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Page No 1.10:

Question 1:

Let A be the set of all human beings in a town at a particular time. Determine whether each of the following relations are reflexive, symmetric and transitive:

(i) R = {(x, y) : x and y work at the same place}
(ii) R = {(x, y) : x and y live in the same locality}
(iii) R = {(x, y) : x is wife of y}
(iv) R = {(x, y) : x is father of and y}

Answer:

(i) Reflexivity:
Let x be an arbitrary element of R. Then,xR  x and x work at the same place is true since they are the same.x, xRSo, R is a reflexive relation.

Symmetry:
Let x, yRx and y work at the same place y and x work at the same placey, xRSo, R is a symmetric relation.

Transitivity:
Let x, yR and y, zR. Then,x and y work at the same place.y and z also work at the same place.x , y and z all work at the same place.x and z work at the same place.x, zRSo, R is a transitive relation.

(ii) Reflexivity:
Let x be an arbitrary element of R. Then,xR x and x live in the same locality is true since they are the same.So, R is a reflexive relation.

Symmetry:
Let x, yRx and y live in the same localityy and x live in the same localityy, xR  So, R is a symmetric relation.

Transitivity:
Let x, yR and y, zR. Then,x and y live in the same locality and y and z live in the same localityx, y and z all live in the same localityx and z live in the same locality x, z RSo, R is a transitive relation.

(iii)
Reflexivity:
Let x be an element of R. Then,x is wife of x cannot be true.x, xRSo, R is not a reflexive relation.

Symmetry:
Let x, yRx is wife of y x is female and y is maley cannot be wife of x as y is husband of xy, xR  So, R is not a symmetric relation.

Transitivity:
Let x, yR, but y, zRSince x is wife of y, but y cannot be the wife of z, y is husband of x.x is not the wife of zx, zRSo, R is a transitive relation.

(iv)
Reflexivity:
Let x be an arbitrary element of R. Then,x is father of x cannot be true since no one can be father of himself.So, R is not a reflexive relation.

Symmetry:
Let x, yRx is father of yy is son/daughter of xy, xR So, R is not a symmetric relation.

Transitivity:
Let x, yR and y, zR. Then, x is father of y and y is father of zx is grandfather of zx, zRSo, R is not a transitive relation.

Page No 1.10:

Question 2:

Three relations R1, R2 and R3 are defined on a set A = {a, b, c} as follows:
R1 = {(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c)}
R2 = {(a, a)}
R3 = {(b, c)}
R4 = {(a, b), (b, c), (c, a)}.

Find whether or not each of the relations R1, R2, R3, R4 on A is (i) reflexive (ii) symmetric and (iii) transitive.

Answer:

(i) R1
Reflexive:
Clearly, (a, a), (b, b) and (c, c)R1
So, R1 is reflexive.

Symmetric:
We see that the ordered pairs obtained by interchanging the components of R1 are also in R1.
So, R1 is symmetric.

Transitive:
Here,
a, bR1, b, cR1 and also a, cR1
So, R1 is transitive.

(ii) R2
Reflexive: Clearly a,aR2. So, R2 is reflexive.
Symmetric: Clearly a,aRa,aR. So, R2 is symmetric.
Transitive: R2 is clearly a transitive relation, since there is only one element in it.

(iii) R3
Reflexive:
Here,
b, bR3 neither c, cR3
So, R3  is not reflexive.

Symmetric:
Here,
b, cR3, but c,bR3So, R3 is not symmetric.

Transitive:
Here, R3 has only two elements. Hence, R3 is transitive.

(iv) R4
Reflexive:
Here,
  a, aR4, b, b R4 c, c R4So, R4 is not reflexive.

Symmetric:
Here,
a, bR4, but b,aR4.So, R4 is not symmetric.

Transitive:
Here,
a, bR4, b, cR4, but a, cR4So, R4 is not transitive.

Page No 1.10:

Question 3:

Test whether the following relations R1, R2, and R3 are (i) reflexive (ii) symmetric and (iii) transitive:
(i) R1 on Q0 defined by (a, b) ∈ R1a = 1/b.
(ii) R2 on Z defined by (a, b) ∈ R2 ⇔ |a – b| ≤ 5
(iii) R3 on R defined by (a, b) ∈ R3a2 – 4ab + 3b2 = 0.

Answer:

(i) Reflexivity:
Let a be an arbitrary element of R1. Then,
aR1a1a for all aQ0So, R1 is not reflexive.

Symmetry:
Let (a, b) R1. Then,

a, bR1a=1bb=1ab, aR1So, R1 is symmetric.

Transitivity:
Here,
a, bR1 and b, cR2a=1b and b=1ca=11c=ca1ca, cR1 So, R1 is not transitive.

(ii)
Reflexivity:
Let a be an arbitrary element of R2. Then,
      aR2 a-a=05So, R1 is reflexive.

Symmetry:
Let a, bR2a-b5b-a5             Since, a-b = b-ab, aR2So, R2 is symmetric.

Transitivity:
Let 1, 3R2 and 3, 7R21-35 and 3-75But 1-75 1,7R2So, R2 is not transitive.

(iii)
Reflexivity: Let a be an arbitrary element of R3. Then,
     aR3a2-4a×a+3a2=0 So, R3  is reflexive.

Symmetry:
Let a, bR3a2-4ab+3b2=0But b2-4ba+3a20 for all a, b RSo, R3 is not symmetric.

Transitivity:
1, 2R3 and 2, 3R31-8+6=0 and 4-24+27=0But 1-12+90So, R3 is not transitive.

Page No 1.10:

Question 4:

Let A = {1, 2, 3}, and let R1 = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}, R2 = {(2, 2), (3, 1), (1, 3)}, R3 = {(1, 3), (3, 3)}. Find whether or not each of the relations R1, R2, R3 on A is (i) reflexive (ii) symmetric (iii) transitive.

Answer:

1 R1
Reflexivity:
Here,
1, 1, 2, 2, 3, 3RSo, R1 is reflexive.

Symmetry:
Here,2, 1R1, but 1, 2R1 So, R1 is not symmetric.

Transitivity:
Here, 2, 1R1 and 1, 3R1, but 2, 3R1So, R1 is not transitive.

2 R2
Reflexivity:

Clearly, 1, 1 and 3, 3R2 So, R2 is not reflexive.

Symmetry:
Here, 1, 3R2 and 3, 1R2So, R2 is symmetric.

Transitivity:
Here, 1, 3R2 and 3, 1R2  But 3, 3R2So, R2 is not transitive.

3 R3
Reflexivity:
Clearly, 1, 1R3 So, R3 is not reflexive.

Symmetry:
Here, 1, 3R3, but 3, 1R3So, R3 is not symmetric.

Transitivity:
Here, 1, 3R3 and 3, 3R3 Also, 1, 3R3So, R3 is transitive.



Page No 1.11:

Question 5:

The following relations are defined on the set of real numbers.
(i) aRb if ab > 0
(ii) aRb if 1 + ab > 0
(iii) aRb if |a| ≤ b

Find whether these relations are reflexive, symmetric or transitive.

Answer:

(i)
Reflexivity: Let a be an arbitrary element of R. Then,

     aRBut a-a = 0 0So, this relation is not reflexive.

Symmetry:

Let a, bRa-b>0-(b-a)>0b-a<0So, the given relation is not symmetric.

