\displaystyle \textbf{Question 1: }\text{Show that the relation }R\text{ defined by}
\displaystyle R=\{(a,b):a-b\text{ is divisible by }3;\ a,b\in\mathbb{Z}\}
\displaystyle \text{is an equivalence relation.}
\displaystyle \text{Answer:}
\displaystyle \text{To prove that }R\text{ is an equivalence relation, we show that it is reflexive,}
\displaystyle \text{symmetric and transitive.}

\displaystyle \text{Reflexivity:}
\displaystyle \text{Let }a\in\mathbb{Z}\text{ be arbitrary. Then}
\displaystyle a-a=0=3\cdot0.
\displaystyle \therefore a-a\text{ is divisible by }3.
\displaystyle \Rightarrow(a,a)\in R.
\displaystyle \therefore R\text{ is reflexive.}

\displaystyle \text{Symmetry:}
\displaystyle \text{Let }a,b\in\mathbb{Z}\text{ such that }(a,b)\in R.
\displaystyle \Rightarrow a-b\text{ is divisible by }3.
\displaystyle \Rightarrow a-b=3p\text{ for some }p\in\mathbb{Z}.
\displaystyle \Rightarrow b-a=-(a-b)=-3p=3(-p).
\displaystyle \text{Since }p\in\mathbb{Z},\ -p\in\mathbb{Z}.
\displaystyle \therefore b-a\text{ is divisible by }3.
\displaystyle \Rightarrow(b,a)\in R.
\displaystyle \therefore R\text{ is symmetric.}

\displaystyle \text{Transitivity:}
\displaystyle \text{Let }a,b,c\in\mathbb{Z}\text{ such that }(a,b)\in R\text{ and }(b,c)\in R.
\displaystyle \Rightarrow a-b\text{ and }b-c\text{ are divisible by }3.
\displaystyle \Rightarrow a-b=3p\text{ and }b-c=3q\text{ for some }p,q\in\mathbb{Z}.
\displaystyle \Rightarrow a-c=(a-b)+(b-c)
\displaystyle \phantom{\Rightarrow a-c}=3p+3q
\displaystyle \phantom{\Rightarrow a-c}=3(p+q).
\displaystyle \text{Since }p,q\in\mathbb{Z},\ p+q\in\mathbb{Z}.
\displaystyle \therefore a-c\text{ is divisible by }3.
\displaystyle \Rightarrow(a,c)\in R.
\displaystyle \therefore R\text{ is transitive.}

\displaystyle \text{Hence, }R\text{ is reflexive, symmetric and transitive.}
\displaystyle \therefore R\text{ is an equivalence relation on }\mathbb{Z}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Show that the relation }R\text{ on the set }\mathbb{Z}\text{ of integers, given by}
\displaystyle R=\{(a,b):2\text{ divides }a-b\},\text{ is an equivalence relation.}
\displaystyle \text{Answer:}
\displaystyle \text{To prove that }R\text{ is an equivalence relation, we show that it is reflexive,}
\displaystyle \text{symmetric and transitive.}

\displaystyle \text{Reflexivity:}
\displaystyle \text{Let }a\in\mathbb{Z}\text{ be arbitrary. Then}
\displaystyle a-a=0=2\cdot0.
\displaystyle \therefore 2\text{ divides }(a-a).
\displaystyle \Rightarrow(a,a)\in R\text{ for all }a\in\mathbb{Z}.
\displaystyle \therefore R\text{ is reflexive on }\mathbb{Z}.

\displaystyle \text{Symmetry:}
\displaystyle \text{Let }(a,b)\in R.
\displaystyle \Rightarrow 2\text{ divides }(a-b).
\displaystyle \Rightarrow a-b=2p\text{ for some }p\in\mathbb{Z}.
\displaystyle \Rightarrow b-a=-(a-b)=-2p=2(-p).
\displaystyle \text{Since }-p\in\mathbb{Z},\ 2\text{ divides }(b-a).
\displaystyle \Rightarrow(b,a)\in R.
\displaystyle \therefore R\text{ is symmetric on }\mathbb{Z}.

\displaystyle \text{Transitivity:}
\displaystyle \text{Let }(a,b)\in R\text{ and }(b,c)\in R.
\displaystyle \Rightarrow a-b=2p\text{ and }b-c=2q\text{ for some }p,q\in\mathbb{Z}.
\displaystyle \Rightarrow a-c=(a-b)+(b-c)
\displaystyle \phantom{\Rightarrow a-c}=2p+2q
\displaystyle \phantom{\Rightarrow a-c}=2(p+q).
\displaystyle \text{Since }p+q\in\mathbb{Z},\ 2\text{ divides }(a-c).
\displaystyle \Rightarrow(a,c)\in R.
\displaystyle \therefore R\text{ is transitive on }\mathbb{Z}.

