\displaystyle \textbf{Question 1: } \text{If }A=\{1,2,3\},\ B=\{4,5,6\},\text{ which of the following are relations from }A
\displaystyle \text{to }B?\text{ Give reasons in support of your answer:}
\displaystyle \text{(i) }\{(1,6),(3,4),(5,2)\}\qquad\text{(ii) }\{(1,5),(2,6),(3,4),(3,6)\}
\displaystyle \text{(iii) }\{(4,2),(4,3),(5,1)\}\qquad\text{(iv) }A\times B
\displaystyle \text{Answer:}
\displaystyle A=\{1,2,3\},\qquad B=\{4,5,6\}
\displaystyle A\times B=\{(1,4),(1,5),(1,6),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6)\}
\displaystyle \text{(i) }\{(1,6),(3,4),(5,2)\}\text{ is not a relation from }A\text{ to }B
\displaystyle \text{since it is not a subset of }A\times B.
\displaystyle \text{(ii) }\{(1,5),(2,6),(3,4),(3,6)\}\text{ is a subset of }A\times B.
\displaystyle \therefore \text{It is a relation from }A\text{ to }B.
\displaystyle \text{(iii) }\{(4,2),(4,3),(5,1)\}\text{ is not a relation from }A\text{ to }B
\displaystyle \text{since it is not a subset of }A\times B.
\displaystyle \text{(iv) }A\times B\text{ is a subset of itself.}
\displaystyle \therefore A\times B\text{ is a relation from }A\text{ to }B.
\displaystyle \\

\displaystyle \textbf{Question 2: } \text{A relation }R\text{ is defined from }A=\{2,3,4,5\}\text{ to }B=\{3,6,7,10\}
\displaystyle \text{as follows: }(x,y)\in R\Leftrightarrow x\text{ is relatively prime to }y.
\displaystyle \text{Express }R\text{ as a set of ordered pairs and determine its domain and range.}
\displaystyle \text{Answer:}
\displaystyle A=\{2,3,4,5\},\qquad B=\{3,6,7,10\}
\displaystyle (x,y)\in R\Leftrightarrow x\text{ is relatively prime to }y.
\displaystyle 2\text{ is relatively prime to }3\text{ and }7.
\displaystyle 3\text{ is relatively prime to }7\text{ and }10.
\displaystyle 4\text{ is relatively prime to }3\text{ and }7.
\displaystyle 5\text{ is relatively prime to }3,6\text{ and }7.
\displaystyle \therefore R=\{(2,3),(2,7),(3,7),(3,10),(4,3),(4,7),(5,3),(5,6),(5,7)\}.
\displaystyle \text{Domain of }R=\{2,3,4,5\}.
\displaystyle \text{Range of }R=\{3,6,7,10\}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Let }A\text{ be the set of the first five natural numbers and let }R\text{ be a relation}
\displaystyle \text{on }A\text{ defined by }(x,y)\in R\Leftrightarrow x\leq y.\text{ Express }R\text{ and }R^{-1}\text{ as sets of}
\displaystyle \text{ordered pairs. Also determine: (i) the domain of }R^{-1}\text{ (ii) the range of }R.