Transitivity:

Let a, bR and b, cR. Then,a-b>0 and b-c>0Adding the two, we geta-b+b-c>0a-c>0 a, cR. So, the given relation is transitive. 

(ii)
Reflexivity: Let a be an arbitrary element of R. Then,

     aR1+a×a>0i.e. 1+a2>0              Since, square of any number is positiveSo, the given relation is reflexive.

Symmetry:

Let a, bR1+ab>01+ba>0b, aRSo, the given relation is symmetric. 

Transitivity:

Let a, bR and b, cR1+ab>0 and 1+bc>0But 1+ac0a, cRSo, the given relation is not transitive. 

(iii)
Reflexivity: Let a be an arbitrary element of R. Then,

     aR aa                Since, a=aSo, R is not reflexive. 

Symmetry:

Let a, bRab  ba for all a, bRb, aR So, R is not symmetric. 

Transitivity:

Let a, bR and b, cRab and bcMultiplying the corresponding sides, we get     a bbcaca, cRThus, R is transitive. 

Page No 1.11:

Question 6:

Check whether the relation R defined on the set A = {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive.

Answer:

Reflexivity:

Letabeanarbitraryelementof R.Then,a=a+1 cannot be true for all aA.a, aR So, R is not reflexive on A.

Symmetry:
Let a, bRb=a+1-a=-b+1a=b-1Thus, b, aRSo, R is not symmetric on A.

Transitivity:
Let 1, 2 and 2, 3R2=1+1 and 3 2+1  is true.But 3  1+11, 3RSo, R is not transitive on A.

Page No 1.11:

Question 7:

Check whether the relation R on R defined by R = {(a, b) : ab3} is reflexive, symmetric or transitive.

Answer:

Reflexivity:

Since 12>123,12, 12RSo, R is not reflexive.

Symmetry:

Since 12, 2R,12<23But 2>1232, 12RSo, R is not symmetric.

Transitivity:
Since 7, 3R and 3, 313R,7<33 and 3=3133But 7>31337, 313RSo, R is not transitive.

Page No 1.11:

Question 8:

Prove that every identity relation on a set is reflexive, but the converse is not necessarily true.

Answer:

Let A be a set. Then,

Identity relation IA=IA is reflexive, since a, aAa

The converse of it need not be necessarily true.
Consider the set A = {1, 2, 3}

Here,
Relation R = {(1, 1), (2, 2) , (3, 3), (2, 1), (1, 3)} is reflexive on A.
However, R is not an identity relation.

Page No 1.11:

Question 9:

If A = {1, 2, 3, 4} define relations on A which have properties of being
(i) reflexive, transitive but not symmetric
(ii) symmetric but neither reflexive nor transitive
(iii) reflexive, symmetric and transitive.

Answer:

(i) The relation on A having properties of being reflexive, transitive, but not symmetric is
R = {(1, 1), (2, 2), (3, 3), (4, 4), (2, 1)}

Relation R satisfies reflexivity and transitivity.1, 1, 2, 2, 3, 3R and 1, 1, 2, 1R 1, 1RHowever, 2, 1R, but 1, 2R

(ii) The relation on A having properties of being symmetric, but neither reflexive nor transitive is
R = {(1, 2), (2, 1)}
The relation R on A is neither reflexive nor transitive, but symmetric.

(iii) The relation on A having properties of being symmetric, reflexive and transitive is
R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1)}
The relation R is an equivalence relation on A.

Page No 1.11:

Question 10:

Let R be a relation defined on the set of natural numbers N as
R = {(x, y) : x, y N, 2x + y = 41}
Find the domain and range of R. Also, verify whether R is (i) reflexive, (ii) symmetric (iii) transitive.

Answer:

Domain of R is the values of x and range of R is the values of y that together should satisfy 2x+y = 41.
So,
Domain of R = {1, 2, 3, 4, ... , 20}
Range of R = {1, 3, 5, ... , 37, 39}

Reflexivity: Let x be an arbitrary element of R. Then,
xR2x+x=41 cannot be true.x, xR  So, R is not reflexive.

Symmetry:
Let x, yR. Then, 2x+y=41 2y+x = 41 y, xRSo, R is not symmetric.

Transitivity:
Let x, y and y, zR2x+y=41 and 2y +z=412x+z=2x+41-2y 41-y-2y=41-3yx, zRThus, R is not transitive.

Page No 1.11:

Question 11:

Is it true that every relation which is symmetric and transitive is also reflexive? Give reasons.

Answer:

No, it is not true.

Consider a set A = {1, 2, 3} and relation R on A such that R = {(1, 2), (2, 1), (2, 3), (1, 3)}
The relation R on A is symmetric and transitive. However, it is not reflexive.

1, 1, 2, 2 and 3, 3 R

Hence, R is not reflexive.

Page No 1.11:

Question 12:

An integer m is said to be related to another integer n if m is a multiple of n.Check if the relation is symmetric, reflexive and transitive.

Answer:

R=m, n : m, nZ, m=kn, where kNReflexivity:Let m be an arbitrary element of R. Then,m=km is true for k=1m, mRThus, R is reflexive.Symmetry: Let m, nRm=kn for some kNn=1kmn, mR Thus, R is not symmetric.Transitivity: Let m, n and n, oRm=kn and n=lo for some k, l Nm=(kl) oHere, klRm, oRThus, R is transitive.

Page No 1.11:

Question 13:

Show that the relation '≥' on the set R of all real numbers is reflexive and transitive but not symmetric.

Answer:

Let R be the set such that R = {(a, b) : a, bR; ab}

Reflexivity:
Let a be an arbitrary element of R. aRa=a aa is true for a=aa, aR Hence, R is reflexive.

Symmetry:
Let a, bRab is same as ba, but not baThus, b, aR Hence, R is not symmetric.

Transitivity:
Let a, b and b, cRab and bcabcaca, cRHence, R is transitive.

Page No 1.11:

Question 14:

Give an example of a relation which is
(i) reflexive and symmetric but not transitive;
(ii) reflexive and transitive but not symmetric;
(iii) symmetric and transitive but not reflexive;
(iv) symmetric but neither reflexive nor transitive.
(v) transitive but neither reflexive nor symmetric.

Answer:

Suppose A be the set such that A = {1, 2, 3}

(i) Let R be the relation on A such that 
R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3)}
Thus,
R is reflexive and symmetric, but not transitive.

(ii) Let R be the relation on A such that 
R = {(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 3)}
Clearly, the relation R on A is reflexive and transitive, but not symmetric.

(iii) Let R be the relation on A such that 
R = {(1, 2), (2, 1), (1, 3), (3, 1), (2, 3)}
We see that the relation R on A is symmetric and transitive, but not reflexive.

(iv) Let R be the relation on A such that 
R = {(1, 2), (2, 1), (1, 3), (3, 1)}
The relation R on A is symmetric, but neither reflexive nor transitive.

(v) Let R be the relation on A such that 
R = {(1, 2), (2, 3), (1, 3)}
The relation R on A is transitive, but neither symmetric nor reflexive.

Page No 1.11:

Question 15:

Given the relation R = {(1, 2), (2, 3)} on the set A = {1, 2, 3}, add a minimum number of ordered pairs so that the enlarged relation is symmeteric, transitive and reflexive.

