\displaystyle \textbf{Question 1: }\text{Let }A\text{ be the set of all human beings in a town at a particular time.}
\displaystyle \text{Determine whether each of the following relations is reflexive, symmetric and transitive:}
\displaystyle \text{(i) }R=\{(x,y):x\text{ and }y\text{ work at the same place}\}
\displaystyle \text{(ii) }R=\{(x,y):x\text{ and }y\text{ live in the same locality}\}
\displaystyle \text{(iii) }R=\{(x,y):x\text{ is the wife of }y\}
\displaystyle \text{(iv) }R=\{(x,y):x\text{ is the father of }y\}
\displaystyle \text{Answer:}

\displaystyle \text{(i) Given }R=\{(x,y):x\text{ and }y\text{ work at the same place}\}
\displaystyle \text{Reflexivity:}
\displaystyle \text{Since }A\text{ contains all human beings, some persons may not work at any place.}
\displaystyle \text{For such a person }x,\ (x,x)\notin R.
\displaystyle \therefore R\text{ is not reflexive.}
\displaystyle \text{Symmetry: Let }x,y\in A\text{ and suppose that }(x,y)\in R.
\displaystyle \Rightarrow x\text{ and }y\text{ work at the same place.}
\displaystyle \Rightarrow y\text{ and }x\text{ work at the same place.}
\displaystyle \Rightarrow (y,x)\in R.
\displaystyle \therefore R\text{ is symmetric.}
\displaystyle \text{Transitivity: Let }x,y,z\in A\text{ such that }(x,y)\in R\text{ and }(y,z)\in R.
\displaystyle \Rightarrow x\text{ and }y\text{ work at the same place, and }y\text{ and }z\text{ work at the same place.}
\displaystyle \Rightarrow x\text{ and }z\text{ work at the same place.}
\displaystyle \Rightarrow (x,z)\in R.
\displaystyle \therefore R\text{ is transitive.}
\displaystyle \text{Hence, }R\text{ is symmetric and transitive but not reflexive.}

\displaystyle \text{(ii) Given }R=\{(x,y):x\text{ and }y\text{ live in the same locality}\}
\displaystyle \text{Reflexivity: For every }x\in A,\ x\text{ lives in the same locality as itself.}
\displaystyle \Rightarrow (x,x)\in R\text{ for every }x\in A.
\displaystyle \therefore R\text{ is reflexive.}
\displaystyle \text{Symmetry: Let }x,y\in A\text{ and suppose that }(x,y)\in R.
\displaystyle \Rightarrow x\text{ and }y\text{ live in the same locality.}
\displaystyle \Rightarrow y\text{ and }x\text{ live in the same locality.}
\displaystyle \Rightarrow (y,x)\in R.
\displaystyle \therefore R\text{ is symmetric.}
\displaystyle \text{Transitivity: Let }x,y,z\in A\text{ such that }(x,y)\in R\text{ and }(y,z)\in R.
\displaystyle \Rightarrow x\text{ and }y\text{ live in the same locality, and }y\text{ and }z\text{ live in the same locality.}
\displaystyle \Rightarrow x\text{ and }z\text{ live in the same locality.}
\displaystyle \Rightarrow (x,z)\in R.
\displaystyle \therefore R\text{ is transitive.}
\displaystyle \text{Hence, }R\text{ is reflexive, symmetric and transitive.}

\displaystyle \text{(iii) Given }R=\{(x,y):x\text{ is the wife of }y\}
\displaystyle \text{Reflexivity: No person can be the wife of oneself.}
\displaystyle \Rightarrow (x,x)\notin R\text{ for every }x\in A.
\displaystyle \therefore R\text{ is not reflexive.}
\displaystyle \text{Symmetry: Let }(x,y)\in R.
\displaystyle \Rightarrow x\text{ is the wife of }y.
\displaystyle \text{However, }y\text{ is not the wife of }x.
\displaystyle \Rightarrow (y,x)\notin R.
\displaystyle \therefore R\text{ is not symmetric.}
\displaystyle \text{Transitivity: Suppose that }(x,y)\in R.
\displaystyle \Rightarrow x\text{ is the wife of }y,\text{ so }y\text{ is the husband of }x.
\displaystyle \text{Therefore, }y\text{ cannot also be the wife of some }z.
\displaystyle \text{Hence, }(x,y)\in R\text{ and }(y,z)\in R\text{ cannot occur simultaneously.}
\displaystyle \therefore R\text{ is transitive vacuously.}
\displaystyle \text{Hence, }R\text{ is transitive but neither reflexive nor symmetric.}

