\displaystyle \textbf{Question 1: }\ \text{Show that the relation }R\text{ on the set }A=\{1,2,3,4,5\},\text{ given by}
\displaystyle R=\{(a,b):|a-b|\text{ is even}\},\text{ is an equivalence relation.} \text{Show that all the elements of } \\ \{1,3,5\}\text{ are related to each other and all the elements of }\{2,4\}\text{ are related to each} 
\displaystyle \text{other. But, no element of }\{1,3,5\}\text{ is related to any element of }\{2,4\}. \text{[CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle R=\{(a,b):|a-b|\text{ is even}\},\text{ where }a,b\in A=\{1,2,3,4,5\}.
\displaystyle \text{We observe the following properties of relation }R.
\displaystyle \textbf{Reflexivity:}\ \text{For any }a\in A,\text{ we have}
\displaystyle |a-a|=0,\text{ which is even}
\displaystyle \therefore (a,a)\in R\text{ for all }a\in A
\displaystyle \text{So, }R\text{ is reflexive.}
\displaystyle \textbf{Symmetry:}\ \text{Let }(a,b)\in R.\text{ Then,}
\displaystyle (a,b)\in R
\displaystyle \Rightarrow |a-b|\text{ is even}
\displaystyle \Rightarrow |b-a|\text{ is even}
\displaystyle \Rightarrow (b,a)\in R
\displaystyle \text{Thus, }(a,b)\in R\Rightarrow (b,a)\in R
\displaystyle \text{So, }R\text{ is symmetric.}
\displaystyle \textbf{Transitivity:}\ \text{Let }(a,b)\in R\text{ and }(b,c)\in R.\text{ Then,}
\displaystyle (a,b)\in R\text{ and }(b,c)\in R
\displaystyle \Rightarrow |a-b|\text{ is even and }|b-c|\text{ is even}
\displaystyle \Rightarrow (a\text{ and }b\text{ both are even or both are odd})\text{ and }(b\text{ and }c\text{ both are even or both are odd})
\displaystyle \text{Now two cases arise:}
\displaystyle \textbf{CASE I:}\ \text{When }b\text{ is even}
\displaystyle \text{In this case,}
\displaystyle (a,b)\in R\text{ and }(b,c)\in R
\displaystyle \Rightarrow |a-b|\text{ is even and }|b-c|\text{ is even}
\displaystyle \Rightarrow a\text{ is even and }c\text{ is even}\qquad[\because b\text{ is even}]
\displaystyle \Rightarrow |a-c|\text{ is even}
\displaystyle \Rightarrow (a,c)\in R
\displaystyle \textbf{CASE II:}\ \text{When }b\text{ is odd}
\displaystyle \text{In this case,}
\displaystyle (a,b)\in R\text{ and }(b,c)\in R
\displaystyle \Rightarrow |a-b|\text{ is even and }|b-c|\text{ is even}
\displaystyle \Rightarrow a\text{ is odd and }c\text{ is odd}\qquad[\because b\text{ is odd}]
\displaystyle \Rightarrow |a-c|\text{ is even}
\displaystyle \Rightarrow (a,c)\in R
\displaystyle \text{Thus, }(a,b)\in R\text{ and }(b,c)\in R\Rightarrow (a,c)\in R
\displaystyle \text{So, }R\text{ is transitive.}
\displaystyle \text{Hence, }R\text{ is an equivalence relation.}
\displaystyle \text{We know that the difference of any two odd (even) natural numbers is always an} \\ \text{even natural number.}\text{Therefore, all the elements of set }\{1,3,5\}\text{ are related to each} \\ \text{other and all the elements of }\{2,4\}\text{ are related to each other.}\text{We know that the} \\ \text{difference of an even natural number and an odd natural number is an odd natural} \\ \text{number. Therefore, no element of }\{1,3,5\}\text{ is related to any element of }\{2,4\}.