Answer:

We have,

R = {(1, 2), (2, 3)}

R can be a transitive only when the elements (1, 3) is added

R can be a reflexive only when the elements (1, 1), (2, 2), (3, 3) are added

R can be a symmetric only when the elements (2, 1), (3, 1) and (3, 2) are added

So, the required enlarged relation, R' = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)} = A × A

Page No 1.11:

Question 16:

Let A = {1, 2, 3} and R = {(1, 2), (1, 1), (2, 3)} be a relation on A. What minimum number of ordered pairs may be added to R so that it may become a transitive relation on A.

Answer:

We have,

A = {1, 2, 3} and R = {(1, 2), (1, 1), (2, 3)}

To make R a transitive relation on A, (1, 3) must be added to it.

So, the minimum number of ordered pairs that may be added to R to make it a transitive relation is 1.

Page No 1.11:

Question 17:

Let A = {a, b, c} and the relation R be defined on A as follows: R = {(a, a), (b, c), (a, b)}. Then, write minimum number of ordered pairs to be added in R to make it reflexive and transitive.                                                                                                                  [NCERT EXEMPLAR]

Answer:

We have,

 A = {abc} and R = {(aa), (bc), (ab)}

R can be a reflexive relation only when elements (b, b) and (c, c) are added to it

R can be a transitive relation only when the element (ac) is added to it 

So, the minmum number of ordered pairs to be added in R is 3.

Page No 1.11:

Question 18:

Each of the following defines a relation on N:

(i) x > y, x, y  N
(ii) x + y = 10, xy  N
(iii) xy is square of an integer, x, y  N
(iv) x + 4y = 10, x, y  N

Determine which of the above relations are reflexive, symmetric and transitive.                                                            [NCERT EXEMPLAR]

Answer:

(i) We have,

R = {(x, y) : x > yxy  N}

As, x=x xNx,xRSo, R is not a reflexive relationLet x,yRx>ybut y<xy,xRSo, R is not a symmeteric relationLet x,yR and y,zRx>y and y>zx>zx,zRSo, R is a transitive relation

(ii) We have,

R = {(x, y) : x + y = 10, xy  N}

R=1,9,2,8,3,7,4,6,5,5,6,4,7,3,8,2,9,1As, 1,1RSo, R is not a reflexive relationLet x,yRx+y=10y+x=10y,xRSo, R is a symmeteric relationAs, 1,9R and 9,1R but 1,1RSo, R is not a transitive relation

(iii) We have,

R = {(x, y) : xy is square of an integer, xy  N}

As, x×x=x2, which is a square of an integer xx,xRSo, R is a reflexive relationLet x,yRxy is square of an integeryx is also a square of an integery,xRSo, R is a symmeteric relationLet x,yR and y,zRxy is square of an integer and yz is also a square of an intergerxz must be a square of an integerx,zRSo, R is a transitive relation

(iv) We have,

R = {(x, y) : x + 4y = 10, xy  N}

R=2,4,6,1As, 2,2RSo, R is not a reflexive relationAs, 2,4R but 4,2RSo, R is not a symmeteric relationAs, 2, 4R but 4 is not related to any natural numberSo, R is a transitive relation



Page No 1.26:

Question 1:

Show that the relation R defined by R = {(a, b) : a – b is divisible by 3; a, b Z} is an equivalence relation.

Answer:

We observe the following relations of relation R.

Reflexivity:
 Let a be an arbitrary element of R. Then,a-a=0=0 × 3a-a is divisible by 3a, aR for all aZSo, R is reflexive on Z.

Symmetry:
Let a, bRa-b is divisible by 3a-b 3p for some pZb-a=3 -p Here, -pZb-a is divisible by 3b, aR for all a, bZSo, R is symmetric on Z.

Transitivity:
Let a, b and b, cRa-b and b-c are divisible by 3a-b=3p for some pZand b-c=3q for some qZAdding the above two, we get     a-b+b-c=3p+3qa-c=3 p+qHere, p+qZa-c is divisible by 3a, cR for all a, c ZSo, R is transitive on Z.

Hence, R is an equivalence relation on Z.

Page No 1.26:

Question 2:

Show that the relation R on the set Z of integers, given by
R = {(a, b) : 2 divides a – b},  is an equivalence relation.

Answer:

We observe the following properties of relation R.

Reflexivity:
Let a be an arbitrary element of the set Z. Then,aRa-a=0=0 × 22 divides a-aa, aR for all aZSo, R is reflexive on Z.

Symmetry:
Let a, bR2 divides a-ba-b2=p for some pZb-a2=-p Here, -pZ2 divides b-ab, aR for all a, b ZSo, R is symmetric on Z.

Transitivity:
Let a, b and b, cR2 divides a-b and 2 divides b-ca-b2=p  and b-c2=q for some p, qZAdding the above two, we geta-b2+b-c2=p+qa-c2=p+q       Here, p+qZ2 divides a-ca, cR for all a, c ZSo, R is transitive on Z.

Hence, R is an equivalence relation on Z.

Page No 1.26:

Question 3:

Prove that the relation R on Z defined by
(a, b) ∈ Rab is divisible by 5
is an equivalence relation on Z.

Answer:

We observe the following properties of relation R.

Reflexivity:
Let a be an arbitrary element of R. Then,a-a = 0 = 0 × 5a-a is divisible by 5a, aR for all aZSo, R is reflexive on Z.

Symmetry:
Let a, bRa-b is divisible by 5a-b = 5p for some pZb-a = 5 -p     Here, -pZ                      [Since pZ]b-a is divisible by 5b, aR for all a, bZSo, R is symmetric on Z.

Transitivity:
Let a, b and b, cRa-b is divisible by 5a-b = 5p for some ZAlso, b-c is divisible by 5b-c = 5q for some ZAdding the above two, we geta-b+b-c = 5p+5qa-c = 5 (p+q)a-c is divisible by 5Here, p+qZa, cR for all a, cZSo, R is transitive on Z.

Hence, R is an equivalence relation on Z.

Page No 1.26:

Question 4:

Let n be a fixed positive integer. Define a relation R on Z as follows:
(a, b) ∈ Rab is divisible by n.
Show that R is an equivalence relation on Z.

Answer:

We observe the following properties of R. Then,
Reflexivity:
Let aNHere,a-a=0=0 × na-a is divisible by na, aRa, aR for all aZSo, R is reflexive on Z.

Symmetry:
Let a, bRHere,a-b is divisible by na-b=np for some pZb-a=n -pb-a is divisible by n         [pZ-pZ]b, aR So, R is symmetric on Z.

Transitivity:
Let a, b and b, cRHere, a-b is divisible by n and b-c is divisible by n.a-b=np for some pZand b-c=nq for some qZAdding the above two, we geta-b+b-c=np+nqa-c=n (p+q)Here, p+qZa, cR for all a, cZSo, R is transitive on Z.

Hence, R is an equivalence relation on Z.

Page No 1.26:

Question 5:

Let Z be the set of integers. Show that the relation
R = {(a, b) : a, bZ and a + b is even}
is an equivalence relation on Z.

Answer:

We observe the following properties of R.

Reflexivity:
Let a be an arbitrary element of Z. Then,     aRClearly, a+a=2a is even for all aZ.a, aR for all aZSo, R is reflexive on Z.

Symmetry:
Let a, bRa+b is evenb+a is evenb, aR for all a, bZSo, R is symmetric on Z.