\displaystyle \text{(iv) Given }R=\{(x,y):x\text{ is the father of }y\}
\displaystyle \text{Reflexivity: No person can be the father of oneself.}
\displaystyle \Rightarrow (x,x)\notin R\text{ for every }x\in A.
\displaystyle \therefore R\text{ is not reflexive.}
\displaystyle \text{Symmetry: Let }(x,y)\in R.
\displaystyle \Rightarrow x\text{ is the father of }y.
\displaystyle \text{Therefore, }y\text{ cannot be the father of }x.
\displaystyle \Rightarrow (y,x)\notin R.
\displaystyle \therefore R\text{ is not symmetric.}
\displaystyle \text{Transitivity: Let }(x,y)\in R\text{ and }(y,z)\in R.
\displaystyle \Rightarrow x\text{ is the father of }y\text{ and }y\text{ is the father of }z.
\displaystyle \Rightarrow x\text{ is the grandfather of }z,\text{ and need not be the father of }z.
\displaystyle \Rightarrow (x,z)\notin R\text{ in general.}
\displaystyle \therefore R\text{ is not transitive.}
\displaystyle \text{Hence, }R\text{ is neither reflexive, symmetric nor transitive.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Relations }R_1,R_2,R_3\text{ and }R_4\text{ are defined on the set}
\displaystyle A=\{a,b,c\}\text{ as follows:}
\displaystyle R_1=\{(a,a),(a,b),(a,c),(b,b),(b,c),(c,a),(c,b),(c,c)\}
\displaystyle R_2=\{(a,a)\}
\displaystyle R_3=\{(b,c)\}
\displaystyle R_4=\{(a,b),(b,c),(c,a)\}
\displaystyle \text{Determine whether each relation is (i) reflexive, (ii) symmetric and (iii) transitive.}
\displaystyle \text{Answer:}
\displaystyle \text{For }R_1=\{(a,a),(a,b),(a,c),(b,b),(b,c),(c,a),(c,b),(c,c)\}:
\displaystyle \text{Since }(a,a),(b,b),(c,c)\in R_1,\ R_1\text{ is reflexive.}
\displaystyle \text{Since }(a,b)\in R_1\text{ but }(b,a)\notin R_1,\ R_1\text{ is not symmetric.}
\displaystyle \text{Also, }(b,c)\in R_1\text{ and }(c,a)\in R_1\text{ but }(b,a)\notin R_1.
\displaystyle \therefore R_1\text{ is not transitive.}
\displaystyle \text{Hence, }R_1\text{ is reflexive but neither symmetric nor transitive.}
\displaystyle \text{For }R_2=\{(a,a)\}:
\displaystyle \text{Since }(b,b)\notin R_2\text{ and }(c,c)\notin R_2,\ R_2\text{ is not reflexive.}
\displaystyle \text{The only ordered pair is }(a,a),\text{ whose reverse is also }(a,a)\in R_2.
\displaystyle \therefore R_2\text{ is symmetric.}
\displaystyle \text{Also, }(a,a)\in R_2\text{ and }(a,a)\in R_2\Rightarrow(a,a)\in R_2.
\displaystyle \therefore R_2\text{ is transitive.}
\displaystyle \text{Hence, }R_2\text{ is symmetric and transitive but not reflexive.}
\displaystyle \text{For }R_3=\{(b,c)\}:
\displaystyle \text{Since }(a,a),(b,b),(c,c)\notin R_3,\ R_3\text{ is not reflexive.}
\displaystyle \text{Since }(b,c)\in R_3\text{ but }(c,b)\notin R_3,\ R_3\text{ is not symmetric.}
\displaystyle \text{There are no pairs }(x,y),(y,z)\in R_3\text{ having the second component of the}
\displaystyle \text{first pair equal to the first component of the second pair.}
\displaystyle \therefore R_3\text{ is transitive vacuously.}
\displaystyle \text{Hence, }R_3\text{ is transitive but neither reflexive nor symmetric.}
\displaystyle \text{For }R_4=\{(a,b),(b,c),(c,a)\}:
\displaystyle \text{Since }(a,a),(b,b),(c,c)\notin R_4,\ R_4\text{ is not reflexive.}
\displaystyle \text{Since }(a,b)\in R_4\text{ but }(b,a)\notin R_4,\ R_4\text{ is not symmetric.}
\displaystyle \text{Also, }(a,b)\in R_4\text{ and }(b,c)\in R_4\text{ but }(a,c)\notin R_4.
\displaystyle \therefore R_4\text{ is not transitive.}
\displaystyle \text{Hence, }R_4\text{ is neither reflexive, symmetric nor transitive.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Test whether the following relations }R_1,R_2\text{ and }R_3\text{ are}
\displaystyle \text{(i) reflexive, (ii) symmetric and (iii) transitive:}
\displaystyle \text{(i) }R_1\text{ on }\mathbb{Q}_0\text{ defined by }(a,b)\in R_1\Leftrightarrow a=\frac{1}{b}
\displaystyle \text{(ii) }R_2\text{ on }\mathbb{Z}\text{ defined by }(a,b)\in R_2\Leftrightarrow |a-b|\leq5
\displaystyle \text{(iii) }R_3\text{ on }\mathbb{R}\text{ defined by }(a,b)\in R_3\Leftrightarrow a^2-4ab+3b^2=0
\displaystyle \text{Answer:}

\displaystyle \text{(i) }R_1\text{ on }\mathbb{Q}_0\text{ is defined by }(a,b)\in R_1\Leftrightarrow a=\frac{1}{b}.
\displaystyle \text{Reflexivity: For }R_1\text{ to be reflexive, }a=\frac{1}{a}\text{ must hold for every }a\in\mathbb{Q}_0.
\displaystyle a=\frac{1}{a}\Rightarrow a^2=1\Rightarrow a=\pm1.
\displaystyle \text{Thus, }(a,a)\in R_1\text{ only for }a=\pm1,\text{ and not for every }a\in\mathbb{Q}_0.
\displaystyle \therefore R_1\text{ is not reflexive.}
\displaystyle \text{Symmetry: Let }(a,b)\in R_1.
\displaystyle \Rightarrow a=\frac{1}{b}\Rightarrow ab=1\Rightarrow b=\frac{1}{a}.
\displaystyle \Rightarrow (b,a)\in R_1.
\displaystyle \therefore R_1\text{ is symmetric.}
\displaystyle \text{Transitivity: We have }\left(2,\frac{1}{2}\right)\in R_1\text{ and }\left(\frac{1}{2},2\right)\in R_1.
\displaystyle \text{But }(2,2)\notin R_1,\text{ since }2\neq\frac{1}{2}.
\displaystyle \therefore R_1\text{ is not transitive.}
\displaystyle \text{Hence, }R_1\text{ is symmetric but neither reflexive nor transitive.}

\displaystyle \text{(ii) }R_2\text{ on }\mathbb{Z}\text{ is defined by }(a,b)\in R_2\Leftrightarrow |a-b|\leq5.
\displaystyle \text{Reflexivity: For every }a\in\mathbb{Z},
\displaystyle |a-a|=0\leq5.
\displaystyle \Rightarrow (a,a)\in R_2\text{ for every }a\in\mathbb{Z}.
\displaystyle \therefore R_2\text{ is reflexive.}
\displaystyle \text{Symmetry: Let }(a,b)\in R_2.
\displaystyle \Rightarrow |a-b|\leq5.
\displaystyle \text{Since }|b-a|=|a-b|,\ |b-a|\leq5.
\displaystyle \Rightarrow (b,a)\in R_2.
\displaystyle \therefore R_2\text{ is symmetric.}
\displaystyle \text{Transitivity: We have }(1,3)\in R_2\text{ and }(3,7)\in R_2,
\displaystyle \text{since }|1-3|=2\leq5\text{ and }|3-7|=4\leq5.
\displaystyle \text{But }|1-7|=6>5.
\displaystyle \Rightarrow (1,7)\notin R_2.
\displaystyle \therefore R_2\text{ is not transitive.}
\displaystyle \text{Hence, }R_2\text{ is reflexive and symmetric but not transitive.}

