text{ and } b=6k\}=\{...,-12,-6,0,6,12,...\}$$.
 * 3) Let $$a T b \Leftrightarrow b|a $$, for all  $$a,b\in \mathbb{Z}$$. Then $$T[3]=\{ b\in \mathbb{Z}: 3 T b \} =\{ b\in \mathbb{Z}: b|3 \} = \{b\in \mathbb{Z}: \exists k\in \mathbb{Z} \text{ and } 3=bk\}=\{-3,-1,1,3\}$$. $$T[-3]=\{ b\in \mathbb{Z}: -3 T b \} =\{ b\in \mathbb{Z}: b|-3 \} = \{b\in \mathbb{Z}: \exists k\in \mathbb{Z} \text{ and } -3=bk\}=\{-3,-1,1,3\}$$. $$T[6]=\{ b\in \mathbb{Z}: 6 T b \} =\{ b\in \mathbb{Z}: b|6 \} = \{b\in \mathbb{Z}: \exists k\in \mathbb{Z} \text{ and } 6=bk\}=\{-6,-3,2,-1,1,2,3,6\}$$.
 * 4) Let $$a Q b \Leftrightarrow a+b=7 $$, 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 $$(x,y)S(z,w)\Leftrightarrow y=3w \text{ for all } (x,y), (z,w) \in\mathbb{R}^{2}$$. $$S[(0,0)]=\{(z,w)\in\mathbb{R}^{2}: (0,0)S(z,w)\}=\{(z,0)\}$$. Because $$y=3w \Rightarrow 0=3w$$, and therefore $$w=0$$. The geometric description of the relation class are: the $$x$$-axis. $$S[(3,4)]=\{(z,w)\in\mathbb{R}^{2}: (3,4)S(z,w)\}=\{(z,\dfrac{4}{3})\}$$. Because $$y=3w \Rightarrow 4=3w$$, and therefore $$w=\dfrac{4}{3}$$. The geometric description of the relation class are: the the line whose equation is $$y=\dfrac{4}{3}$$.
 * 2) Let $$T$$ be the relation defined by $$(x,y)T(z,w)\Leftrightarrow x^{2}+3y^{2}=7z^{2}+w^{2}\text{, for all } (x,y), (z,w) \in\mathbb{R}^{2}$$. $$T[(0,0)]=\{(z,w)\in\mathbb{R}^{2}: (0,0)T(z,w)\}=(0,0). Because 0^{2}+3\cdot 0^{2}=7z^{2}+w^{2}\Rightarrow Z=0 and w=o $$. $$T[(3,4)]=\{(z,w)\in\mathbb{R}^{2}: (3,4)T(z,w)\}=\{ (z,\sqrt{57-7z^{2}}) \}$$. 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 $$(x,y)T(z,w)\Leftrightarrow x=z \vee y=w\text{, for all } (x,y), (z,w) \in\mathbb{R}^{2}$$. $$Z[(0,0)]=\{(z,w)\in\mathbb{R}^{2}: (0,0)Z(z,w)\}=\{(0, w) \vee (z,0)\}$$. $$Z[(3,4)]=\{(z,w)\in\mathbb{R}^{2}: (3,4)Z(z,w)\}=\{ (3,w) \vee (z,4) \}$$.

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