\displaystyle \text{Answer:}
\displaystyle A=\{1,2,3,4,5\}
\displaystyle \left[\because A\text{ is the set of the first five natural numbers}\right]
\displaystyle (x,y)\in R\Leftrightarrow x\leq y
\displaystyle \therefore R=\{(1,1),(1,2),(1,3),(1,4),(1,5),(2,2),(2,3),(2,4),
\displaystyle (2,5),(3,3),(3,4),(3,5),(4,4),(4,5),(5,5)\}
\displaystyle \text{Interchanging the components of every ordered pair in }R,
\displaystyle R^{-1}=\{(1,1),(2,1),(3,1),(4,1),(5,1),(2,2),(3,2),(4,2),
\displaystyle (5,2),(3,3),(4,3),(5,3),(4,4),(5,4),(5,5)\}
\displaystyle \text{(i) Domain of }R^{-1}=\{1,2,3,4,5\}
\displaystyle \text{(ii) Range of }R=\{1,2,3,4,5\}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the inverse relation }R^{-1}\text{ in each of the following cases:}
\displaystyle \text{(i) }R=\{(1,2),(1,3),(2,3),(3,2),(5,6)\}
\displaystyle \text{(ii) }R=\{(x,y):x,y\in N,\ x+2y=8\}
\displaystyle \text{(iii) }R\text{ is a relation from }\{11,12,13\}\text{ to }\{8,10,12\}\text{ defined by }y=x-3.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given, }R=\{(1,2),(1,3),(2,3),(3,2),(5,6)\}
\displaystyle \therefore R^{-1}=\{(2,1),(3,1),(3,2),(2,3),(6,5)\}
\displaystyle \text{(ii) Given, }R=\{(x,y):x,y\in N,\ x+2y=8\}
\displaystyle x+2y=8\Rightarrow x=8-2y
\displaystyle \text{For }y=1,2,3,\text{ the corresponding values of }x\text{ are }6,4,2.
\displaystyle \text{For }y=4,\ x=0\notin N,\text{ and for }y>4,\ x\notin N.
\displaystyle \therefore R=\{(6,1),(4,2),(2,3)\}
\displaystyle \therefore R^{-1}=\{(1,6),(2,4),(3,2)\}
\displaystyle \text{(iii) Given, }R\text{ is a relation from }\{11,12,13\}\text{ to }\{8,10,12\}
\displaystyle \text{defined by }y=x-3.
\displaystyle \text{For }x=11,\ y=8;\qquad \text{for }x=12,\ y=9;\qquad \text{for }x=13,\ y=10.
\displaystyle \text{Since }9\notin\{8,10,12\},\text{ the pair }(12,9)\text{ does not belong to }R.
\displaystyle \therefore R=\{(11,8),(13,10)\}
\displaystyle \therefore R^{-1}=\{(8,11),(10,13)\}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Write the following relations as sets of ordered pairs:}
\displaystyle \text{(i) A relation }R\text{ from }\{2,3,4,5,6\}\text{ to }\{1,2,3\}\text{ defined by }x=2y.
\displaystyle \text{(ii) A relation }R\text{ on }\{1,2,3,4,5,6,7\}\text{ defined by}
\displaystyle (x,y)\in R\Leftrightarrow x\text{ is relatively prime to }y.
\displaystyle \text{(iii) A relation }R\text{ on }\{0,1,2,\ldots,10\}\text{ defined by }2x+3y=12.
\displaystyle \text{(iv) A relation }R\text{ from }A=\{5,6,7,8\}\text{ to }B=\{10,12,15,16,18\}
\displaystyle \text{defined by }(x,y)\in R\Leftrightarrow x\text{ divides }y.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given, }x=2y.
\displaystyle \text{For }y=1,2,3,\text{ the corresponding values of }x\text{ are }2,4,6.
\displaystyle \therefore R=\{(2,1),(4,2),(6,3)\}.
\displaystyle \text{(ii) Given, }(x,y)\in R\Leftrightarrow x\text{ is relatively prime to }y.
\displaystyle 1\text{ is relatively prime to }1,2,3,4,5,6\text{ and }7.
\displaystyle 2\text{ is relatively prime to }1,3,5\text{ and }7.
\displaystyle 3\text{ is relatively prime to }1,2,4,5\text{ and }7.
\displaystyle 4\text{ is relatively prime to }1,3,5\text{ and }7.
\displaystyle 5\text{ is relatively prime to }1,2,3,4,6\text{ and }7.
\displaystyle 6\text{ is relatively prime to }1,5\text{ and }7.
\displaystyle 7\text{ is relatively prime to }1,2,3,4,5\text{ and }6.
\displaystyle \therefore R=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(1,7),
\displaystyle (2,1),(2,3),(2,5),(2,7),(3,1),(3,2),(3,4),(3,5),(3,7),
\displaystyle (4,1),(4,3),(4,5),(4,7),(5,1),(5,2),(5,3),(5,4),(5,6),(5,7),
\displaystyle (6,1),(6,5),(6,7),(7,1),(7,2),(7,3),(7,4),(7,5),(7,6)\}.