\displaystyle \text{(iii) }R_3\text{ on }\mathbb{R}\text{ is defined by }(a,b)\in R_3
\displaystyle \Leftrightarrow a^2-4ab+3b^2=0.
\displaystyle \text{Reflexivity: For every }a\in\mathbb{R},
\displaystyle a^2-4a^2+3a^2=0.
\displaystyle \Rightarrow (a,a)\in R_3\text{ for every }a\in\mathbb{R}.
\displaystyle \therefore R_3\text{ is reflexive.}
\displaystyle \text{Symmetry: We have }(3,1)\in R_3,\text{ since }3^2-4(3)(1)+3(1)^2=0.
\displaystyle \text{But }1^2-4(1)(3)+3(3)^2=16\neq0.
\displaystyle \Rightarrow (1,3)\notin R_3.
\displaystyle \therefore R_3\text{ is not symmetric.}
\displaystyle \text{Transitivity: We have }(9,3)\in R_3,\text{ since }9^2-4(9)(3)+3(3)^2=0.
\displaystyle \text{Also, }(3,1)\in R_3,\text{ since }3^2-4(3)(1)+3(1)^2=0.
\displaystyle \text{But }9^2-4(9)(1)+3(1)^2=48\neq0.
\displaystyle \Rightarrow (9,1)\notin R_3.
\displaystyle \therefore R_3\text{ is not transitive.}
\displaystyle \text{Hence, }R_3\text{ is reflexive but neither symmetric nor transitive.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Let }A=\{1,2,3\}\text{ and let}
\displaystyle R_1=\{(1,1),(1,3),(3,1),(2,2),(2,1),(3,3)\},
\displaystyle R_2=\{(2,2),(3,1),(1,3)\}\text{ and }R_3=\{(1,3),(3,3)\}.
\displaystyle \text{Determine whether each relation on }A\text{ is (i) reflexive, (ii) symmetric and}
\displaystyle \text{(iii) transitive.}
\displaystyle \text{Answer:}

\displaystyle \text{For }R_1=\{(1,1),(1,3),(3,1),(2,2),(2,1),(3,3)\}:
\displaystyle \text{Since }(1,1),(2,2),(3,3)\in R_1,\ R_1\text{ is reflexive.}
\displaystyle \text{Since }(2,1)\in R_1\text{ but }(1,2)\notin R_1,\ R_1\text{ is not symmetric.}
\displaystyle \text{Also, }(2,1)\in R_1\text{ and }(1,3)\in R_1\text{ but }(2,3)\notin R_1.
\displaystyle \therefore R_1\text{ is not transitive.}
\displaystyle \text{Hence, }R_1\text{ is reflexive but neither symmetric nor transitive.}

\displaystyle \text{For }R_2=\{(2,2),(3,1),(1,3)\}:
\displaystyle \text{Since }(1,1)\notin R_2\text{ and }(3,3)\notin R_2,\ R_2\text{ is not reflexive.}
\displaystyle \text{The reverse of }(1,3)\text{ is }(3,1),\text{ and both belong to }R_2.
\displaystyle \text{Also, the reverse of }(2,2)\text{ is }(2,2)\in R_2.
\displaystyle \therefore R_2\text{ is symmetric.}
\displaystyle \text{Since }(1,3)\in R_2\text{ and }(3,1)\in R_2\text{ but }(1,1)\notin R_2,
\displaystyle R_2\text{ is not transitive.}
\displaystyle \text{Hence, }R_2\text{ is symmetric but neither reflexive nor transitive.}

\displaystyle \text{For }R_3=\{(1,3),(3,3)\}:
\displaystyle \text{Since }(1,1)\notin R_3\text{ and }(2,2)\notin R_3,\ R_3\text{ is not reflexive.}
\displaystyle \text{Since }(1,3)\in R_3\text{ but }(3,1)\notin R_3,\ R_3\text{ is not symmetric.}
\displaystyle \text{For transitivity, the possible composable pairs are:}
\displaystyle (1,3)\in R_3\text{ and }(3,3)\in R_3\Rightarrow(1,3)\in R_3,
\displaystyle (3,3)\in R_3\text{ and }(3,3)\in R_3\Rightarrow(3,3)\in R_3.
\displaystyle \therefore R_3\text{ is transitive.}
\displaystyle \text{Hence, }R_3\text{ is transitive but neither reflexive nor symmetric.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The following relations are defined on the set of real numbers:}
\displaystyle \text{(i) }aRb\text{ if }a-b>0\qquad\text{(ii) }aRb\text{ if and only if }1+ab>0
\displaystyle \text{(iii) }aRb\text{ if }|a|\leq b.
\displaystyle \text{Determine whether these relations are reflexive, symmetric or transitive.}
\displaystyle \text{Answer:}

\displaystyle \text{(i) The relation is defined by }aRb\Leftrightarrow a-b>0.
\displaystyle \text{Reflexivity: For any }a\in\mathbb{R},
\displaystyle a-a=0\ngtr0.
\displaystyle \Rightarrow (a,a)\notin R.
\displaystyle \therefore R\text{ is not reflexive.}
\displaystyle \text{Symmetry: Let }(a,b)\in R.
\displaystyle \Rightarrow a-b>0\Rightarrow b-a<0.
\displaystyle \Rightarrow (b,a)\notin R.
\displaystyle \therefore R\text{ is not symmetric.}
\displaystyle \text{Transitivity: Let }(a,b)\in R\text{ and }(b,c)\in R.
\displaystyle \Rightarrow a-b>0\text{ and }b-c>0.
\displaystyle \text{Adding, }a-b+b-c>0.
\displaystyle \Rightarrow a-c>0\Rightarrow(a,c)\in R.
\displaystyle \therefore R\text{ is transitive.}
\displaystyle \text{Hence, the relation is transitive but neither reflexive nor symmetric.}