\displaystyle \text{Hence, }R\text{ is reflexive, symmetric and transitive.}
\displaystyle \therefore R\text{ is an equivalence relation on }\mathbb{Z}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Prove that the relation }R\text{ on }\mathbb{Z}\text{ defined by}
\displaystyle (a,b)\in R\iff a-b\text{ is divisible by }5\text{ is an equivalence relation on }\mathbb{Z}.
\displaystyle \text{Answer:}
\displaystyle \text{To prove that }R\text{ is an equivalence relation, we show that it is reflexive,}
\displaystyle \text{symmetric and transitive.}

\displaystyle \text{Reflexivity:}
\displaystyle \text{Let }a\in\mathbb{Z}\text{ be arbitrary. Then}
\displaystyle a-a=0=5\cdot0.
\displaystyle \therefore a-a\text{ is divisible by }5.
\displaystyle \Rightarrow(a,a)\in R\text{ for all }a\in\mathbb{Z}.
\displaystyle \therefore R\text{ is reflexive on }\mathbb{Z}.

\displaystyle \text{Symmetry:}
\displaystyle \text{Let }(a,b)\in R.
\displaystyle \Rightarrow a-b\text{ is divisible by }5.
\displaystyle \Rightarrow a-b=5p\text{ for some }p\in\mathbb{Z}.
\displaystyle \Rightarrow b-a=-(a-b)=-5p=5(-p).
\displaystyle \text{Since }-p\in\mathbb{Z},\ b-a\text{ is divisible by }5.
\displaystyle \Rightarrow(b,a)\in R.
\displaystyle \therefore R\text{ is symmetric on }\mathbb{Z}.

\displaystyle \text{Transitivity:}
\displaystyle \text{Let }(a,b)\in R\text{ and }(b,c)\in R.
\displaystyle \Rightarrow a-b=5p\text{ and }b-c=5q\text{ for some }p,q\in\mathbb{Z}.
\displaystyle \Rightarrow a-c=(a-b)+(b-c)
\displaystyle \phantom{\Rightarrow a-c}=5p+5q
\displaystyle \phantom{\Rightarrow a-c}=5(p+q).
\displaystyle \text{Since }p+q\in\mathbb{Z},\ a-c\text{ is divisible by }5.
\displaystyle \Rightarrow(a,c)\in R.
\displaystyle \therefore R\text{ is transitive on }\mathbb{Z}.

\displaystyle \text{Hence, }R\text{ is reflexive, symmetric and transitive.}
\displaystyle \therefore R\text{ is an equivalence relation on }\mathbb{Z}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Let }n\text{ be a fixed positive integer. Define a relation }R\text{ on }\mathbb{Z}\text{ by}
\displaystyle (a,b)\in R\iff a-b\text{ is divisible by }n.\text{ Show that }R\text{ is an equivalence relation on }\mathbb{Z}.
\displaystyle \text{Answer:}
\displaystyle \text{To prove that }R\text{ is an equivalence relation, we show that it is reflexive,}
\displaystyle \text{symmetric and transitive.}

\displaystyle \text{Reflexivity:}
\displaystyle \text{Let }a\in\mathbb{Z}\text{ be arbitrary. Then}
\displaystyle a-a=0=n\cdot0.
\displaystyle \therefore a-a\text{ is divisible by }n.
\displaystyle \Rightarrow(a,a)\in R\text{ for all }a\in\mathbb{Z}.
\displaystyle \therefore R\text{ is reflexive on }\mathbb{Z}.

\displaystyle \text{Symmetry:}
\displaystyle \text{Let }(a,b)\in R.
\displaystyle \Rightarrow a-b\text{ is divisible by }n.
\displaystyle \Rightarrow a-b=np\text{ for some }p\in\mathbb{Z}.
\displaystyle \Rightarrow b-a=-(a-b)=-np=n(-p).
\displaystyle \text{Since }p\in\mathbb{Z},\ -p\in\mathbb{Z}.
\displaystyle \therefore b-a\text{ is divisible by }n.
\displaystyle \Rightarrow(b,a)\in R.
\displaystyle \therefore R\text{ is symmetric on }\mathbb{Z}.

\displaystyle \text{Transitivity:}
\displaystyle \text{Let }(a,b)\in R\text{ and }(b,c)\in R.
\displaystyle \Rightarrow a-b=np\text{ and }b-c=nq\text{ for some }p,q\in\mathbb{Z}.
\displaystyle \Rightarrow a-c=(a-b)+(b-c)
\displaystyle \phantom{\Rightarrow a-c}=np+nq
\displaystyle \phantom{\Rightarrow a-c}=n(p+q).
\displaystyle \text{Since }p+q\in\mathbb{Z},\ a-c\text{ is divisible by }n.
\displaystyle \Rightarrow(a,c)\in R.
\displaystyle \therefore R\text{ is transitive on }\mathbb{Z}.