\displaystyle \text{(iii) Given, }2x+3y=12.
\displaystyle 2x=12-3y\Rightarrow x=\frac{12-3y}{2}.
\displaystyle \text{For }y=0,2,4,\text{ the corresponding values of }x\text{ are }6,3,0.
\displaystyle \therefore R=\{(6,0),(3,2),(0,4)\}.
\displaystyle \text{(iv) Given, }(x,y)\in R\Leftrightarrow x\text{ divides }y.
\displaystyle 5\text{ divides }10\text{ and }15.
\displaystyle 6\text{ divides }12\text{ and }18.
\displaystyle 7\text{ does not divide any element of }B.
\displaystyle 8\text{ divides }16.
\displaystyle \therefore R=\{(5,10),(5,15),(6,12),(6,18),(8,16)\}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Let }R\text{ be a relation in }N\text{ defined by }(x,y)\in R\Leftrightarrow x+2y=8.
\displaystyle \text{Express }R\text{ and }R^{-1}\text{ as sets of ordered pairs.}
\displaystyle \text{Answer:}
\displaystyle (x,y)\in R\Leftrightarrow x+2y=8
\displaystyle x+2y=8\Rightarrow x=8-2y
\displaystyle \text{For }y=1,2,3,\text{ the corresponding values of }x\text{ are }6,4,2.
\displaystyle \text{For }y=4,\ x=0\notin N,\text{ and for }y>4,\ x\notin N.
\displaystyle \therefore R=\{(6,1),(4,2),(2,3)\}
\displaystyle \therefore R^{-1}=\{(1,6),(2,4),(3,2)\}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Let }A=\{3,5\}\text{ and }B=\{7,11\}. \text{ Let }R=\{(a,b):a\in A,
\displaystyle \text{ }b\in B,\ a-b\text{ is odd}\}. \text{ Show that }R\text{ is an empty relation from }A\text{ to }B.
\displaystyle \text{Answer:}
\displaystyle A=\{3,5\},\qquad B=\{7,11\}
\displaystyle R=\{(a,b):a\in A,\ b\in B,\ a-b\text{ is odd}\}
\displaystyle 3-7=-4,\qquad 3-11=-8,
\displaystyle 5-7=-2,\qquad 5-11=-6
\displaystyle \text{All the above differences are even.}
\displaystyle \therefore \text{There is no ordered pair }(a,b)\text{ such that }a-b\text{ is odd.}
\displaystyle \therefore R=\phi.
\displaystyle \therefore R\text{ is an empty relation from }A\text{ to }B.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Let }A=\{1,2\}\text{ and }B=\{3,4\}. \text{ Find the total number of}
\displaystyle \text{relations from }A\text{ to }B.
\displaystyle \text{Answer:}
\displaystyle A=\{1,2\},\qquad B=\{3,4\}
\displaystyle n(A)=2,\qquad n(B)=2
\displaystyle n(A\times B)=n(A)\times n(B)=2\times2=4
\displaystyle \text{Every relation from }A\text{ to }B\text{ is a subset of }A\times B.
\displaystyle \therefore \text{Total number of relations}=2^{\,n(A\times B)}=2^4=16.
\displaystyle \left[\because \text{If }n(A)=a\text{ and }n(B)=b,\text{ then the total number of relations is }2^{ab}\right]
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Determine the domain and range of the relation }R\text{ defined by:}
\displaystyle \text{(i) }R=\{(x,x+5):x\in\{0,1,2,3,4,5\}\}
\displaystyle \text{(ii) }R=\{(x,x^3):x\text{ is a prime number less than }10\}
\displaystyle \text{Answer:}
\displaystyle \text{(i) }R=\{(x,x+5):x\in\{0,1,2,3,4,5\}\}
\displaystyle \therefore R=\{(0,5),(1,6),(2,7),(3,8),(4,9),(5,10)\}
\displaystyle \text{Domain}(R)=\{0,1,2,3,4,5\}
\displaystyle \text{Range}(R)=\{5,6,7,8,9,10\}
\displaystyle \text{(ii) }x\text{ is a prime number less than }10.