\displaystyle \text{(ii) The relation is defined by }aRb\Leftrightarrow1+ab>0.
\displaystyle \text{Reflexivity: For every }a\in\mathbb{R},
\displaystyle 1+a^2>0,\text{ since }a^2\geq0.
\displaystyle \Rightarrow(a,a)\in R\text{ for every }a\in\mathbb{R}.
\displaystyle \therefore R\text{ is reflexive.}
\displaystyle \text{Symmetry: Let }(a,b)\in R.
\displaystyle \Rightarrow1+ab>0.
\displaystyle \text{Since }ab=ba,\ 1+ba>0.
\displaystyle \Rightarrow(b,a)\in R.
\displaystyle \therefore R\text{ is symmetric.}
\displaystyle \text{Transitivity: Take }a=1,\ b=0\text{ and }c=-2.
\displaystyle 1+ab=1+(1)(0)=1>0\Rightarrow(1,0)\in R.
\displaystyle 1+bc=1+(0)(-2)=1>0\Rightarrow(0,-2)\in R.
\displaystyle \text{But }1+ac=1+(1)(-2)=-1\ngtr0.
\displaystyle \Rightarrow(1,-2)\notin R.
\displaystyle \therefore R\text{ is not transitive.}
\displaystyle \text{Hence, the relation is reflexive and symmetric but not transitive.}

\displaystyle \text{(iii) The relation is defined by }aRb\Leftrightarrow|a|\leq b.
\displaystyle \text{Reflexivity: For }a=-1,
\displaystyle |-1|=1\nleq-1.
\displaystyle \Rightarrow(-1,-1)\notin R.
\displaystyle \therefore R\text{ is not reflexive.}
\displaystyle \text{Symmetry: We have }|1|=1\leq2.
\displaystyle \Rightarrow(1,2)\in R.
\displaystyle \text{But }|2|=2\nleq1.
\displaystyle \Rightarrow(2,1)\notin R.
\displaystyle \therefore R\text{ is not symmetric.}
\displaystyle \text{Transitivity: Let }(a,b)\in R\text{ and }(b,c)\in R.
\displaystyle \Rightarrow|a|\leq b\text{ and }|b|\leq c.
\displaystyle \text{Since }|a|\geq0\text{ and }|a|\leq b,\text{ we have }b\geq0.
\displaystyle \Rightarrow|b|=b.
\displaystyle \text{Therefore, }|a|\leq b=|b|\leq c.
\displaystyle \Rightarrow|a|\leq c\Rightarrow(a,c)\in R.
\displaystyle \therefore R\text{ is transitive.}
\displaystyle \text{Hence, the relation is transitive but neither reflexive nor symmetric.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Check whether the relation }R\text{ defined on the set}
\displaystyle A=\{1,2,3,4,5,6\}\text{ by }R=\{(a,b):b=a+1\}\text{ is reflexive, symmetric}
\displaystyle \text{or transitive.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }A=\{1,2,3,4,5,6\}\text{ and }R=\{(a,b):b=a+1\}.
\displaystyle \therefore R=\{(1,2),(2,3),(3,4),(4,5),(5,6)\}.
\displaystyle \text{Reflexivity:}
\displaystyle \text{For every }a\in A,\ (a,a)\notin R,\text{ since }a\neq a+1.
\displaystyle \text{In particular, }(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\notin R.
\displaystyle \therefore R\text{ is not reflexive.}
\displaystyle \text{Symmetry:}
\displaystyle \text{We have }(1,2)\in R\text{ but }(2,1)\notin R.
\displaystyle \therefore R\text{ is not symmetric.}
\displaystyle \text{Transitivity:}
\displaystyle \text{We have }(1,2)\in R\text{ and }(2,3)\in R.
\displaystyle \text{But }(1,3)\notin R.
\displaystyle \therefore R\text{ is not transitive.}
\displaystyle \text{Hence, }R\text{ is neither reflexive, symmetric nor transitive.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Check whether the relation }R\text{ on }\mathbb{R}\text{ defined by}
\displaystyle R=\{(a,b):a\leq b^3\}\text{ is reflexive, symmetric or transitive.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }R=\{(a,b):a\leq b^3\}.
\displaystyle \text{Reflexivity:}
\displaystyle \text{For }a=\frac{1}{2},
\displaystyle \frac{1}{2}>\left(\frac{1}{2}\right)^3=\frac{1}{8}.
\displaystyle \Rightarrow\left(\frac{1}{2},\frac{1}{2}\right)\notin R.
\displaystyle \therefore R\text{ is not reflexive.}
\displaystyle \text{Symmetry:}
\displaystyle (1,2)\in R,\text{ since }1\leq2^3=8.
\displaystyle \text{But }(2,1)\notin R,\text{ since }2\nleq1^3=1.
\displaystyle \therefore R\text{ is not symmetric.}
\displaystyle \text{Transitivity:}
\displaystyle \left(3,\frac{3}{2}\right)\in R,\text{ since }3\leq\left(\frac{3}{2}\right)^3=\frac{27}{8}.
\displaystyle \left(\frac{3}{2},\frac{6}{5}\right)\in R,\text{ since }\frac{3}{2}\leq\left(\frac{6}{5}\right)^3=\frac{216}{125}.
\displaystyle \text{But }\left(3,\frac{6}{5}\right)\notin R,\text{ since }3>\left(\frac{6}{5}\right)^3=\frac{216}{125}.
\displaystyle \therefore R\text{ is not transitive.}
\displaystyle \text{Hence, }R\text{ is neither reflexive, symmetric nor transitive.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Prove that every identity relation on a set is reflexive, but the}
\displaystyle \text{converse is not necessarily true.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A\text{ be any set.}
\displaystyle \text{The identity relation on }A\text{ is defined as}
\displaystyle I_A=\{(x,x):x\in A\}.
\displaystyle \text{For every }x\in A,\ (x,x)\in I_A.
\displaystyle \therefore I_A\text{ is reflexive.}
\displaystyle \text{Hence, every identity relation on a set is reflexive.}
\displaystyle \text{Now, consider the converse.}
\displaystyle \text{Let }A=\{a,b,c\}\text{ and define}
\displaystyle R=\{(a,a),(b,b),(c,c),(a,b),(c,a)\}.
\displaystyle \text{Since }(a,a),(b,b),(c,c)\in R,\ R\text{ is reflexive on }A.
\displaystyle \text{However, }(a,b)\in R\text{ and }(c,a)\in R,\text{ where the two components are unequal.}
\displaystyle \text{Therefore, }R\neq I_A.
\displaystyle \therefore R\text{ is reflexive but is not an identity relation.}
\displaystyle \text{Hence, every identity relation is reflexive, but the converse is not necessarily true.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If }A=\{1,2,3,4\},\text{ define relations on }A\text{ which are}
\displaystyle \text{(i) reflexive and transitive but not symmetric,}
\displaystyle \text{(ii) symmetric but neither reflexive nor transitive,}
\displaystyle \text{(iii) reflexive, symmetric and transitive.}
\displaystyle \text{Answer:}