\displaystyle \text{Hence, }R\text{ is reflexive, symmetric and transitive.}
\displaystyle \therefore R\text{ is an equivalence relation on }\mathbb{Z}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Let }\mathbb{Z}\text{ be the set of integers. Show that the relation}
\displaystyle R=\{(a,b):a,b\in\mathbb{Z}\text{ and }a+b\text{ is even}\}
\displaystyle \text{is an equivalence relation on }\mathbb{Z}.
\displaystyle \text{Answer:}
\displaystyle \text{To prove that }R\text{ is an equivalence relation, we show that it is reflexive,}
\displaystyle \text{symmetric and transitive.}

\displaystyle \text{Reflexivity:}
\displaystyle \text{Let }a\in\mathbb{Z}\text{ be arbitrary. Then}
\displaystyle a+a=2a.
\displaystyle \text{Since }a\in\mathbb{Z},\ 2a\text{ is even.}
\displaystyle \Rightarrow(a,a)\in R.
\displaystyle \therefore R\text{ is reflexive on }\mathbb{Z}.

\displaystyle \text{Symmetry:}
\displaystyle \text{Let }(a,b)\in R.
\displaystyle \Rightarrow a+b\text{ is even.}
\displaystyle \text{Since }a+b=b+a,\ b+a\text{ is also even.}
\displaystyle \Rightarrow(b,a)\in R.
\displaystyle \therefore R\text{ is symmetric on }\mathbb{Z}.

\displaystyle \text{Transitivity:}
\displaystyle \text{Let }(a,b)\in R\text{ and }(b,c)\in R.
\displaystyle \Rightarrow a+b\text{ and }b+c\text{ are even.}
\displaystyle \Rightarrow a+b=2x\text{ and }b+c=2y\text{ for some }x,y\in\mathbb{Z}.
\displaystyle \text{Adding the two equations, we get}
\displaystyle a+2b+c=2x+2y.
\displaystyle \Rightarrow a+c=2x+2y-2b
\displaystyle \phantom{\Rightarrow a+c}=2(x+y-b).
\displaystyle \text{Since }x,y,b\in\mathbb{Z},\ x+y-b\in\mathbb{Z}.
\displaystyle \therefore a+c\text{ is even.}
\displaystyle \Rightarrow(a,c)\in R.
\displaystyle \therefore R\text{ is transitive on }\mathbb{Z}.

\displaystyle \text{Hence, }R\text{ is reflexive, symmetric and transitive.}
\displaystyle \therefore R\text{ is an equivalence relation on }\mathbb{Z}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{An integer }m\text{ is said to be related to an integer }n\text{ if }m-n\text{ is divisible by }13.
\displaystyle \text{Does this define an equivalence relation?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }R=\{(m,n):m,n\in\mathbb{Z},\ m-n\text{ is divisible by }13\}.
\displaystyle \text{To prove that }R\text{ is an equivalence relation, we show that it is reflexive,}
\displaystyle \text{symmetric and transitive.}

\displaystyle \text{Reflexivity:}
\displaystyle \text{Let }m\in\mathbb{Z}\text{ be arbitrary. Then}
\displaystyle m-m=0=13\cdot0.
\displaystyle \therefore m-m\text{ is divisible by }13.
\displaystyle \Rightarrow(m,m)\in R\text{ for all }m\in\mathbb{Z}.
\displaystyle \therefore R\text{ is reflexive on }\mathbb{Z}.

\displaystyle \text{Symmetry:}
\displaystyle \text{Let }(m,n)\in R.
\displaystyle \Rightarrow m-n\text{ is divisible by }13.
\displaystyle \Rightarrow m-n=13p\text{ for some }p\in\mathbb{Z}.
\displaystyle \Rightarrow n-m=-(m-n)=-13p=13(-p).
\displaystyle \text{Since }-p\in\mathbb{Z},\ n-m\text{ is divisible by }13.
\displaystyle \Rightarrow(n,m)\in R.
\displaystyle \therefore R\text{ is symmetric on }\mathbb{Z}.