\displaystyle \therefore x=2,3,5,7
\displaystyle R=\{(2,8),(3,27),(5,125),(7,343)\}
\displaystyle \text{Domain}(R)=\{2,3,5,7\}
\displaystyle \text{Range}(R)=\{8,27,125,343\}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Determine the domain and range of the following relations:}
\displaystyle \text{(i) }R=\{(a,b):a\in N,\ a<5,\ b=4\}
\displaystyle \text{(ii) }S=\{(a,b):b=|a-1|,\ a\in Z\text{ and }|a|\leq3\}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given, }R=\{(a,b):a\in N,\ a<5,\ b=4\}
\displaystyle a=1,2,3,4,\qquad b=4
\displaystyle \therefore R=\{(1,4),(2,4),(3,4),(4,4)\}
\displaystyle \text{Domain}(R)=\{1,2,3,4\}
\displaystyle \text{Range}(R)=\{4\}
\displaystyle \text{(ii) Given, }S=\{(a,b):b=|a-1|,\ a\in Z\text{ and }|a|\leq3\}
\displaystyle a=-3,-2,-1,0,1,2,3
\displaystyle \therefore b=4,3,2,1,0,1,2\text{ respectively}
\displaystyle \therefore S=\{(-3,4),(-2,3),(-1,2),(0,1),(1,0),(2,1),(3,2)\}
\displaystyle \text{Domain}(S)=\{-3,-2,-1,0,1,2,3\}
\displaystyle \text{Range}(S)=\{0,1,2,3,4\}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Let }A=\{a,b\}. \text{ List all relations on }A\text{ and find their number.}
\displaystyle \text{Answer:}
\displaystyle A=\{a,b\}
\displaystyle A\times A=\{(a,a),(a,b),(b,a),(b,b)\}
\displaystyle \text{Every relation on }A\text{ is a subset of }A\times A.
\displaystyle \therefore \text{Total number of relations}=2^4=16.
\displaystyle \text{The }16\text{ relations are:}
\displaystyle \phi,
\displaystyle \{(a,a)\},\ \{(a,b)\},\ \{(b,a)\},\ \{(b,b)\},
\displaystyle \{(a,a),(a,b)\},\ \{(a,a),(b,a)\},\ \{(a,a),(b,b)\},
\displaystyle \{(a,b),(b,a)\},\ \{(a,b),(b,b)\},\ \{(b,a),(b,b)\},
\displaystyle \{(a,a),(a,b),(b,a)\},
\displaystyle \{(a,a),(a,b),(b,b)\},
\displaystyle \{(a,a),(b,a),(b,b)\},
\displaystyle \{(a,b),(b,a),(b,b)\},
\displaystyle \{(a,a),(a,b),(b,a),(b,b)\}.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Let }A=\{x,y,z\}\text{ and }B=\{a,b\}. \text{ Find the total number of}
\displaystyle \text{relations from }A\text{ to }B.
\displaystyle \text{Answer:}
\displaystyle A=\{x,y,z\},\qquad B=\{a,b\}
\displaystyle n(A)=3,\qquad n(B)=2
\displaystyle n(A\times B)=n(A)\times n(B)=3\times2=6
\displaystyle \text{Every relation from }A\text{ to }B\text{ is a subset of }A\times B.
\displaystyle \therefore \text{Total number of relations}=2^{\,6}=64.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Let }R\text{ be a relation from }N\text{ to }N\text{ defined by}
\displaystyle R=\{(a,b):a,b\in N\text{ and }a=b^2\}. \text{ Are the following statements true?}
\displaystyle \text{(i) }(a,a)\in R\text{ for all }a\in N
\displaystyle \text{(ii) }(a,b)\in R\Rightarrow(b,a)\in R
\displaystyle \text{(iii) }(a,b)\in R\text{ and }(b,c)\in R\Rightarrow(a,c)\in R
\displaystyle \text{Answer:}
\displaystyle \text{Given }R=\{(a,b):a,b\in N\text{ and }a=b^2\}
\displaystyle \text{(i) }(a,a)\in R\text{ for all }a\in N
\displaystyle \text{False, since }(2,2)\notin R\text{ because }2\neq2^2.