\displaystyle \text{(i) Let }R_1=\{(1,1),(2,2),(3,3),(4,4),(2,1)\}.
\displaystyle \text{Since }(1,1),(2,2),(3,3),(4,4)\in R_1,\ R_1\text{ is reflexive.}
\displaystyle \text{The only non-identity pair in }R_1\text{ is }(2,1).
\displaystyle (2,2)\in R_1\text{ and }(2,1)\in R_1\Rightarrow(2,1)\in R_1.
\displaystyle (2,1)\in R_1\text{ and }(1,1)\in R_1\Rightarrow(2,1)\in R_1.
\displaystyle \text{All other possible compositions involve identity pairs and remain in }R_1.
\displaystyle \therefore R_1\text{ is transitive.}
\displaystyle \text{However, }(2,1)\in R_1\text{ but }(1,2)\notin R_1.
\displaystyle \therefore R_1\text{ is not symmetric.}
\displaystyle \text{Hence, }R_1\text{ is reflexive and transitive but not symmetric.}

\displaystyle \text{(ii) Let }R_2=\{(1,2),(2,1),(2,3),(3,2)\}.
\displaystyle \text{Since }(1,1)\notin R_2,\ R_2\text{ is not reflexive.}
\displaystyle \text{The reverse of }(1,2)\text{ is }(2,1),\text{ and the reverse of }(2,3)\text{ is }(3,2).
\displaystyle \therefore R_2\text{ is symmetric.}
\displaystyle \text{However, }(1,2)\in R_2\text{ and }(2,3)\in R_2\text{ but }(1,3)\notin R_2.
\displaystyle \therefore R_2\text{ is not transitive.}
\displaystyle \text{Hence, }R_2\text{ is symmetric but neither reflexive nor transitive.}