\displaystyle \text{Transitivity:}
\displaystyle \text{Let }(m,n)\in R\text{ and }(n,o)\in R.
\displaystyle \Rightarrow m-n=13p\text{ and }n-o=13q\text{ for some }p,q\in\mathbb{Z}.
\displaystyle \Rightarrow m-o=(m-n)+(n-o)
\displaystyle \phantom{\Rightarrow m-o}=13p+13q
\displaystyle \phantom{\Rightarrow m-o}=13(p+q).
\displaystyle \text{Since }p+q\in\mathbb{Z},\ m-o\text{ is divisible by }13.
\displaystyle \Rightarrow(m,o)\in R.
\displaystyle \therefore R\text{ is transitive on }\mathbb{Z}.

\displaystyle \text{Hence, }R\text{ is reflexive, symmetric and transitive.}
\displaystyle \therefore R\text{ is an equivalence relation on }\mathbb{Z}.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Let }R\text{ be a relation on the set }A\text{ of ordered pairs of non-zero integers}
\displaystyle \text{defined by }(x,y)R(u,v)\iff xv=yu.\text{ Show that }R\text{ is an equivalence relation.}
\displaystyle \text{Answer:}
\displaystyle \text{To prove that }R\text{ is an equivalence relation, we show that it is reflexive,}
\displaystyle \text{symmetric and transitive.}

\displaystyle \text{Reflexivity:}
\displaystyle \text{Let }(x,y)\in A.
\displaystyle xy=yx.
\displaystyle \Rightarrow(x,y)R(x,y).
\displaystyle \therefore R\text{ is reflexive.}

\displaystyle \text{Symmetry:}
\displaystyle \text{Let }(x,y)R(u,v).
\displaystyle \Rightarrow xv=yu.
\displaystyle \Rightarrow uy=vx.
\displaystyle \Rightarrow(u,v)R(x,y).
\displaystyle \therefore R\text{ is symmetric.}

\displaystyle \text{Transitivity:}
\displaystyle \text{Let }(x,y)R(u,v)\text{ and }(u,v)R(p,q).
\displaystyle \Rightarrow xv=yu\quad\text{and}\quad uq=vp.
\displaystyle \text{Multiplying }xv=yu\text{ by }q,\text{ we get}
\displaystyle xvq=yuq.
\displaystyle \text{Multiplying }uq=vp\text{ by }y,\text{ we get}
\displaystyle yuq=yvp.
\displaystyle \therefore xvq=yvp.
\displaystyle \Rightarrow v(xq)=v(yp).
\displaystyle \text{Since }v\neq0,\ xq=yp.
\displaystyle \Rightarrow(x,y)R(p,q).
\displaystyle \therefore R\text{ is transitive.}

\displaystyle \text{Hence, }R\text{ is reflexive, symmetric and transitive.}
\displaystyle \therefore R\text{ is an equivalence relation on }A.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Show that the relation }R\text{ on the set}
\displaystyle A=\{x\in\mathbb{Z}:0\leq x\leq12\},\text{ given by }R=\{(a,b):a=b\},
\displaystyle \text{is an equivalence relation. Find the set of all elements related to }1.
\displaystyle \text{Answer:}
\displaystyle \text{To prove that }R\text{ is an equivalence relation, we show that it is reflexive,}
\displaystyle \text{symmetric and transitive.}

\displaystyle \text{Reflexivity:}
\displaystyle \text{Let }a\in A.
\displaystyle \text{Since }a=a,\ (a,a)\in R.
\displaystyle \therefore R\text{ is reflexive on }A.

\displaystyle \text{Symmetry:}
\displaystyle \text{Let }(a,b)\in R.
\displaystyle \Rightarrow a=b.
\displaystyle \Rightarrow b=a.
\displaystyle \Rightarrow(b,a)\in R.
\displaystyle \therefore R\text{ is symmetric on }A.

\displaystyle \text{Transitivity:}
\displaystyle \text{Let }(a,b)\in R\text{ and }(b,c)\in R.
\displaystyle \Rightarrow a=b\text{ and }b=c.
\displaystyle \Rightarrow a=c.
\displaystyle \Rightarrow(a,c)\in R.
\displaystyle \therefore R\text{ is transitive on }A.
\displaystyle \text{Hence, }R\text{ is reflexive, symmetric and transitive.}
\displaystyle \therefore R\text{ is an equivalence relation on }A.

\displaystyle \text{Now, the set of all elements related to }1\text{ is}
\displaystyle \{x\in A:(x,1)\in R\}.
\displaystyle \text{Since }(x,1)\in R\iff x=1,
\displaystyle \{x\in A:(x,1)\in R\}=\{1\}.
\displaystyle \therefore \text{ the set of all elements related to }1\text{ is }\{1\}.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Let }L\text{ be the set of all lines in the }XY\text{-plane and let }R\text{ be the relation on }L
\displaystyle \text{defined by }R=\{(L_1,L_2):L_1\text{ is parallel to }L_2\}.\text{ Show that }R\text{ is an equivalence relation.}
\displaystyle \text{Find the set of all lines related to the line }y=2x+4.
\displaystyle \text{Answer:}
\displaystyle \text{To prove that }R\text{ is an equivalence relation, we show that it is reflexive,}
\displaystyle \text{symmetric and transitive.}

\displaystyle \text{Reflexivity:}
\displaystyle \text{Let }L_1\in L.
\displaystyle \text{Since every line is parallel to itself, }L_1\text{ is parallel to }L_1.
\displaystyle \Rightarrow(L_1,L_1)\in R.
\displaystyle \therefore R\text{ is reflexive on }L.