Transitivity:
Let a, b and b, cRa+b and b+c are evenNow, let a+b=2x for some xZand b+c=2y for some yZAdding the above two, we get     a+2b+c=2x+2ya+c=2(x+y-b), which is even for all x, y, bZThus, a, cRSo, R is transitive on Z.

Hence, R is an equivalence relation on Z.

Page No 1.26:

Question 6:

m is said to be related to n if m and n are integers and mn is divisible by 13. Does this define an equivalence relation?

Answer:

We observe the following properties of relation R.

Let  R={m, n : m, nZ : m-n is divisible by 13}Relexivity: Let m be an arbitrary element of Z. Then,mRm-m=0=0 × 13m-m is divisible by 13m, m is reflexive on Z.Symmetry: Let m, nR. Then,m-n is divisible by 13m-n=13pHere, pZn-m=13 -p Here, -pZn-m is divisible by 13n, mR for all m, nZSo, R is symmetric on Z.Transitivity: Let m, n and n, oRm-n and n-o are divisible by 13m-n=13p and n-o=13q for some p, qZAdding the above two, we get     m-n+n-o=13p+13qm-o=13 p+qHere, p+qZm-o is divisible by 13m, oR for all m, oZSo, R is transitive on Z.

Hence, R is an equivalence relation on Z.

Page No 1.26:

Question 7:

Let R be a relation on the set A of ordered pair of integers defined by (x, y) R (u, v) if xv = yu. Show that R is an equivalence relation.

Answer:

We observe the following properties of R.

Reflexivity: Let a, b be an arbitrary element of the set A. Then,    a, bAab=ba   a, b R a, bThus, R is reflexive on A.Symmetry: Let x, y and u, vA such that x, y R u, v. Then,     xv=yuvx=uyuy=vxu, v R x, ySo, R is symmetric on A.Transitivity:  Let x, y, u, v and p, qR such that x, y R u, v and u, v R p, q.xv=yu and uq=vpMultiplying the corresponding sides, we  getxv × uq=yu × vpxq=ypx, y R p, qSo, R is transitive on A.

Hence, R is an equivalence relation on A.

Page No 1.26:

Question 8:

Show that the relation R on the set A = {x Z ; 0 ≤ x ≤ 12}, given by R = {(a, b) : a = b}, is an equivalence relation. Find the set of all elements related to 1.

Answer:


We observe the following properties of R.

Reflexivity: Let a be an arbitrary element of A. Then,
aRa=a               Since, every element is equal to itselfa, aR for all aASo, R is reflexive on A.Symmetry: Let a, b Ra bb=ab, aR for all a, bASo, R is symmetric on A.Transitivity: Let a, b and b, cRa=b and b=ca=b ca=ca, cRSo, R is transitive on A.

Hence, R is an equivalence relation on A.

The set of all elements related to 1 is {1}.



Page No 1.27:

Question 9:

Let L be the set of all lines in XY-plane and R be the relation in L defined as R = {L1, L2) : L1 is parallel to L2}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.

Answer:

We observe the following properties of R.

Reflexivity: Let L1 be an arbitrary element of the set L. Then,L1LL1 is parallel to L1               Every line is parallel to itselfL1, L1R  for all L1LSo, R is reflexive on L.Symmetry: Let L1, L2RL1  is parallel to L2L2  is parallel to L1L2, L1R  for all L1 and L2LSo, R is symmetric on L.Transitivity: Let L1, L2 and L2, L3RL1  is parallel to L2  and L2  is parallel to L3L1, L2 and L3 are all parallel to each otherL1 is parallel to L3L1, L3RSo, R is transitive on L.

Hence, R is an equivalence relation on L.

Set of all the lines related to y = 2x+4
                                                 = L' = {(x, y) : y = 2x+c, where cR}

Page No 1.27:

Question 10:

Show that the relation R, defined on the set A of all polygons as
R = {(P1, P2) : P1 and P2 have same number of sides},
is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?

Answer:

We observe the following properties on R.

Reflexivity: Let P1 be an arbitrary element of A.Then, polygon P1 and P1 have the same number of sides, since they are one and the same.P1, P1R for all P1ASo, R is reflexive on A.Symmetry: Let P1, P2RP1 and P2 have the same number of sides.P2 and P1 have the same number of sides.P2, P1R for all P1, P2ASo, R is symmetric on A.Transitivity: Let P1, P2, P2, P3RP1 and P2 have the same number of sides and P2 and P3 have the same number of sides.P1, P2 and P3 have the same number of sides.P1 and P3 have the same number of sides.P1, P3R for all P1, P3 ASo, R is transitive on A.

Hence, R is an equivalence relation on the set A.

Also, the set of all the trianglesA is related to the right angle triangle T with the sides 3, 4, 5.

Page No 1.27:

Question 11:

Let O be the origin. We define a relation between two points P and Q in a plane if OP = OQ. Show that the relation, so defined is an equivalence relation.

Answer:

Let A be the set of all points in a plane such that

A={P : P is a point in the plane}Let R be the relation such that R=P, Q : P, QA and OP=OQ, where O is the origin

We observe the following properties of R.

Reflexivity: Let P be an arbitrary element of R.
The distance of a point P will remain the same from the origin.
So, OP = OP
P, PRSo, R is reflexive on A.Symmetry: Let P, QROP=OQOQ=OPQ, PRSo, R is symmetric on A.Transitivity: Let P, Q, Q, RROP=OQ and OQ=OROP=OQ=OROP=ORP, RRSo, R is transitive on A.

Hence, R is an equivalence relation on A.

Page No 1.27:

Question 12:

Let R be the relation defined on the set A = {1, 2, 3, 4, 5, 6, 7} by R = {(a, b) : both a and b are either odd or even}. Show that R is an equivalence relation. Further, show that all the elements of the subset {1, 3, 5, 7} are related to each other and all the elements of the subset {2, 4, 6} are related to each other, but no element of the subset {1, 3, 5, 7} is related to any element of the subset {2, 4, 6}.

Answer:

We observe the following properties of R.

Reflexivity:

Let a be an arbitrary element of R. Then,aRa, aR for all aASo, R is reflexive on A.Symmetry: Let a, bRBoth a and b are either even or odd.Both b and a are either even or odd.b, aR for all a, bASo, R is symmetric on A.Transitivity: Let a, b and b, cRBoth a and b are either even or odd and both b and c are either even or odd.a, b and c are either even or odd.a and c both are either even or odd.a, c R for all a, cASo, R is transitive on A.

Thus, R is an equivalence relation on A.

We observe that all the elements of the subset {1, 3, 5, 7} are odd. Thus, they are related to each other.
This is because the relation R on A is an equivalence relation.

Similarly, the elements of the subset {2, 4, 6} are even. Thus, they are related to each other because every element is even.

Hence proved.

Page No 1.27:

Question 13:

Let S be a relation on the set R of all real numbers defined by
S = {(a, b) ∈ R × R : a2 + b2 = 1}
Prove that S is not an equivalence relation on R.

Answer:

We observe the following properties of S.

Reflexivity:Let a be an arbitrary element of R. Then,     aRa2+a21aRa, aSSo, S is not reflexive on R.Symmetry: Let a, bRa2+b2=1b2+a2=1b, aS for all a, bRSo, S is symmetric on R.Transitivity: Let a, b and b, cSa2+b2=1 and b2+c2=1Adding the above two, we geta2+c2=2-2b21 for all a, b, cRSo, S is not transitive on R.

Hence, S is not an equivalence relation on R.