\displaystyle \textbf{Question 2: }\ \text{Show that the relation }R\text{ on the set }A=\{x\in Z:0\leq x\leq12\},\text{ given by }R=\{(a,b):|a-b|\text{ is a multiple of }4\}\text{ is an equivalence relation.}
\displaystyle \text{Find the set of all elements related to }1\text{ i.e. equivalence class }[1]. \ \ \   \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle\text{We have,}
\displaystyle R=\{(a,b):|a-b|\text{ is a multiple of }4\},\text{ where }a,b\in A=\{x\in Z:0\leq x\leq12\}=\{0,1,2,\ldots,12\}.
\displaystyle \text{We observe the following properties of relation }R.
\displaystyle \textbf{Reflexivity:}\ \text{For any }a\in A,\text{ we have}
\displaystyle |a-a|=0,\text{ which is a multiple of }4.
\displaystyle \Rightarrow (a,a)\in R
\displaystyle \text{Thus, }(a,a)\in R\text{ for all }a\in A.
\displaystyle \text{So, }R\text{ is reflexive.}
\displaystyle \textbf{Symmetry:}\ \text{Let }(a,b)\in R.\text{ Then,}
\displaystyle (a,b)\in R
\displaystyle \Rightarrow |a-b|\text{ is a multiple of }4
\displaystyle \Rightarrow |a-b|=4\lambda\text{ for some }\lambda\in N
\displaystyle \Rightarrow |b-a|=4\lambda\text{ for some }\lambda\in N\qquad[\because |a-b|=|b-a|]
\displaystyle \Rightarrow (b,a)\in R
\displaystyle \text{So, }R\text{ is symmetric.}
\displaystyle \textbf{Transitivity:}\ \text{Let }(a,b)\in R\text{ and }(b,c)\in R.\text{ Then,}
\displaystyle (a,b)\in R\text{ and }(b,c)\in R
\displaystyle \Rightarrow |a-b|\text{ is a multiple of }4\text{ and }|b-c|\text{ is a multiple of }4
\displaystyle \Rightarrow |a-b|=4\lambda\text{ and }|b-c|=4\mu\text{ for some }\lambda,\mu\in N
\displaystyle \Rightarrow a-b=\pm4\lambda\text{ and }b-c=\pm4\mu
\displaystyle \Rightarrow a-c=\pm4\lambda\pm4\mu
\displaystyle \Rightarrow a-c\text{ is a multiple of }4
\displaystyle \Rightarrow |a-c|\text{ is a multiple of }4
\displaystyle \Rightarrow (a,c)\in R
\displaystyle \text{Thus, }(a,b)\in R\text{ and }(b,c)\in R\Rightarrow (a,c)\in R
\displaystyle \text{So, }R\text{ is transitive.}
\displaystyle \text{Hence, }R\text{ is an equivalence relation.}
\displaystyle \text{Let }x\text{ be an element of }A\text{ such that }(x,1)\in R.\text{ Then,}
\displaystyle |x-1|\text{ is a multiple of }4
\displaystyle \Rightarrow |x-1|=0,4,8,12
\displaystyle \Rightarrow x-1=0,4,8,12
\displaystyle \Rightarrow x=1,5,9
\displaystyle \text{Hence, the set of all elements of }A\text{ which are related to }1\text{ is }\{1,5,9\}\text{ i.e. }[1]=\{1,5,9\}.

\displaystyle \textbf{Question 3: }\ \text{Prove that the relation }R\text{ on the set }N\times N\text{ defined by}
\displaystyle (a,b)\ R\ (c,d)\iff a+d=b+c\text{ for all }(a,b),(c,d)\in N\times N \text{is an equivalence}
\displaystyle \text{relation. Also, find the equivalence classes }[(2,3)]\text{ and }[(1,3)]. \ \ \ \ \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{We observe the following properties of relation }R.