\displaystyle \text{(iii) Let }R_3=\{(1,1),(2,2),(3,3),(4,4),(1,2),(2,1)\}.
\displaystyle \text{Since }(1,1),(2,2),(3,3),(4,4)\in R_3,\ R_3\text{ is reflexive.}
\displaystyle \text{Also, }(1,2)\in R_3\text{ and }(2,1)\in R_3.
\displaystyle \text{Every ordered pair in }R_3\text{ has its reverse in }R_3.
\displaystyle \therefore R_3\text{ is symmetric.}
\displaystyle (1,2)\in R_3\text{ and }(2,1)\in R_3\Rightarrow(1,1)\in R_3.
\displaystyle (2,1)\in R_3\text{ and }(1,2)\in R_3\Rightarrow(2,2)\in R_3.
\displaystyle \text{All other possible compositions also belong to }R_3.
\displaystyle \therefore R_3\text{ is transitive.}
\displaystyle \text{Hence, }R_3\text{ is reflexive, symmetric and transitive.}
\displaystyle \therefore R_3\text{ is an equivalence relation on }A.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Let }R\text{ be a relation defined on the set of natural numbers }\mathbb{N}
\displaystyle \text{as }R=\{(x,y):x,y\in\mathbb{N},\ 2x+y=41\}.
\displaystyle \text{Find the domain and range of }R.\text{ Also, verify whether }R\text{ is}
\displaystyle \text{(i) reflexive, (ii) symmetric and (iii) transitive.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }R=\{(x,y):x,y\in\mathbb{N},\ 2x+y=41\}.
\displaystyle 2x+y=41\Rightarrow y=41-2x.
\displaystyle \text{Since }x,y\in\mathbb{N},\ y\geq1.
\displaystyle 41-2x\geq1
\displaystyle \Rightarrow 2x\leq40\Rightarrow x\leq20.
\displaystyle \text{Also, }x\geq1.
\displaystyle \therefore \text{Domain of }R=\{1,2,3,\ldots,20\}.
\displaystyle \text{For }x=1,2,3,\ldots,20,\ y=41-2x.
\displaystyle \therefore \text{Range of }R=\{39,37,35,33,\ldots,5,3,1\}.
\displaystyle \text{Reflexivity:}
\displaystyle \text{For }x=2,\ 2(2)+2=6\neq41.
\displaystyle \Rightarrow(2,2)\notin R.
\displaystyle \therefore R\text{ is not reflexive.}
\displaystyle \text{Symmetry:}
\displaystyle (1,39)\in R,\text{ since }2(1)+39=41.
\displaystyle \text{But }(39,1)\notin R,\text{ since }2(39)+1=79\neq41.
\displaystyle \therefore R\text{ is not symmetric.}
\displaystyle \text{Transitivity:}
\displaystyle (15,11)\in R,\text{ since }2(15)+11=41.
\displaystyle (11,19)\in R,\text{ since }2(11)+19=41.
\displaystyle \text{But }(15,19)\notin R,\text{ since }2(15)+19=49\neq41.
\displaystyle \therefore R\text{ is not transitive.}
\displaystyle \text{Hence, }R\text{ is neither reflexive, symmetric nor transitive.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Is it true that every relation which is symmetric and transitive is also}
\displaystyle \text{reflexive? Give reasons.}
\displaystyle \text{Answer:}
\displaystyle \text{No, every relation which is symmetric and transitive need not be reflexive.}
\displaystyle \text{Consider the set }A=\{1,2\}\text{ and the relation }R=\{(1,1)\}\text{ on }A.
\displaystyle \text{Symmetry:}
\displaystyle \text{The only ordered pair in }R\text{ is }(1,1),\text{ whose reverse is also }(1,1)\in R.
\displaystyle \therefore R\text{ is symmetric.}
\displaystyle \text{Transitivity:}
\displaystyle (1,1)\in R\text{ and }(1,1)\in R\Rightarrow(1,1)\in R.
\displaystyle \therefore R\text{ is transitive.}
\displaystyle \text{Reflexivity:}
\displaystyle \text{For }R\text{ to be reflexive on }A,\text{ both }(1,1)\text{ and }(2,2)\text{ must belong to }R.
\displaystyle \text{But }(2,2)\notin R.
\displaystyle \therefore R\text{ is not reflexive.}
\displaystyle \text{Hence, a relation may be symmetric and transitive without being reflexive.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{An integer }m\text{ is said to be related to another integer }n\text{ if }m\text{ is a}
\displaystyle \text{multiple of }n.\text{ Check whether the relation is symmetric, reflexive and transitive.}
\displaystyle \text{Answer:}
\displaystyle \text{The relation }R\text{ on }\mathbb{Z}\text{ is defined by}
\displaystyle R=\{(m,n):m,n\in\mathbb{Z},\ m=kn\text{ for some }k\in\mathbb{Z}\}.
\displaystyle \text{Reflexivity:}
\displaystyle \text{For every }m\in\mathbb{Z},
\displaystyle m=1\cdot m,\text{ where }1\in\mathbb{Z}.
\displaystyle \Rightarrow(m,m)\in R.
\displaystyle \therefore R\text{ is reflexive.}
\displaystyle \text{Symmetry:}
\displaystyle (2,1)\in R,\text{ since }2=2\cdot1.
\displaystyle \text{But }(1,2)\notin R,\text{ since there is no }k\in\mathbb{Z}\text{ such that }1=2k.
\displaystyle \therefore R\text{ is not symmetric.}
\displaystyle \text{Transitivity:}
\displaystyle \text{Let }(m,n)\in R\text{ and }(n,p)\in R.
\displaystyle \Rightarrow m=kn\text{ and }n=lp\text{ for some }k,l\in\mathbb{Z}.
\displaystyle \Rightarrow m=k(lp)=(kl)p.
\displaystyle \text{Since }k,l\in\mathbb{Z},\ kl\in\mathbb{Z}.
\displaystyle \Rightarrow(m,p)\in R.
\displaystyle \therefore R\text{ is transitive.}
\displaystyle \text{Hence, }R\text{ is reflexive and transitive but not symmetric.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Show that the relation }``\geq"\text{ on the set }\mathbb{R}\text{ of all real numbers is}
\displaystyle \text{reflexive and transitive but not symmetric.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }S=\{(a,b):a,b\in\mathbb{R},\ a\geq b\}.
\displaystyle \text{Reflexivity:}
\displaystyle \text{For every }a\in\mathbb{R},\ a\geq a.
\displaystyle \Rightarrow(a,a)\in S.
\displaystyle \therefore S\text{ is reflexive.}
\displaystyle \text{Symmetry:}
\displaystyle \text{We have }(2,1)\in S,\text{ since }2\geq1.
\displaystyle \text{But }(1,2)\notin S,\text{ since }1\ngeq2.
\displaystyle \therefore S\text{ is not symmetric.}
\displaystyle \text{Transitivity:}
\displaystyle \text{Let }(a,b)\in S\text{ and }(b,c)\in S.
\displaystyle \Rightarrow a\geq b\text{ and }b\geq c.
\displaystyle \Rightarrow a\geq c.
\displaystyle \Rightarrow(a,c)\in S.
\displaystyle \therefore S\text{ is transitive.}
\displaystyle \text{Hence, the relation }``\geq"\text{ is reflexive and transitive but not symmetric.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Give an example of a relation which is}
\displaystyle \text{(i) reflexive and symmetric but not transitive,}
\displaystyle \text{(ii) reflexive and transitive but not symmetric,}
\displaystyle \text{(iii) symmetric and transitive but not reflexive,}
\displaystyle \text{(iv) symmetric but neither reflexive nor transitive,}
\displaystyle \text{(v) transitive but neither reflexive nor symmetric.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=\{1,2,3\}.

\displaystyle \text{(i) Reflexive and symmetric but not transitive:}
\displaystyle R_1=\{(1,1),(2,2),(3,3),(1,3),(3,1),(2,3),(3,2)\}.
\displaystyle \text{Since }(1,1),(2,2),(3,3)\in R_1,\ R_1\text{ is reflexive.}
\displaystyle \text{Also, }(1,3),(3,1)\in R_1\text{ and }(2,3),(3,2)\in R_1.
\displaystyle \text{Thus, the reverse of every ordered pair in }R_1\text{ also belongs to }R_1.
\displaystyle \therefore R_1\text{ is symmetric.}
\displaystyle \text{However, }(2,3)\in R_1\text{ and }(3,1)\in R_1,
\displaystyle \text{but }(2,1)\notin R_1.
\displaystyle \therefore R_1\text{ is not transitive.}

\displaystyle \text{(ii) Reflexive and transitive but not symmetric:}
\displaystyle R_2=\{(1,1),(2,2),(3,3),(1,2),(1,3),(2,3)\}.
\displaystyle \text{Since }(1,1),(2,2),(3,3)\in R_2,\ R_2\text{ is reflexive.}
\displaystyle \text{Also, }(1,2)\in R_2\text{ and }(2,3)\in R_2\Rightarrow(1,3)\in R_2.
\displaystyle \text{All other possible compositions involve identity pairs and remain in }R_2.
\displaystyle \therefore R_2\text{ is transitive.}
\displaystyle \text{However, }(1,2)\in R_2\text{ but }(2,1)\notin R_2.
\displaystyle \therefore R_2\text{ is not symmetric.}