Page No 1.27:

Question 14:

Let Z be the set of all integers and Z0 be the set of all non-zero integers. Let a relation R on Z × Z0 be defined as
(a, b) R (c, d) ⇔ ad = bc for all (a, b), (c, d) ∈ Z × Z0,
Prove that R is an equivalence relation on Z × Z0.

Answer:

We observe the following properties of R.

Reflexivity:
Let a, b be an arbitrary element of Z × Z0. Then,a, bZ × Z0a, bZ, Z0ab=baa, bR for all a, bZ × Z0So, R is reflexive on Z × Z0.

Symmetry:
Let a, b, c, dZ×Z0 such that a, b R c, d. Then,a, b R c, dad=bccb=dac, d R a, bThus, a, b R c, dc, d R a, b for all a, b, c, dZ×Z0So, R is symmetric on Z×Z0.

Transitivity:
Let a, b, c, d, e, fN×N0 such that a, b R c, d and c, d  R e, f. Then,a, b R c, dad=bcc, d R e, fcf=dead cf=bc deaf=bea, b R e, fThus, a, b R c, d and c, d  R e, fa, b R e, fa, b R e, f for all values a, b, c, d, e, fN×N0So, R is transitive on N×N0.

Page No 1.27:

Question 15:

If R and S are relations on a set A, then prove that
(i) R and S are symmetric ⇒ RS and RS are symmetric
(ii) R is reflexive and S is any relation ⇒ RS is reflexive.

Answer:

(i) R and S are symmetric relations on the set A.

RA×A and SA×ARSA×AThus, RS is a relation on A.Let a, bA such that a, bRS. Then,a, bRSa, bR and a, bSb, aR and b, aS                  Since R and S are symmetricb, aRSThus, a, bRSb, aRS for all a, bASo, RS is symmetric on A.

Also,
Let a, bA such that a, bRSa, bR or a, bSb, aR or b, aS                 Since R and S are symmetricb, aRSSo, RS is symmetric on A.

(ii) R is reflexive and S is any relation.
Suppose aA. Then,     a, aR                               Since R is reflexivea, aRSRS is reflexive on A.

Page No 1.27:

Question 16:

If R and S are transitive relations on a set A, then prove that RS may not be a transitive relation on A.

Answer:

Let  A = {a, b, c} and R and S be two relations on A, given by

R = {(a, a), (a, b), (b, a), (b, b)} and
S = {(b, b), (b, c), (c, b), (c, c)}

Here, the relations R and S are transitive on A.

a, bRS and b, cRSBut a, cRS 

Hence, RS is not a transitive relation on A.

Page No 1.27:

Question 17:

Let C be the set of all complex numbers and Cbe the set of all no-zero complex numbers. Let a relation R on Cbe defined as

                                             z1 z2   z1-z2z1+z2 is real for all z1, z2 C0.

Show that R is an equivalence relation.

Answer:

(i) Test for reflexivity:

Since, z1-z1z1+z1=0, which is a real number.

So, z1, z1R

Hence, R is relexive relation.

(ii) Test for symmetric:

Let z1, z2R.

Then, z1-z2z1+z2=x, where x is real

-z1-z2z1+z2=-xz2-z1z2+z1=-x, is also a real number

So, z2, z1R

Hence, R is symmetric relation.

(iii) Test for transivity:

Let z1, z2R and z2, z3R.

Then, 

z1-z2z1+z2=x, where x is a real number.z1-z2=xz1+xz2z1-xz1=z2+xz2z11-x=z21+xz1z2=1+x1-x              ...1

Also, 

z2-z3z2+z3=y, where y is a real number.z2-z3=yz2+Yz3z2-yz2=z3+yz3z21-y=z31+yz2z3=1+y1-y              ...2

Dividing (1) and (2), we get

z1z3=1+x1-x×1-y1+y=z, where z is a real number.z1-z3z1+z3=z-1z+1, which is realz1, z3R

Hence, R is transitive relation.

From (i), (ii), and (iii),

R is an equivalenve relation.



Page No 1.29:

Question 1:

Write the domain of the relation R defined on the set Z of integers as follows:
(a, b) ∈ Ra2 + b2 = 25

Answer:

Domain of R is the set of values satisfying the relation R.
As a should be an integer, we get the given values of a:

0, ±3, ±4, ±5Thus,Domain of R=0, ±3, ±4, ±5



Page No 1.30:

Question 18:

Let R = {(aa3) : a is a prime number less than 5} be a relation. Find the range of R.                                                                    [CBSE 2014]

Answer:

We have,
R = {(aa3) : a is a prime number less than 5}
Or,
R = {(2, 8), (3, 27)}

So, the range of R is {8, 27}.

Page No 1.30:

Question 19:

Let R be the equivalence relation on the set Z of the integers given by R = {(a, b) : 2 divides a - b}. Write the equivalence class [0].
                                                                                                                                                                                           [NCERT EXEMPLAR]

Answer:

We have,
An equivalence relation, R = {(ab) : 2 divides a - b}

If b=0, then a-b=a-0=aAs, 2 divides a-bAnd, the set of integers which are divided by 2 is 0,±2,±4,±6,...So, the equivalence class [0]=0,±2,±4,±6,...

Page No 1.30:

Question 20:

For the set A = {1, 2, 3}, define a relation R on the set A as follows:
R = {(1, 1), (2, 2), (3, 3), (1, 3)}
Write the ordered pairs to be added to R to make the smallest equivalence relation.

Answer:

We have,
R = {(1, 1), (2, 2), (3, 3), (1, 3)}

As, (a, a R, for all values of a  A
So, R is a reflexive relation

R can be a symmetric and transitive relation only when element (3, 1) is added

Hence, the ordered pairs to be added to R to make the smallest equivalence relation is (3, 1).

Page No 1.30:

Question 21:

Let A = {0, 1, 2, 3} and R be a relation on A defined as
R = {(0, 0), (0, 1), (0, 3), (1, 0), (1, 1), (2, 2), (3, 0), (3, 3)}
Is R reflexive? symmetric? transitive?

Answer:

We have,
R = {(0, 0), (0, 1), (0, 3), (1, 0), (1, 1), (2, 2), (3, 0), (3, 3)}

As, a,aRaASo, R is a reflexive relationAlso, a,bR and b,aRSo, R is a symmetric relation as wellAnd, 0,1R but 1,2R and 2,3RSo, R is not a transitive relation

Page No 1.30:

Question 22:

Let the relation R be defined on the set A = {1, 2, 3, 4, 5} by R = {(a, b) : |a2 - b2| < 8}. Write as a set of ordered pairs.

Answer:

As, R = {(ab) : |a2 - b2| < 8}
So, R = {(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (3, 3), (3, 4), (4, 3), (4, 4), (5, 5)}

Page No 1.30:

Question 23:

Let the relation R be defined on N by aRb iff 2a + 3b = 30. Then write R as a set of ordered pairs.

Answer:

As, R = {(a, b) : 2a + 3b = 30; a, b  N}

So, R = {(3, 8), (6, 6), (9, 4), (12, 2)}

Page No 1.30:

Question 24:

Write the smallest equivalence relation on the set A = {1, 2, 3}.

Answer:

The smallest equivalence relation on the set A = {1, 2, 3} is R = {(1, 1), (2, 2), (3, 3)}



Page No 1.30.:

Question 2:

If R = {(x, y) : x2 + y2 ≤ 4; x, yZ} is a relation on Z, write the domain of R.