\displaystyle \text{Symmetry:}
\displaystyle \text{Let }(L_1,L_2)\in R.
\displaystyle \Rightarrow L_1\text{ is parallel to }L_2.
\displaystyle \Rightarrow L_2\text{ is parallel to }L_1.
\displaystyle \Rightarrow(L_2,L_1)\in R.
\displaystyle \therefore R\text{ is symmetric on }L.

\displaystyle \text{Transitivity:}
\displaystyle \text{Let }(L_1,L_2)\in R\text{ and }(L_2,L_3)\in R.
\displaystyle \Rightarrow L_1\text{ is parallel to }L_2\text{ and }L_2\text{ is parallel to }L_3.
\displaystyle \Rightarrow L_1\text{ is parallel to }L_3.
\displaystyle \Rightarrow(L_1,L_3)\in R.
\displaystyle \therefore R\text{ is transitive on }L.

\displaystyle \text{Hence, }R\text{ is reflexive, symmetric and transitive.}
\displaystyle \therefore R\text{ is an equivalence relation on }L.
\displaystyle \text{The line }y=2x+4\text{ has slope }2.
\displaystyle \text{Therefore, every line related to it must also have slope }2.
\displaystyle \therefore \text{ the set of all lines related to }y=2x+4\text{ is}
\displaystyle \{y=2x+c:c\in\mathbb{R}\}.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Show that the relation }R\text{ defined on the set }A\text{ of all polygons by}
\displaystyle R=\{(P_1,P_2):P_1\text{ and }P_2\text{ have the same number of sides}\}
\displaystyle \text{is an equivalence relation. What is the set of all elements in }A\text{ related to the}
\displaystyle \text{right-angled triangle }T\text{ with sides }3,4\text{ and }5?
\displaystyle \text{Answer:}
\displaystyle \text{To prove that }R\text{ is an equivalence relation, we show that it is reflexive,}
\displaystyle \text{symmetric and transitive.}

\displaystyle \text{Reflexivity:}
\displaystyle \text{Let }P\in A.
\displaystyle \text{The polygon }P\text{ has the same number of sides as itself.}
\displaystyle \Rightarrow(P,P)\in R.
\displaystyle \therefore R\text{ is reflexive on }A.

\displaystyle \text{Symmetry:}
\displaystyle \text{Let }(P_1,P_2)\in R.
\displaystyle \Rightarrow P_1\text{ and }P_2\text{ have the same number of sides.}
\displaystyle \Rightarrow P_2\text{ and }P_1\text{ have the same number of sides.}
\displaystyle \Rightarrow(P_2,P_1)\in R.
\displaystyle \therefore R\text{ is symmetric on }A.

\displaystyle \text{Transitivity:}
\displaystyle \text{Let }(P_1,P_2)\in R\text{ and }(P_2,P_3)\in R.
\displaystyle \Rightarrow P_1\text{ and }P_2\text{ have the same number of sides,}
\displaystyle \text{and }P_2\text{ and }P_3\text{ have the same number of sides.}
\displaystyle \Rightarrow P_1\text{ and }P_3\text{ have the same number of sides.}
\displaystyle \Rightarrow(P_1,P_3)\in R.
\displaystyle \therefore R\text{ is transitive on }A.

\displaystyle \text{Hence, }R\text{ is reflexive, symmetric and transitive.}
\displaystyle \therefore R\text{ is an equivalence relation on }A.
\displaystyle \text{The right-angled triangle }T\text{ has three sides.}
\displaystyle \text{Therefore, every polygon related to }T\text{ must also have three sides.}
\displaystyle \therefore \text{ the set of all elements in }A\text{ related to }T\text{ is the set of all triangles.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Let }O\text{ be the origin. A relation is defined between two points }P\text{ and }Q\text{ in a plane}
\displaystyle \text{by }P\,R\,Q\iff OP=OQ.\text{ Show that the relation so defined is an equivalence relation.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A\text{ be the set of all points in the plane and let}
\displaystyle R=\{(P,Q):OP=OQ\}\text{ be a relation on }A.
\displaystyle \text{To prove that }R\text{ is an equivalence relation, we show that it is reflexive,}
\displaystyle \text{symmetric and transitive.}

\displaystyle \text{Reflexivity:}
\displaystyle \text{Let }P\in A.
\displaystyle \text{Since }OP=OP,\ (P,P)\in R.
\displaystyle \therefore R\text{ is reflexive on }A.