\displaystyle \text{(iii) Symmetric and transitive but not reflexive:}
\displaystyle R_3=\{(1,1),(3,3),(1,3),(3,1)\}.
\displaystyle \text{Since }(2,2)\notin R_3,\ R_3\text{ is not reflexive on }A.
\displaystyle \text{Also, }(1,3)\in R_3\text{ and }(3,1)\in R_3.
\displaystyle \text{The diagonal pairs }(1,1)\text{ and }(3,3)\text{ are their own reverses.}
\displaystyle \therefore R_3\text{ is symmetric.}
\displaystyle (1,3)\in R_3\text{ and }(3,1)\in R_3\Rightarrow(1,1)\in R_3.
\displaystyle (3,1)\in R_3\text{ and }(1,3)\in R_3\Rightarrow(3,3)\in R_3.
\displaystyle \text{All other possible compositions also belong to }R_3.
\displaystyle \therefore R_3\text{ is transitive.}

\displaystyle \text{(iv) Symmetric but neither reflexive nor transitive:}
\displaystyle R_4=\{(1,1),(1,3),(3,1),(2,3),(3,2)\}.
\displaystyle \text{Since }(2,2),(3,3)\notin R_4,\ R_4\text{ is not reflexive.}
\displaystyle \text{Also, }(1,3),(3,1)\in R_4\text{ and }(2,3),(3,2)\in R_4.
\displaystyle \therefore R_4\text{ is symmetric.}
\displaystyle \text{However, }(2,3)\in R_4\text{ and }(3,1)\in R_4,
\displaystyle \text{but }(2,1)\notin R_4.
\displaystyle \therefore R_4\text{ is not transitive.}

\displaystyle \text{(v) Transitive but neither reflexive nor symmetric:}
\displaystyle R_5=\{(1,2),(2,3),(1,3)\}.
\displaystyle \text{Since }(1,1),(2,2),(3,3)\notin R_5,\ R_5\text{ is not reflexive.}
\displaystyle \text{Also, }(1,2)\in R_5\text{ but }(2,1)\notin R_5.
\displaystyle \therefore R_5\text{ is not symmetric.}
\displaystyle \text{The only non-trivial composable pairs are }(1,2)\text{ and }(2,3).
\displaystyle (1,2)\in R_5\text{ and }(2,3)\in R_5\Rightarrow(1,3)\in R_5.
\displaystyle \therefore R_5\text{ is transitive.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Given the relation }R=\{(1,2),(2,3)\}\text{ on the set }A=\{1,2,3\},
\displaystyle \text{add a minimum number of ordered pairs so that the enlarged relation is symmetric,}
\displaystyle \text{transitive and reflexive.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }A=\{1,2,3\}\text{ and }R=\{(1,2),(2,3)\}.
\displaystyle \text{To make }R\text{ reflexive, add }(1,1),(2,2)\text{ and }(3,3).
\displaystyle \text{Since }(1,2),(2,3)\in R,\text{ symmetry requires }(2,1)\text{ and }(3,2).
\displaystyle \text{Also, }(1,2)\in R\text{ and }(2,3)\in R.
\displaystyle \text{Therefore, transitivity requires }(1,3).
\displaystyle \text{Since }(1,3)\text{ is added, symmetry further requires }(3,1).
\displaystyle \text{Thus, the ordered pairs to be added are}
\displaystyle (1,1),(2,2),(3,3),(2,1),(3,2),(1,3),(3,1).
\displaystyle \text{The enlarged relation is}
\displaystyle R'=\{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)\}.
\displaystyle \text{Thus, }R'=A\times A.
\displaystyle \text{Hence, }R'\text{ is reflexive, symmetric and transitive.}
\displaystyle \therefore \text{The minimum number of ordered pairs to be added is }7.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Let }A=\{1,2,3\}\text{ and }R=\{(1,2),(1,1),(2,3)\}\text{ be a relation}
\displaystyle \text{on }A.\text{ What minimum number of ordered pairs may be added to }R\text{ so that it becomes}
\displaystyle \text{a transitive relation on }A?
\displaystyle \text{Answer:}
\displaystyle \text{Given }R=\{(1,2),(1,1),(2,3)\}.
\displaystyle \text{For transitivity, if }(a,b)\in R\text{ and }(b,c)\in R,\text{ then }(a,c)\in R.
\displaystyle \text{Here, }(1,2)\in R\text{ and }(2,3)\in R.
\displaystyle \therefore (1,3)\text{ must belong to }R.
\displaystyle \text{Add the ordered pair }(1,3)\text{ to }R.
\displaystyle R'=\{(1,1),(1,2),(1,3),(2,3)\}.
\displaystyle \text{Now, }(1,1)\in R'\text{ and }(1,2)\in R'\Rightarrow(1,2)\in R'.
\displaystyle (1,1)\in R'\text{ and }(1,3)\in R'\Rightarrow(1,3)\in R'.
\displaystyle (1,2)\in R'\text{ and }(2,3)\in R'\Rightarrow(1,3)\in R'.
\displaystyle \text{Thus, all possible compositions satisfy the condition of transitivity.}
\displaystyle \therefore R'\text{ is transitive.}
\displaystyle \text{Hence, the minimum number of ordered pairs to be added is }1.
\displaystyle \text{The ordered pair to be added is }(1,3).
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Let }A=\{a,b,c\}\text{ and the relation }R\text{ be defined on }A\text{ as}
\displaystyle R=\{(a,a),(b,c),(a,b)\}.\text{ Write the minimum number of ordered pairs to be added}
\displaystyle \text{in }R\text{ to make it reflexive and transitive.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }R=\{(a,a),(b,c),(a,b)\}\text{ on }A=\{a,b,c\}.
\displaystyle \text{To make }R\text{ reflexive, all diagonal pairs must belong to }R.
\displaystyle \text{Since }(b,b)\text{ and }(c,c)\notin R,\text{ add }(b,b)\text{ and }(c,c).
\displaystyle \text{For transitivity, }(a,b)\in R\text{ and }(b,c)\in R.
\displaystyle \therefore (a,c)\text{ must belong to }R.
\displaystyle \text{After adding }(a,c),\text{ all other possible compositions involve existing}
\displaystyle \text{or newly added reflexive pairs and satisfy transitivity.}
\displaystyle \text{Hence, the ordered pairs to be added are }(b,b),(c,c)\text{ and }(a,c).
\displaystyle \therefore \text{The minimum number of ordered pairs to be added is }3.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Each of the following defines a relation on }\mathbb{N}:
\displaystyle \text{(i) }x>y,\quad x,y\in\mathbb{N}
\displaystyle \text{(ii) }x+y=10,\quad x,y\in\mathbb{N}
\displaystyle \text{(iii) }xy\text{ is the square of an integer},\quad x,y\in\mathbb{N}
\displaystyle \text{(iv) }x+4y=10,\quad x,y\in\mathbb{N}.
\displaystyle \text{Determine which of the above relations are reflexive, symmetric and transitive.}
\displaystyle \text{Answer:}