Answer:

Domain of R is the set of values of x satisfying the relation R.
As x must be an integer, we get the given values of x:

0, ±1, ±2Thus, Domain of R=0, ±1, ±2

Page No 1.30.:

Question 3:

Write the identity relation on set A = {a, b, c}.

Answer:

Identity set of A is
I = {(a, a), (b, b), (c, c)}

Every element of this relation is related to itself.

Page No 1.30.:

Question 4:

Write the smallest reflexive relation on set A = {1, 2, 3, 4}.

Answer:

Here,
A
= {1, 2, 3, 4}
Also, a relation is reflexive iff every element of the set is related to itself.

So, the smallest reflexive relation on the set A is
R = {(1, 1), (2, 2), (3, 3), (4, 4)}

Page No 1.30.:

Question 5:

If R = {(x, y) : x + 2y = 8} is a relation on N by, then write the range of R.

Answer:

R = {(x, y) : x + 2y = 8, x, yN}
Then, the values of y can be 1, 2, 3 only.
Also, y = 4 cannot result in x = 0 because x is a natural number.

Therefore, range of R is {1, 2, 3}.

Page No 1.30.:

Question 6:

If R is a symmetric relation on a set A, then write a relation between R and R−1.

Answer:

Here, R is symmetric on the set A.

Let a, bRb, aR           Since R is symmetrica, bR-1       By definition of inverse relationRR-1Let x, yR-1y, xR            By definition of inverse relationx, yR            Since R is symmetric R-1RThus, R=R-1

Page No 1.30.:

Question 7:

Let R = {(x, y) : |x2y2| <1) be a relation on set A = {1, 2, 3, 4, 5}. Write R as a set of ordered pairs.

Answer:

R is the set of ordered pairs satisfying the above relation. Also, no two different elements can satisfy the relation; only the same elements can satisfy the given relation.

So, R = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5)}

Page No 1.30.:

Question 8:

If A = {2, 3, 4}, B = {1, 3, 7} and R = {(x, y) : xA, yB and x < y} is a relation from A to B, then write R−1.

Answer:

Since R = {(x, y) : x A, y A and x < y},
R = {(2, 3), (2, 7), (3, 7), (4, 7)}

So, R-1 = {(3, 2), (7, 2), (7, 3), (7, 4)}

Page No 1.30.:

Question 9:

Let A = {3, 5, 7}, B = {2, 6, 10} and R be a relation from A to B defined by R = {(x, y) : x and y are relatively prime}. Then, write R and R−1.

Answer:

R = {(x, y) : x and y are relatively prime}
Then,

R = {(3, 2), (5, 2), (7, 2), (3, 10), (7, 10), (5, 6), (7, 6)}

So, R-1 = {(2, 3), (2, 5), (2, 7), (10, 3), (10, 7), (6, 5), (6, 7)}

Page No 1.30.:

Question 10:

Define a reflexive relation.

Answer:

A relation R on A is said to be reflexive iff every element of A is related to itself.

i.e. R is reflexive a, aR for all aA

Page No 1.30.:

Question 11:

Define a symmetric relation.

Answer:

A relation R on a set A is said to be symmetric iff

a, bRb, aR for all a, bAi.e. aRbbRa for all a, bA

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Question 12:

Define a transitive relation.

Answer:

A relation R on a set A is said to be transitive iff

     a, bR and b, cRa, cR for all a, b, cRi.e. aRb and bRcaRc for all a, b, cR

Page No 1.30.:

Question 13:

Define an equivalence relation.

Answer:

A relation R on set A is said to be an equivalence relation iff
(i) it is reflexive,
(ii) it is symmetric and
(iii) it is transitive.

Relation R on set A satisfying all the above three properties is an equivalence relation.

Page No 1.30.:

Question 14:

If A = {3, 5, 7} and B = {2, 4, 9} and R is a relation given by "is less than", write R as a set ordered pairs.

Answer:

 Since, R=x, y : x, yN and x<y,R = {(3, 4), (3, 9), (5, 9), (7,9)}

Page No 1.30.:

Question 15:

A = {1, 2, 3, 4, 5, 6, 7, 8} and if R = {(x, y) : y is one half of x; x, yA} is a relation on A, then write R as a set of ordered pairs.

Answer:

Since R = {(x, y) : y is one half of x; x, yA}

So, R = {(2, 1), (4, 2), (6, 3), (8, 4)}

Page No 1.30.:

Question 16:

Let A = {2, 3, 4, 5} and B = {1, 3, 4}. If R is the relation from A to B given by a R b if "a is a divisor of b". Write R as a set of ordered pairs.

Answer:

Since R = a, b : a, bN : a is a divisor of b

So, R = {(2, 4), (3, 3), (4, 4)}

Page No 1.30.:

Question 17:

State the reason for the relation R on the set {1, 2, 3} given by R = {(1, 2), (2, 1)} to be transitive.

Answer:

Since 1, 2R, 2, 1R but 1, 1R, R is not transitive on the set 1, 2, 3.For R to be in a transitive relation, we must have 1, 1R.



Page No 1.31:

Question 1:

Let R be a relation on the set N given by
R = {(a, b) : a = b − 2, b > 6}. Then,
(a) (2, 4) ∈ R
(b) (3, 8) ∈ R
(c) (6, 8) ∈ R
(d) (8, 7) ∈ R

Answer:

(c) (6, 8) ∈ R

6, 8R Then, a=b-26=8-2and b=8 > 6Hence, 6, 8R

Page No 1.31:

Question 2:

If a relation R is defined on the set Z of integers as follows:
(a, b) ∈ Ra2 + b2 = 25. Then, domain (R) is
(a) {3, 4, 5}
(b) {0, 3, 4, 5}
(c) {0, ± 3, ± 4, ± 5}
(d) none of these

Answer:

(c) {0, ± 3, ± 4, ± 5}

     R=a, b : a2+b2=25, a, bZa-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 and      b-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5

So, domain (R)=0, ± 3,± 4, ±5

Page No 1.31:

Question 3:

R is a relation on the set Z of integers and it is given by
(x, y) ∈ R ⇔ | xy | ≤ 1. Then, R is
(a) reflexive and transitive
(b) reflexive and symmetric
(c) symmetric and transitive
(d) an equivalence relation

Answer:

(b) reflexive and symmetric

Reflexivity: Let xR. Then,x-x=0 < 1x-x1x, xR for all xZSo, R is reflexive on Z.Symmetry: Let x, yR. Then,x-y  0-(y-x)  1y-x  1             Since x-y=y-xy, xR for all x, yZSo, R is symmetric on Z.Transitivity: Let x, yR and y, zR. Then,x-y  1 and y-z  1It is not always true that x-y  1.x, zRSo, R is not transitive on Z.

Page No 1.31:

Question 4:

The relation R defined on the set A = {1, 2, 3, 4, 5} by
R = {(a, b) : | a2b2 | < 16} is given by
(a) {(1, 1), (2, 1), (3, 1), (4, 1), (2, 3)}
(b) {(2, 2), (3, 2), (4, 2), (2, 4)}
(c) {(3, 3), (4, 3), (5, 4), (3, 4)}
(d) none of these

Answer:

(d) none of these

R is given by {(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3), (4, 5), (5, 4), (1, 3), (3, 1), (1, 4), (4, 1) ,(2, 4), (4, 2)}, which is not mentioned in (a), (b) or (c).