\displaystyle \text{Symmetry:}
\displaystyle \text{Let }(P,Q)\in R.
\displaystyle \Rightarrow OP=OQ.
\displaystyle \Rightarrow OQ=OP.
\displaystyle \Rightarrow(Q,P)\in R.
\displaystyle \therefore R\text{ is symmetric on }A.

\displaystyle \text{Transitivity:}
\displaystyle \text{Let }(P,Q)\in R\text{ and }(Q,S)\in R.
\displaystyle \Rightarrow OP=OQ\text{ and }OQ=OS.
\displaystyle \Rightarrow OP=OS.
\displaystyle \Rightarrow(P,S)\in R.
\displaystyle \therefore R\text{ is transitive on }A.

\displaystyle \text{Hence, }R\text{ is reflexive, symmetric and transitive.}
\displaystyle \therefore R\text{ is an equivalence relation on }A.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Let }R\text{ be the relation defined on the set }A=\{1,2,3,4,5,6,7\}\text{ by}
\displaystyle R=\{(a,b):a\text{ and }b\text{ are either both odd or both even}\}.\text{ Show that }R\text{ is an}
\displaystyle \text{equivalence relation. Further, show that all the elements of }\{1,3,5,7\}\text{ are related to each other}
\displaystyle \text{and all the elements of }\{2,4,6\}\text{ are related to each other, but no element of }\{1,3,5,7\}
\displaystyle \text{is related to any element of }\{2,4,6\}.
\displaystyle \text{Answer:}
\displaystyle \text{To prove that }R\text{ is an equivalence relation, we show that it is reflexive,}
\displaystyle \text{symmetric and transitive.}

\displaystyle \text{Reflexivity:}
\displaystyle \text{Let }a\in A.
\displaystyle \text{The integers }a\text{ and }a\text{ are either both odd or both even.}
\displaystyle \Rightarrow(a,a)\in R.
\displaystyle \therefore R\text{ is reflexive on }A.

\displaystyle \text{Symmetry:}
\displaystyle \text{Let }(a,b)\in R.
\displaystyle \Rightarrow a\text{ and }b\text{ are either both odd or both even.}
\displaystyle \Rightarrow b\text{ and }a\text{ are either both odd or both even.}
\displaystyle \Rightarrow(b,a)\in R.
\displaystyle \therefore R\text{ is symmetric on }A.

\displaystyle \text{Transitivity:}
\displaystyle \text{Let }(a,b)\in R\text{ and }(b,c)\in R.
\displaystyle \text{Then }a\text{ and }b\text{ have the same parity, and }b\text{ and }c\text{ have the same parity.}
\displaystyle \text{If }b\text{ is odd, then }a,b\text{ and }c\text{ are all odd.}
\displaystyle \text{If }b\text{ is even, then }a,b\text{ and }c\text{ are all even.}
\displaystyle \text{Thus, in either case, }a\text{ and }c\text{ are either both odd or both even.}
\displaystyle \Rightarrow(a,c)\in R.
\displaystyle \therefore R\text{ is transitive on }A.

\displaystyle \text{Hence, }R\text{ is reflexive, symmetric and transitive.}
\displaystyle \therefore R\text{ is an equivalence relation on }A.
\displaystyle \text{Now, every element of }\{1,3,5,7\}\text{ is odd.}
\displaystyle \text{Therefore, any two elements of }\{1,3,5,7\}\text{ are related to each other.}
\displaystyle \text{Similarly, every element of }\{2,4,6\}\text{ is even.}
\displaystyle \text{Therefore, any two elements of }\{2,4,6\}\text{ are related to each other.}
\displaystyle \text{However, an odd integer and an even integer are neither both odd nor both even.}
\displaystyle \therefore \text{ no element of }\{1,3,5,7\}\text{ is related to any element of }\{2,4,6\}.
\displaystyle \text{Thus, the two equivalence classes are }\{1,3,5,7\}\text{ and }\{2,4,6\}.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Let }S\text{ be a relation on the set }\mathbb{R}\text{ of all real numbers defined by}
\displaystyle S=\{(a,b)\in\mathbb{R}\times\mathbb{R}:a^2+b^2=1\}. \text{ Prove that }S\text{ is not an equivalence relation on }\mathbb{R}.
\displaystyle \text{Answer:}
\displaystyle \text{To be an equivalence relation, }S\text{ must be reflexive.}
\displaystyle \text{For reflexivity, }(a,a)\in S\text{ for every }a\in\mathbb{R}.
\displaystyle \text{Take }a=1.
\displaystyle \text{Then }1^2+1^2=2\neq1.
\displaystyle \Rightarrow(1,1)\notin S.
\displaystyle \therefore S\text{ is not reflexive on }\mathbb{R}.
\displaystyle \text{Since every equivalence relation must be reflexive, }S\text{ is not an equivalence relation on }\mathbb{R}.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Let }\mathbb{Z}\text{ be the set of all integers and }\mathbb{Z}_0\text{ be the set of all non-zero integers.}
\displaystyle \text{Define a relation }R\text{ on }\mathbb{Z}\times\mathbb{Z}_0\text{ by}
\displaystyle (a,b)\,R\,(c,d)\iff ad=bc,\text{ for all }(a,b),(c,d)\in\mathbb{Z}\times\mathbb{Z}_0.
\displaystyle \text{Prove that }R\text{ is an equivalence relation on }\mathbb{Z}\times\mathbb{Z}_0.
\displaystyle \text{Answer:}
\displaystyle \text{To prove that }R\text{ is an equivalence relation, we show that it is reflexive,}
\displaystyle \text{symmetric and transitive.}