\displaystyle \textbf{Reflexivity:}\ \text{Let }(a,b)\text{ be an arbitrary element of }N\times N.\text{ Then,}
\displaystyle (a,b)\in N\times N
\displaystyle \Rightarrow a,b\in N
\displaystyle \Rightarrow a+b=b+a\qquad[\text{By commutativity of addition on }N]
\displaystyle \Rightarrow (a,b)\ R\ (a,b)
\displaystyle \text{Thus, }(a,b)\ R\ (a,b)\text{ for all }(a,b)\in N\times N.\text{ So, }R\text{ is reflexive on }N\times N.
\displaystyle \textbf{Symmetry:}\ \text{Let }(a,b),(c,d)\in N\times N\text{ be such that }(a,b)\ R\ (c,d).\text{ Then,}
\displaystyle (a,b)\ R\ (c,d)
\displaystyle \Rightarrow a+d=b+c
\displaystyle \Rightarrow c+b=d+a\qquad[\text{By commutativity of addition on }N]
\displaystyle \Rightarrow (c,d)\ R\ (a,b)\qquad[\text{By definition of }R]
\displaystyle \text{Thus, }(a,b)\ R\ (c,d)\Rightarrow (c,d)\ R\ (a,b)\text{ for all }(a,b),(c,d)\in N\times N.
\displaystyle \text{So, }R\text{ is symmetric on }N\times N.
\displaystyle \textbf{Transitivity:}\ \text{Let }(a,b),(c,d),(e,f)\in N\times N\text{ such that }(a,b)\ R\ (c,d)\text{ and }(c,d)\ R\ (e,f).\text{ Then,}
\displaystyle (a,b)\ R\ (c,d)\Rightarrow a+d=b+c\Rightarrow (a+d)+(c+f)=(b+c)+(d+e)
\displaystyle (c,d)\ R\ (e,f)\Rightarrow c+f=d+e
\displaystyle \Rightarrow a+f=b+e\Rightarrow (a,b)\ R\ (e,f)
\displaystyle \text{Thus, }(a,b)\ R\ (c,d)\text{ and }(c,d)\ R\ (e,f)\Rightarrow (a,b)\ R\ (e,f)\text{ for all }(a,b),(c,d),(e,f)\in N\times N.
\displaystyle \text{So, }R\text{ is transitive on }N\times N.
\displaystyle \text{Hence, }R,\text{ being reflexive, symmetric and transitive, is an equivalence relation on }N\times N.
\displaystyle [(2,3)]=\{(x,y)\in N\times N:(x,y)\ R\ (2,3)\}
\displaystyle \Rightarrow [(2,3)]=\{(x,y)\in N\times N:x+3=y+2\}=\{(x,y)\in N\times N:x-y=1\}
\displaystyle =\{(x,y)\in N\times N:y=x+1\}
\displaystyle =\{(x,x+1):x\in N\}
\displaystyle =\{(1,2),(2,3),(3,4),(4,5),\ldots\}
\displaystyle [(7,3)]=\{(x,y)\in N\times N:(x,y)\ R\ (7,3)\}
\displaystyle =\{(x,y)\in N\times N:x+3=y+7\}
\displaystyle =\{(x,y)\in N\times N:y=x-4\}
\displaystyle =\{(x,x-4)\in N\times N:x\in N\}
\displaystyle =\{(5,1),(6,2),(7,3),(8,4),(9,5),\ldots\}

\displaystyle \textbf{Question 4: }\ \text{Let }A=\{1,2,3,\ldots,9\}\text{ and }R\text{ be the relation on }A\times A \text{ defined by} \\ (a,b)\ R\ (c,d)\text{ if}  a+d=b+c\text{ for all }(a,b),(c,d)\in A\times A.\text{ Prove that }R\text{ is an} \\ \text{equivalence} \text{ relation and also obtain } \text{the equivalence class }[(2,5)]. \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{We observe the following properties of relation }R.