Page No 1.31:

Question 5:

Let R be the relation over the set of all straight lines in a plane such that  l1 R l2l 1l2. Then, R is
(a) symmetric
(b) reflexive
(c) transitive
(d) an equivalence relation

Answer:

(a) symmetric

A = Set of all straight lines in the plane

R=l1, l2 : l1, l2A : l1l2Reflexivity: l1 is not  l1l1, l1RSo, R is not reflexive on A.Symmetry: Let l1, l2Rl1l2l2l1l2, l1RSo, R is symmetric on A.Transitivity: Let l1, l2R, l2, l3Rl1 l2 and l2 l3But l1 is not  l3l1, l3RSo, R is not transitive on A.

Page No 1.31:

Question 6:

If A = {a, b, c}, then the relation R = {(b, c)} on A is
(a) reflexive only
(b) symmetric only
(c) transitive only
(d) reflexive and transitive only

Answer:

(c) transitive only

The relation R = {(b,c)} is neither reflexive nor symmetric because every element of A is not related to itself. Also, the ordered pair of R obtained by interchanging its elements is not contained in R.

We observe that R is transitive on A because there is only one pair.

Page No 1.31:

Question 7:

Let A = {2, 3, 4, 5, ..., 17, 18}. Let '≃' be the equivalence relation on A × A, cartesian product of A with itself, defined by (a, b) ≃ (c, d) if ad = bc. Then, the number of ordered pairs of the equivalence class of (3, 2) is
(a) 4
(b) 5
(c) 6
(d) 7

Answer:

(c) 6

The ordered pairs of the equivalence class of (3, 2) are {(3, 2), (6, 4), (9, 6), (12, 8), (15, 10), (18, 12)}.
We observe that these are 6 pairs.

Page No 1.31:

Question 8:

Let A = {1, 2, 3}. Then, the number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is
(a) 1
(b) 2
(c) 3
(d) 4

Answer:

(a) 1

The required relation is R.
R = {(1, 2), (1, 3), (1, 1), (2, 2), (3, 3), (2, 1), (3, 1)} 

Hence, there is only 1 such relation that is reflexive and symmetric, but not transitive.

Page No 1.31:

Question 9:

The relation 'R' in N × N such that
(a, b) R (c, d) ⇔ a + d = b + c is
(a) reflexive but not symmetric
(b) reflexive and transitive but not symmetric
(c) an equivalence relation
(d) none of the these

Answer:

(c) an equivalence relation

We observe the following properties of relation R.

Reflexivity: Let (a, b)N×Na, bNa+b=b+aa, bR So, R is reflexive on N×N.Symmetry: Let a, b, c, dN×N such that a, b R c, da+d=b+cd+a=c+bd, c, b, aR  So, R is symmetric on N×N.Transitivity: Let a, b, c, d, e, fN×N such that a, b R c, d and c, d R e, fa+d=b+c and c+f=d+ea+d+c+f=b+c+d+ea+f=b+ea, b R e, fSo, R is transitive on N×N.

Hence, R is an equivalence relation on N.



Page No 1.32:

Question 10:

If A = {1, 2, 3}, B = {1, 4, 6, 9} and R is a relation from A to B defined by 'x is greater than y'. The range of R is
(a) {1, 4, 6, 9}
(b) {4, 6, 9}
(c) {1}
(d) none of these

Answer:

(c) {1}

Here,
R=x,  y : xA and yB : x > yR=2, 1, 3, 1

Thus,
Range of R = {1} 

Page No 1.32:

Question 11:

A relation R is defined from {2, 3, 4, 5} to {3, 6, 7, 10} by : x R yx is relatively prime to y. Then, domain of R is
(a) {2, 3, 5}
(b) {3, 5}
(c) {2, 3, 4}
(d) {2, 3, 4, 5}

Answer:

(d) {2, 3, 4, 5}

The relation R is defined as
     R = x, y : x2, 3, 4, 5, y3, 6, 7, 10 : x is relatively prime to yR= 2, 3, 2, 7, 3, 7, 3, 10, 4, 7, 5, 3, 5, 7

Hence, the domain of R includes all the values of x, i.e. {2, 3, 4, 5}.

Page No 1.32:

Question 12:

A relation ϕ from C to R is defined by x ϕ y ⇔ | x | = y. Which one is correct?
(a) (2 + 3 i) ϕ 13
(b) 3 ϕ (−3)
(c) (1 + i) ϕ 2
(d) i ϕ 1

Answer:

(d)  i ϕ 1

 2+3i=1313     3-3     1+i=22and  i =1So, i, 1ϕ

Page No 1.32:

Question 13:

Let R be a relation on N defined by x + 2y = 8. The domain of R is
(a) {2, 4, 8}
(b) {2, 4, 6, 8}
(c) {2, 4, 6}
(d) {1, 2, 3, 4}

Answer:

(c) {2,4,6}

The relation R is defined as
     R=x, y : x, yN and x+2y = 8R=x, y : x, yN and y = 8-x2

Domain of R is all values of xN satisfying the relation R. Also, there are only three values of x that result in y, which is a natural number. These are {2, 6, 4}.

Page No 1.32:

Question 14:

R is a relation from {11, 12, 13} to {8, 10, 12} defined by y = x − 3. Then, R−1 is
(a) {(8, 11), (10, 13)}
(b) {(11, 8), (13, 10)}
(c) {(10, 13), (8, 11)}
(d) none of these

Answer:

(a) {(8, 11), (10, 13)}

The relation R is defined by
     R=x, y : x11, 12, 13, y8, 10, 12 : y = x-3R=11, 8, (13, 10)So, R-1=8, 11, 10, 13

Page No 1.32:

Question 15:

Let R = {(a, a), (b, b), (c, c), (a, b)} be a relation on set A = a, b, c. Then, R is
(a) identify relation
(b) reflexive
(c) symmetric
(d) antisymmetric

Answer:

(b) reflexive

Reflexivity: Since a, aR aA, R is reflexive on A.Symmetry: Since a, bR but b, aR, R is not symmetric on A.R is not antisymmetric on A.Also, R is not an identity relation on A.

Page No 1.32:

Question 16:

Let A = {1, 2, 3} and B = {(1, 2), (2, 3), (1, 3)} be a relation on A. Then, R is
(a) neither reflexive nor transitive
(b) neither symmetric nor transitive
(c) transitive
(d) none of these

Answer:

(c)  transitive

Reflexivity: Since (1, 1)B, B is not reflexive on A.Symmetry: Since 1, 2B but 2, 1B, B is not symmetric on A.Transitivity: Since 1, 2B, 2, 3B and 1, 3B, B is transitive on A.