\displaystyle \text{Reflexivity:}
\displaystyle \text{Let }(a,b)\in\mathbb{Z}\times\mathbb{Z}_0.
\displaystyle ab=ba.
\displaystyle \Rightarrow(a,b)\,R\,(a,b).
\displaystyle \therefore R\text{ is reflexive on }\mathbb{Z}\times\mathbb{Z}_0.

\displaystyle \text{Symmetry:}
\displaystyle \text{Let }(a,b)\,R\,(c,d).
\displaystyle \Rightarrow ad=bc.
\displaystyle \Rightarrow cb=da.
\displaystyle \Rightarrow(c,d)\,R\,(a,b).
\displaystyle \therefore R\text{ is symmetric on }\mathbb{Z}\times\mathbb{Z}_0.

\displaystyle \text{Transitivity:}
\displaystyle \text{Let }(a,b)\,R\,(c,d)\text{ and }(c,d)\,R\,(e,f).
\displaystyle \Rightarrow ad=bc\quad\text{and}\quad cf=de.
\displaystyle \text{Multiplying }ad=bc\text{ by }f,\text{ we get}
\displaystyle adf=bcf.
\displaystyle \text{Multiplying }cf=de\text{ by }b,\text{ we get}
\displaystyle bcf=bde.
\displaystyle \therefore adf=bde.
\displaystyle \Rightarrow d(af)=d(be).
\displaystyle \text{Since }d\in\mathbb{Z}_0,\ d\neq0.
\displaystyle \therefore af=be.
\displaystyle \Rightarrow(a,b)\,R\,(e,f).
\displaystyle \therefore R\text{ is transitive on }\mathbb{Z}\times\mathbb{Z}_0.

\displaystyle \text{Hence, }R\text{ is reflexive, symmetric and transitive.}
\displaystyle \therefore R\text{ is an equivalence relation on }\mathbb{Z}\times\mathbb{Z}_0.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{If }R\text{ and }S\text{ are relations on a set }A,\text{ prove the following:}
\displaystyle \text{(i) If }R\text{ and }S\text{ are symmetric, then }R\cap S\text{ and }R\cup S\text{ are symmetric.}
\displaystyle \text{(ii) If }R\text{ is reflexive and }S\text{ is any relation, then }R\cup S\text{ is reflexive.}
\displaystyle \text{Answer:}

\displaystyle \text{(i) Let }R\text{ and }S\text{ be symmetric relations on }A.
\displaystyle \text{First, we prove that }R\cap S\text{ is symmetric.}
\displaystyle \text{Let }(a,b)\in R\cap S.
\displaystyle \Rightarrow(a,b)\in R\text{ and }(a,b)\in S.
\displaystyle \text{Since }R\text{ and }S\text{ are symmetric,}
\displaystyle (b,a)\in R\text{ and }(b,a)\in S.
\displaystyle \Rightarrow(b,a)\in R\cap S.
\displaystyle \therefore R\cap S\text{ is symmetric.}
\displaystyle \text{Now, we prove that }R\cup S\text{ is symmetric.}
\displaystyle \text{Let }(a,b)\in R\cup S.
\displaystyle \Rightarrow(a,b)\in R\text{ or }(a,b)\in S.
\displaystyle \text{If }(a,b)\in R,\text{ then }(b,a)\in R\text{ since }R\text{ is symmetric.}
\displaystyle \text{If }(a,b)\in S,\text{ then }(b,a)\in S\text{ since }S\text{ is symmetric.}
\displaystyle \text{Thus, }(b,a)\in R\text{ or }(b,a)\in S.
\displaystyle \Rightarrow(b,a)\in R\cup S.
\displaystyle \therefore R\cup S\text{ is symmetric.}