\displaystyle \textbf{Reflexivity:}\ \text{Let }(a,b)\text{ be an arbitrary element of }A\times A.\text{ Then,}
\displaystyle (a,b)\in A\times A
\displaystyle \Rightarrow a,b\in A
\displaystyle \Rightarrow a+b=b+a\qquad[\text{By commutativity of addition on }N]
\displaystyle \Rightarrow (a,b)\ R\ (a,b)
\displaystyle \text{Thus, }(a,b)\ R\ (a,b)\text{ for all }(a,b)\in A\times A.\text{ So, }R\text{ is reflexive on }A\times A.
\displaystyle \textbf{Symmetry:}\ \text{Let }(a,b),(c,d)\in A\times A\text{ be such that }(a,b)\ R\ (c,d).\text{ Then,}
\displaystyle (a,b)\ R\ (c,d)
\displaystyle \Rightarrow a+d=b+c
\displaystyle \Rightarrow c+b=d+a\qquad[\text{By commutativity of addition on }N]
\displaystyle \Rightarrow (c,d)\ R\ (a,b)
\displaystyle \text{Thus, }(a,b)\ R\ (c,d)\Rightarrow (c,d)\ R\ (a,b)\text{ for all }(a,b),(c,d)\in A\times A.
\displaystyle \text{So, }R\text{ is symmetric on }A\times A.
\displaystyle \textbf{Transitivity:}\ \text{Let }(a,b),(c,d),(e,f)\in A\times A\text{ such that }(a,b)\ R\ (c,d)\text{ and }(c,d)\ R\ (e,f).\text{ Then,}
\displaystyle (a,b)\ R\ (c,d)\Rightarrow a+d=b+c
\displaystyle (c,d)\ R\ (e,f)\Rightarrow c+f=d+e\Rightarrow (a+d)+(c+f)=(b+c)+(d+e)
\displaystyle \Rightarrow a+f=b+e
\displaystyle \Rightarrow (a,b)\ R\ (e,f)
\displaystyle \text{Thus, }(a,b)\ R\ (c,d)\text{ and }(c,d)\ R\ (e,f)\Rightarrow (a,b)\ R\ (e,f)\text{ for all }(a,b),(c,d),(e,f)\in A\times A.
\displaystyle \text{So, }R\text{ is a transitive relation on }A\times A.
\displaystyle \text{Hence, }R\text{ is an equivalence relation on }A\times A.
\displaystyle \text{Now,}
\displaystyle [(2,5)]=\{(x,y)\in A\times A:(x,y)\ R\ (2,5)\}
\displaystyle =\{(x,y)\in A\times A:x+5=y+2\}=\{(x,y)\in A\times A:y=x+3\}
\displaystyle =\{(x,x+3):x\in A\text{ and }x+3\in A\}=\{(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)\}

\displaystyle \textbf{Question 5: }\ \text{Let }N\text{ denote the set of all natural numbers and }R\text{ be the relation on }\\ N\times N\text{ defined by}
\displaystyle (a,b)\ R\ (c,d)\iff ad(b+c)=bc(a+d).\text{ Check whether }R\text{ is an equivalence relation on } \\ N\times N. \ \ \ \  \text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{We observe the following properties of relation }R.
\displaystyle \textbf{Reflexivity:}\ \text{Let }(a,b)\text{ be an arbitrary element of }N\times N.\text{ Then,}
\displaystyle (a,b)\in N\times N
\displaystyle \Rightarrow a,b\in N
\displaystyle \Rightarrow ab(b+a)=ba(a+b)\qquad[\text{By commutativity of addition and multiplication on }N]
\displaystyle \Rightarrow (a,b)\ R\ (a,b)
\displaystyle \text{Thus, }(a,b)\ R\ (a,b)\text{ for all }(a,b)\in N\times N.\text{ So, }R\text{ is reflexive on }N\times N.