\displaystyle \text{(ii) }(a,b)\in R\Rightarrow(b,a)\in R
\displaystyle \text{False, since }(4,2)\in R\text{ but }(2,4)\notin R.
\displaystyle \text{(iii) }(a,b)\in R\text{ and }(b,c)\in R\Rightarrow(a,c)\in R
\displaystyle \text{False, since }(16,4)\in R\text{ and }(4,2)\in R,
\displaystyle \text{but }(16,2)\notin R\text{ because }16\neq2^2.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Let }A=\{1,2,3,\ldots,14\}.\text{ Define a relation on }A\text{ by}
\displaystyle R=\{(x,y):3x-y=0,\text{ where }x,y\in A\}.\text{ Depict this relation using an}
\displaystyle \text{arrow diagram. Write down its domain, co-domain and range.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }3x-y=0
\displaystyle \Rightarrow y=3x
\displaystyle \text{For }x=1,2,3,4,\text{ the corresponding values of }y\text{ are }3,6,9,12.
\displaystyle \text{For }x\geq5,\ y=3x\geq15,\text{ which does not belong to }A.
\displaystyle \therefore R=\{(1,3),(2,6),(3,9),(4,12)\}2019-08-23_21-45-06\displaystyle \text{The arrow diagram is represented by:}
\displaystyle 1\longrightarrow3,\qquad2\longrightarrow6,\qquad3\longrightarrow9,\qquad4\longrightarrow12
\displaystyle \therefore \text{Domain}(R)=\{1,2,3,4\}
\displaystyle \text{Co-domain}(R)=A=\{1,2,3,\ldots,14\}
\displaystyle \text{Range}(R)=\{3,6,9,12\}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Define a relation }R\text{ on the set }N\text{ of natural numbers by}
\displaystyle R=\{(x,y):y=x+5,\ x\text{ is a natural number less than }4,\ x,y\in N\}.
\displaystyle \text{Depict this relation using (i) roster form (ii) an arrow diagram.}
\displaystyle \text{Write down the domain and range of }R.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }R=\{(x,y):y=x+5,\ x<4,\ x,y\in N\}
\displaystyle \text{(i) Roster form:}
\displaystyle \text{For }x=1,2,3,\text{ the corresponding values of }y\text{ are }6,7,8\text{ respectively.}
\displaystyle \therefore R=\{(1,6),(2,7),(3,8)\}
\displaystyle \text{(ii) Arrow diagram:}2019-08-23_21-47-50\displaystyle 1\longrightarrow6,\qquad2\longrightarrow7,\qquad3\longrightarrow8
\displaystyle \text{Domain}(R)=\{1,2,3\}
\displaystyle \text{Range}(R)=\{6,7,8\}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Let }A=\{1,2,3,5\}\text{ and }B=\{4,6,9\}. \text{ Define a relation }R
\displaystyle \text{from }A\text{ to }B\text{ by }R=\{(x,y):\text{ the difference between }x\text{ and }y\text{ is odd,}
\displaystyle x\in A,\ y\in B\}. \text{ Write }R\text{ in roster form.}
\displaystyle \text{Answer:}
\displaystyle A=\{1,2,3,5\},\qquad B=\{4,6,9\}
\displaystyle R=\{(x,y):\text{ the difference between }x\text{ and }y\text{ is odd},\ x\in A,\ y\in B\}
\displaystyle \text{For }x=1,\ y=4,6
\displaystyle \text{For }x=2,\ y=9
\displaystyle \text{For }x=3,\ y=4,6
\displaystyle \text{For }x=5,\ y=4,6
\displaystyle \therefore R=\{(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)\}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Write the relation }R=\{(x,x^3):x\text{ is a prime number less than }10\}
\displaystyle \text{in roster form.}
\displaystyle \text{Answer:}
\displaystyle \text{The prime numbers less than }10\text{ are }2,3,5,7.