\displaystyle \text{(ii) Let }R\text{ be reflexive on }A\text{ and let }S\text{ be any relation on }A.
\displaystyle \text{Let }a\in A.
\displaystyle \text{Since }R\text{ is reflexive, }(a,a)\in R.
\displaystyle \text{Since }R\subseteq R\cup S,\ (a,a)\in R\cup S.
\displaystyle \text{Thus, }(a,a)\in R\cup S\text{ for every }a\in A.
\displaystyle \therefore R\cup S\text{ is reflexive on }A.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{If }R\text{ and }S\text{ are transitive relations on a set }A,\text{ prove that }R\cup S
\displaystyle \text{may not be a transitive relation on }A.
\displaystyle \text{Answer:}
\displaystyle \text{We prove the result by giving a counterexample.}
\displaystyle \text{Let }A=\{a,b,c\}\text{ and define}
\displaystyle R=\{(a,a),(b,b),(c,c),(a,b)\}
\displaystyle \text{and}
\displaystyle S=\{(a,a),(b,b),(c,c),(b,c)\}.
\displaystyle \text{First, we show that }R\text{ is transitive.}
\displaystyle \text{The only non-identity ordered pair in }R\text{ is }(a,b).
\displaystyle \text{The possible compositions involving }(a,b)\text{ give }(a,b),\text{ which belongs to }R.
\displaystyle \therefore R\text{ is transitive.}
\displaystyle \text{Similarly, the only non-identity ordered pair in }S\text{ is }(b,c).
\displaystyle \text{The possible compositions involving }(b,c)\text{ give }(b,c),\text{ which belongs to }S.
\displaystyle \therefore S\text{ is transitive.}
\displaystyle \text{Now,}
\displaystyle R\cup S=\{(a,a),(b,b),(c,c),(a,b),(b,c)\}.
\displaystyle \text{Here, }(a,b)\in R\cup S\text{ and }(b,c)\in R\cup S,
\displaystyle \text{but }(a,c)\notin R\cup S.
\displaystyle \therefore R\cup S\text{ is not transitive on }A.
\displaystyle \text{Hence, the union of two transitive relations need not be transitive.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Let }\mathbb{C}\text{ be the set of all complex numbers and }\mathbb{C}_0\text{ be the set of all}
\displaystyle \text{non-zero complex numbers. Let a relation }R\text{ on }\mathbb{C}_0\text{ be defined by}
\displaystyle z_1\,R\,z_2\iff\frac{z_1-z_2}{z_1+z_2}\text{ is real, for }z_1,z_2\in\mathbb{C}_0.
\displaystyle \text{Show that }R\text{ is an equivalence relation.}
\displaystyle \text{Answer:}
\displaystyle \text{The given relation is not an equivalence relation as stated.}
\displaystyle \text{For }z_1=1\text{ and }z_2=-1,\text{ we have}
\displaystyle z_1+z_2=1+(-1)=0.
\displaystyle \text{Therefore, }\frac{z_1-z_2}{z_1+z_2}\text{ is not defined.}
\displaystyle \text{Hence, the given condition does not define a relation for every pair in }\mathbb{C}_0\times\mathbb{C}_0.
\displaystyle \text{Even if the relation is interpreted to hold only when the quotient is defined and real,}
\displaystyle \text{it is not transitive.}
\displaystyle \text{Take }z_1=1,\quad z_2=2\quad\text{and}\quad z_3=-1.
\displaystyle \frac{z_1-z_2}{z_1+z_2}=\frac{1-2}{1+2}=-\frac13\in\mathbb{R}.
\displaystyle \Rightarrow 1\,R\,2.
\displaystyle \frac{z_2-z_3}{z_2+z_3}=\frac{2-(-1)}{2+(-1)}=3\in\mathbb{R}.
\displaystyle \Rightarrow 2\,R\,(-1).
\displaystyle \text{However,}
\displaystyle \frac{z_1-z_3}{z_1+z_3}=\frac{1-(-1)}{1+(-1)}=\frac{2}{0},
\displaystyle \text{which is not defined.}
\displaystyle \Rightarrow 1\not R(-1).
\displaystyle \text{Thus, }1\,R\,2\text{ and }2\,R\,(-1),\text{ but }1\not R(-1).
\displaystyle \therefore R\text{ is not transitive.}
\displaystyle \text{Hence, }R\text{ is not an equivalence relation on }\mathbb{C}_0.
\displaystyle \\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.