\displaystyle \textbf{Symmetry:}\ \text{Let }(a,b),(c,d)\in N\times N\text{ be such that }(a,b)\ R\ (c,d).\text{ Then,}
\displaystyle (a,b)\ R\ (c,d)
\displaystyle \Rightarrow ad(b+c)=bc(a+d)
\displaystyle \Rightarrow cb(d+a)=da(c+b)\qquad[\text{By commutativity of addition and multiplication on }N]
\displaystyle \Rightarrow (c,d)\ R\ (a,b)
\displaystyle \text{Thus, }(a,b)\ R\ (c,d)\Rightarrow (c,d)\ R\ (a,b)\text{ for all }(a,b),(c,d)\in N\times N.
\displaystyle \text{So, }R\text{ is symmetric on }N\times N.
\displaystyle \textbf{Transitivity:}\ \text{Let }(a,b),(c,d),(e,f)\in N\times N\text{ such that }(a,b)\ R\ (c,d)\text{ and }(c,d)\ R\ (e,f).\text{ Then,}
\displaystyle (a,b)\ R\ (c,d) \\ \\ \Rightarrow ad(b+c)=bc(a+d) \\ \\ \Rightarrow \frac{b+c}{bc}=\frac{a+d}{ad}=\frac{1}{b}+\frac{1}{c}=\frac{1}{a}+\frac{1}{d}\qquad\text{(i)}
\displaystyle \text{and,}
\displaystyle (c,d)\ R\ (e,f) \\ \\ \Rightarrow cf(d+e)=de(c+f) \\ \\ \Rightarrow \frac{d+e}{de}=\frac{c+f}{cf}=\frac{1}{d}+\frac{1}{e}=\frac{1}{c}+\frac{1}{f}\qquad\text{(ii)}
\displaystyle \text{Adding (i) and (ii), we get}
\displaystyle \left(\frac{1}{b}+\frac{1}{c}\right)+\left(\frac{1}{d}+\frac{1}{e}\right)=\left(\frac{1}{a}+\frac{1}{d}\right)+\left(\frac{1}{c}+\frac{1}{f}\right)
\displaystyle \Rightarrow \frac{1}{b}+\frac{1}{e}=\frac{1}{a}+\frac{1}{f} \\ \\ \Rightarrow \frac{b+e}{be}=\frac{a+f}{af} \\ \\ \Rightarrow af(b+e)=be(a+f) \\ \\ \Rightarrow (a,b)\ R\ (e,f)
\displaystyle \text{Thus, }(a,b)\ R\ (c,d)\text{ and }(c,d)\ R\ (e,f) \\ \\ \Rightarrow (a,b)\ R\ (e,f)\text{ for all }(a,b),(c,d),(e,f)\in N\times N.
\displaystyle \text{So, }R\text{ is transitive on }N\times N.
\displaystyle \text{Hence, }R,\text{ being reflexive, symmetric and transitive, is an equivalence relation on }N\times N.

\displaystyle \textbf{Question 6: }\ \text{Let }R\text{ be the equivalence relation in the set }A=\{0,1,2,3,4,5\}\text{ given by} 
\displaystyle R=\{(a,b):2\text{ divides }(a-b)\}.\text{ Write the equivalence class }[0]. \ \ \ \  \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Clearly, the equivalence class }[0]\text{ is the set of those elements in }A\text{ which are related}
\displaystyle \text{to }0\text{ under the relation }R\text{ i.e. }[0]=\{(a,0)\in R:a\in A\}.
\displaystyle \text{Now,}\ (a,0)\in R
\displaystyle \Rightarrow a-0\text{ is divisible by }2\text{ and }a\in A
\displaystyle \Rightarrow a\in A\text{ such that }2\text{ divides }a
\displaystyle \Rightarrow a=0,2,4
\displaystyle \text{Thus, }[0]=\{0,2,4\}.

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

Subscribe to get the latest posts sent to your email.