\displaystyle \text{(i) Let }R_1=\{(x,y):x,y\in\mathbb{N},\ x>y\}.
\displaystyle \text{For every }x\in\mathbb{N},\ x\nobreak>x\text{ is false.}
\displaystyle \Rightarrow(x,x)\notin R_1.
\displaystyle \therefore R_1\text{ is not reflexive.}
\displaystyle \text{Also, }(2,1)\in R_1\text{ but }(1,2)\notin R_1.
\displaystyle \therefore R_1\text{ is not symmetric.}
\displaystyle \text{Let }(x,y)\in R_1\text{ and }(y,z)\in R_1.
\displaystyle \Rightarrow x>y\text{ and }y>z.
\displaystyle \Rightarrow x>z.
\displaystyle \Rightarrow(x,z)\in R_1.
\displaystyle \therefore R_1\text{ is transitive.}
\displaystyle \text{Hence, }R_1\text{ is transitive but neither reflexive nor symmetric.}

\displaystyle \text{(ii) Let }R_2=\{(x,y):x,y\in\mathbb{N},\ x+y=10\}.
\displaystyle R_2=\{(1,9),(2,8),(3,7),(4,6),(5,5),(6,4),(7,3),(8,2),(9,1)\}.
\displaystyle \text{Since }(1,1)\notin R_2,\ R_2\text{ is not reflexive.}
\displaystyle \text{Let }(x,y)\in R_2.
\displaystyle \Rightarrow x+y=10.
\displaystyle \Rightarrow y+x=10.
\displaystyle \Rightarrow(y,x)\in R_2.
\displaystyle \therefore R_2\text{ is symmetric.}
\displaystyle \text{However, }(1,9)\in R_2\text{ and }(9,1)\in R_2,
\displaystyle \text{but }(1,1)\notin R_2.
\displaystyle \therefore R_2\text{ is not transitive.}
\displaystyle \text{Hence, }R_2\text{ is symmetric but neither reflexive nor transitive.}

\displaystyle \text{(iii) Let }R_3=\{(x,y):x,y\in\mathbb{N},\ xy\text{ is a perfect square}\}.
\displaystyle \text{For every }x\in\mathbb{N},\ xx=x^2\text{ is a perfect square.}
\displaystyle \Rightarrow(x,x)\in R_3.
\displaystyle \therefore R_3\text{ is reflexive.}
\displaystyle \text{Let }(x,y)\in R_3.
\displaystyle \Rightarrow xy\text{ is a perfect square.}
\displaystyle \text{Since }xy=yx,\ yx\text{ is also a perfect square.}
\displaystyle \Rightarrow(y,x)\in R_3.
\displaystyle \therefore R_3\text{ is symmetric.}
\displaystyle \text{Let }(x,y)\in R_3\text{ and }(y,z)\in R_3.
\displaystyle \Rightarrow xy\text{ and }yz\text{ are perfect squares.}
\displaystyle \text{For every prime }p,\text{ let the exponents of }p\text{ in }x,y,z\text{ be }\alpha,\beta,\gamma.
\displaystyle \text{Since }xy\text{ is a perfect square, }\alpha+\beta\text{ is even.}
\displaystyle \text{Since }yz\text{ is a perfect square, }\beta+\gamma\text{ is even.}
\displaystyle \Rightarrow\alpha\text{ and }\gamma\text{ have the same parity.}
\displaystyle \Rightarrow\alpha+\gamma\text{ is even.}
\displaystyle \text{Thus, the exponent of every prime in }xz\text{ is even.}
\displaystyle \Rightarrow xz\text{ is a perfect square.}
\displaystyle \Rightarrow(x,z)\in R_3.
\displaystyle \therefore R_3\text{ is transitive.}
\displaystyle \text{Hence, }R_3\text{ is reflexive, symmetric and transitive.}

\displaystyle \text{(iv) Let }R_4=\{(x,y):x,y\in\mathbb{N},\ x+4y=10\}.
\displaystyle \text{For }y=1,\ x=6,\text{ and for }y=2,\ x=2.
\displaystyle \therefore R_4=\{(6,1),(2,2)\}.
\displaystyle \text{Since }(1,1)\notin R_4,\ R_4\text{ is not reflexive.}
\displaystyle \text{Also, }(6,1)\in R_4\text{ but }(1,6)\notin R_4.
\displaystyle \therefore R_4\text{ is not symmetric.}
\displaystyle \text{The only composable ordered pairs in }R_4\text{ are }(2,2)\text{ and }(2,2).
\displaystyle (2,2)\in R_4\text{ and }(2,2)\in R_4\Rightarrow(2,2)\in R_4.
\displaystyle \text{There is no ordered pair in }R_4\text{ whose first component is }1.
\displaystyle \text{Hence, }(6,1)\text{ forms no composable chain with another ordered pair.}
\displaystyle \therefore R_4\text{ is transitive.}
\displaystyle \text{Hence, }R_4\text{ is transitive but neither reflexive nor symmetric.}
\displaystyle \\

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