\displaystyle \therefore R=\{(2,8),(3,27),(5,125),(7,343)\}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Let }A=\{1,2,3,4,5,6\}. \text{ Let }R\text{ be a relation on }A
\displaystyle \text{defined by }R=\{(a,b):a,b\in A,\ b\text{ is exactly divisible by }a\}.
\displaystyle \text{(i) Write }R\text{ in roster form. (ii) Find the domain of }R.
\displaystyle \text{(iii) Find the range of }R.
\displaystyle \text{Answer:}
\displaystyle A=\{1,2,3,4,5,6\}
\displaystyle R=\{(a,b):a,b\in A,\ b\text{ is exactly divisible by }a\}
\displaystyle \text{For }a=1,\ b=1,2,3,4,5,6
\displaystyle \text{For }a=2,\ b=2,4,6
\displaystyle \text{For }a=3,\ b=3,6
\displaystyle \text{For }a=4,\ b=4
\displaystyle \text{For }a=5,\ b=5
\displaystyle \text{For }a=6,\ b=6
\displaystyle \text{(i) }R=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,2),(2,4),(2,6),
\displaystyle (3,3),(3,6),(4,4),(5,5),(6,6)\}
\displaystyle \text{(ii) Domain}(R)=\{1,2,3,4,5,6\}
\displaystyle \text{(iii) Range}(R)=\{1,2,3,4,5,6\}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The figure below shows a relationship between the sets }P\text{ and }Q.
\displaystyle \text{Write this relation in (i) set-builder form (ii) roster form.}
\displaystyle \text{What are its domain and range?} 2019-08-24_6-27-40.jpg\displaystyle \text{Answer:}
\displaystyle P=\{5,6,7\},\qquad Q=\{3,4,5\}
\displaystyle \text{(i) Set-builder form:}
\displaystyle R=\{(x,y):y=x-2,\ x\in P,\ y\in Q\}
\displaystyle \text{For }x=5,\ y=3
\displaystyle \text{For }x=6,\ y=4
\displaystyle \text{For }x=7,\ y=5
\displaystyle \text{(ii) Roster form:}
\displaystyle R=\{(5,3),(6,4),(7,5)\}
\displaystyle \text{Domain}(R)=\{5,6,7\}
\displaystyle \text{Range}(R)=\{3,4,5\}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Let }R\text{ be the relation on }Z\text{ defined by}
\displaystyle R=\{(a,b):a,b\in Z,\ a-b\text{ is an integer}\}.
\displaystyle \text{Find the domain and range of }R.
\displaystyle \text{Answer:}
\displaystyle R=\{(a,b):a,b\in Z,\ a-b\text{ is an integer}\}
\displaystyle \text{The difference of any two integers is always an integer.}
\displaystyle \therefore R=Z\times Z
\displaystyle \therefore \text{Domain}(R)=Z
\displaystyle \text{Range}(R)=Z
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{For the relation }R_1\text{ defined on }R\text{ by the rule}
\displaystyle (a,b)\in R_1\Leftrightarrow1+ab>0,\text{ prove that}
\displaystyle (a,b)\in R_1\text{ and }(b,c)\in R_1\Rightarrow(a,c)\in R_1
\displaystyle \text{is not true for all }a,b,c\in R.
\displaystyle \text{Answer:}
\displaystyle R_1=\{(a,b):1+ab>0\}
\displaystyle \text{Let }a=1,\qquad b=-0.5,\qquad c=-4.
\displaystyle 1+ab=1+(1)(-0.5)=0.5>0
\displaystyle \therefore (1,-0.5)\in R_1
\displaystyle 1+bc=1+(-0.5)(-4)=3>0
\displaystyle \therefore (-0.5,-4)\in R_1
\displaystyle \text{However, }1+ac=1+(1)(-4)=-3<0
\displaystyle \therefore (1,-4)\notin R_1
\displaystyle \therefore (a,b)\in R_1\text{ and }(b,c)\in R_1\Rightarrow(a,c)\in R_1
\displaystyle \text{is not true for all }a,b,c\in R.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Let }R\text{ be a relation on }N\times N\text{ defined by}
\displaystyle (a,b)\,R\,(c,d)\Leftrightarrow a+d=b+c,\text{ for all }(a,b),(c,d)\in N\times N.
\displaystyle \text{Show that:}
\displaystyle \text{(i) }(a,b)\,R\,(a,b)\text{ for all }(a,b)\in N\times N
\displaystyle \text{(ii) }(a,b)\,R\,(c,d)\Rightarrow(c,d)\,R\,(a,b)
\displaystyle \text{for all }(a,b),(c,d)\in N\times N
\displaystyle \text{(iii) }(a,b)\,R\,(c,d)\text{ and }(c,d)\,R\,(e,f)\Rightarrow(a,b)\,R\,(e,f)
\displaystyle \text{for all }(a,b),(c,d),(e,f)\in N\times N.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }(a,b)\,R\,(c,d)\Leftrightarrow a+d=b+c.
\displaystyle \text{(i) For any }(a,b)\in N\times N,
\displaystyle a+b=b+a
\displaystyle \therefore (a,b)\,R\,(a,b).
\displaystyle \therefore R\text{ is reflexive.}
\displaystyle \text{(ii) Let }(a,b)\,R\,(c,d).
\displaystyle \Rightarrow a+d=b+c
\displaystyle \Rightarrow c+b=d+a
\displaystyle \therefore (c,d)\,R\,(a,b).
\displaystyle \therefore R\text{ is symmetric.}
\displaystyle \text{(iii) Let }(a,b)\,R\,(c,d)\text{ and }(c,d)\,R\,(e,f).
\displaystyle \Rightarrow a+d=b+c\qquad\text{and}\qquad c+f=d+e
\displaystyle \text{Adding the two equations,}
\displaystyle a+d+c+f=b+c+d+e
\displaystyle \Rightarrow a+f=b+e
\displaystyle \therefore (a,b)\,R\,(e,f).
\displaystyle \therefore R\text{ is transitive.}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Find the linear relation between the components of the ordered}
\displaystyle \text{pairs of the relation }R,\text{ where:}
\displaystyle \text{(i) }R=\{(0,2),(-1,5),(2,-4),\ldots\}
\displaystyle \text{(ii) }R=\{(-1,-1),(0,2),(1,5),\ldots\}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let the linear relation be }y=mx+c.
\displaystyle \text{Using the point }(0,2),\ c=2.
\displaystyle \text{Using the point }(-1,5),\ 5=-m+2\Rightarrow m=-3.
\displaystyle \therefore y=-3x+2=2-3x.
\displaystyle \text{(ii) Let the linear relation be }y=mx+c.
\displaystyle \text{Using the point }(0,2),\ c=2.
\displaystyle \text{Using the point }(-1,-1),\ -1=-m+2\Rightarrow m=3.
\displaystyle \therefore y=3x+2.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{If }A=\{1,3,5,6\}\text{ and }B=\{3,4,5\},\text{ write the relation }R
\displaystyle \text{as a set of ordered pairs, if:}
\displaystyle \text{(i) }R=\{(x,y):(x,y)\in A\times B,\ x+y\text{ is even}\}
\displaystyle \text{(ii) }R=\{(x,y):(x,y)\in A\times B,\ xy\text{ is odd}\}
\displaystyle \text{Answer:}
\displaystyle A=\{1,3,5,6\},\qquad B=\{3,4,5\}
\displaystyle A\times B=\{(1,3),(1,4),(1,5),(3,3),(3,4),(3,5),
\displaystyle (5,3),(5,4),(5,5),(6,3),(6,4),(6,5)\}
\displaystyle \text{(i) Since }x+y\text{ is even when both }x\text{ and }y\text{ are of the same parity,}
\displaystyle R=\{(1,3),(1,5),(3,3),(3,5),(5,3),(5,5),(6,4)\}
\displaystyle \text{(ii) Since }xy\text{ is odd only when both }x\text{ and }y\text{ are odd,}
\displaystyle R=\{(1,3),(1,5),(3,3),(3,5),(5,3),(5,5)\}
\displaystyle \\

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