Page No 1.32:

Question 17:

If R is the largest equivalence relation on a set A and S is any relation on A, then
(a) RS
(b) SR
(c) R = S
(d) none of these

Answer:

(b) SR

Since R is the largest equivalence relation on set A,
R  A × A


Since S is any relation on A,
S  A × A


So, SR

Page No 1.32:

Question 18:

If R is a relation on the set A = {1, 2, 3, 4, 5, 6, 7, 8, 9} given by x R yy = 3 x, then R =
(a) {(3, 1), (6, 2), (8, 2), (9, 3)}
(b) {(3, 1), (6, 2), (9, 3)}
(c) {(3, 1), (2, 6), (3, 9)}
(d) none of these

Answer:

(d) none of these

The relation R is defined as
     R = x, y : x, yA : y = 3xR = 1, 3, 2, 6, 3, 9

Page No 1.32:

Question 19:

If R is a relation on the set A = {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3)}, then R is
(a) reflexive
(b) symmetric
(c) transitive
(d) all the three options

Answer:

(d) all the three options

R=a, b : a=b and a, bAReflexivity: Let aA. Then,a=aa, aR for all aASo, R is reflexive on A.Symmetry: Let a, bA such that a, bR. Then,a, bRa=bb=ab, aR for all aASo, R is symmetric on A.Transitivity: Let a, b, cA such that a, bR and b, cR. Then,a, bRa=band b, cRb=ca=ca, cR for all aASo, R is transitive on A.

Hence, R is an equivalence relation on A.

Page No 1.32:

Question 20:

If A = {a, b, c, d}, then a relation R = {(a, b), (b, a), (a, a)} on A is
(a) symmetric and transitive only
(b) reflexive and transitive only
(c) symmetric only
(d) transitive only

Answer:

(a) symmetric and transitive only

Reflexivity: Since b, bR, R is not reflexive on A.Symmetry: Since a, bR and b, aR, R is symmetric on A.Transitivity: Since a, bR,  b, aR and a, aR, R is transitive on A.

Page No 1.32:

Question 21:

If A = {1, 2, 3}, then a relation R = {(2, 3)} on A is
(a) symmetric and transitive only
(b) symmetric only
(c) transitive only
(d) none of these

Answer:

(c) transitive only

The relation R is not reflexive because every element of A is not related to itself. Also, R is not symmetric since on interchanging the elements, the ordered pair in R is not contained in it.

R is transitive by default because there is only one element in it.

Page No 1.32:

Question 22:

Let R be the relation on the set A = {1, 2, 3, 4} given by
R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Then,
(a) R is reflexive and symmetric but not transitive
(b) R is reflexive and transitive but not symmetric
(c) R is symmetric and transitive but not reflexive
(d) R is an equivalence relation

Answer:

(b) R is reflexive and transitive but not symmetric.

Reflexivity: Clearly, (a, a)R  aASo, R is reflexive on A.Symmetry: Since 1, 2R, but 2, 1R,R is not symmetric on A.Transitivity: Since, 1, 3, 3, 2R and 1, 2R,R is transitive on A.

Page No 1.32:

Question 23:

Let A = {1, 2, 3}. Then, the number of equivalence relations containing (1, 2) is
(a) 1
(b) 2
(c) 3
(d) 4

Answer:

(b) 2

There are 2 equivalence relations containing {1, 2}.
R = {(1, 2)}
S = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2), (1, 3), (3, 1)}

Page No 1.32:

Question 24:

The relation R = {(1, 1), (2, 2), (3, 3)} on the set {1, 2, 3} is
(a) symmetric only
(b) reflexive only
(c) an equivalence relation
(d) transitive only

Answer:

(c) an equivalence relation

R=a, b : a=b and a, bAReflexivity: Let aA Here,a=aa, aR for all aASo, R is reflexive on A.Symmetry: Let a, bA such that a, bR. Then,a, bRa=bb=ab, aR for all aASo, R is symmetric on A.Transitive: Let a, b, cA such that a, bR and b, cR. Then, a, bRa=band b, cRb=ca=ca, cR for all aASo, R is transitive on A.

Hence, R is an equivalence relation on A.



Page No 1.33:

Question 25:

S is a relation over the set R of all real numbers and it is given by
(a, b) ∈ Sab ≥ 0. Then, S is
(a) symmetric and transitive only
(b) reflexive and symmetric only
(c) antisymmetric relation
(d) an equivalence relation

Answer:

(d) an equivalence relation

Reflexivity: Let aR
Then,

aa=a2>0a, aR  aR

So, S is reflexive on R.

Symmetry: Let (a, b)S
Then,

a, bSab0 ba0 b, aS a, bR

So, S is symmetric on R.

Transitivity:

If a, b, b, cSab0 and bc0ab×bc0ac0                        b2  0a, cS for all a, b, cset R

Hence, S is an equivalence relation on R.

Page No 1.33:

Question 26:

In the set Z of all integers, which of the following relation R is not an equivalence relation?
(a) x R y : if xy
(b) x R y : if x = y
(c) x R y : if xy is an even integer
(d) x R y : if xy (mod 3)

Answer:

(a) x R y : if xy

Clearly, R is not symmetric because x < y does not imply y < x.

Hence, (a) is not an equivalence relation.

Page No 1.33:

Question 27:

Mark the correct alternative in the following question:

Let A = {1, 2, 3} and consider the relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}. Then, R is

(a) reflexive but not symmetric                                                               (b) reflexive but not transitive
(c) symmetric and transitive                                                                    (d) neither symmetric nor transitive

Answer:

We have,

R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}

As, a,aR aASo, R is reflexive relationAlso, 1,2R but 2,1RSo, R is not symmetric relationAnd, 1,2R,2,3R and 1,3RSo, R is transitive relation

Hence, the correct alternative is option (a).

Page No 1.33:

Question 28:

Mark the correct alternative in the following question:

The relation S defined on the set R of all real number by the rule aSb iff a b is

(a) an equivalence relation
(b) reflexive, transitive but not symmetric
(c) symmetric, transitive but not reflexive
(d) neither transitive nor reflexive but symmetric

Answer:

We have,

S = {(a, b) : a  b; a, b  R}

As, a=a aRa,aSSo, S is reflexive relationLet a,bSabBut bab,aSSo, S is not symmetric relationLet a,bS and b,cSab and bcaca,cSSo, S is transitive relation

Hence, the correct alternative is option (b).

Page No 1.33:

Question 29:

Mark the correct alternative in the following question:

The maximum number of equivalence relations on the set A = {1, 2, 3} is

(a) 1                                  (b) 2                                  (c) 3                                  (d) 5

Answer:

Consider the relation R1=1,1It is clearly reflexive, symmetric and transitiveSimilarly, R2=2,2 and R3=3,3 are reflexive, symmetric and transitiveAlso, R4=1,1,2,2,3,3,1,2,2,1It is reflexive as a,aR4 for all a1,2,3It is symmetric as a,bR4b,aR4 for all a1,2,3Also, it is transitive as 1,2R4,2,1R4(1,1)R4The relation defined by R5=1,1,2,2,3,3,1,2,1,3,2,1,2,3,3,1,3,2 is reflexive, symmetric and transitive as well.Thus, the maximum number of equivalence relation on set A={1,2,3} is 5.

Hence, the correct alternative is option (d).

Page No 1.33:

Question 30:

Mark the correct alternative in the following question:

Let R be a relation on the set N of natural numbers defined by nRm iff n divides m. Then, R is

(a) Reflexive and symmetric                                                         (b) Transitive and symmetric
(c) Equivalence                                                                          (d) Reflexive, transitive but not symmetric                  [NCERT EXEMPLAR]

Answer:

We have,

R = {(m, n) : n divides m; m, n  N}

As, m divides mm,mR mNSo, R is reflexiveSince, 2,1R i.e. 1 divides 2but 2 cannot divide 1 i.e. 2,1RSo, R is not symmetricLet m,nR and n,pR. Then,n divides m and p divides np divides mm,pRSo, R is transitive

Hence, the correct alternative is option (d).

Page No 1.33:

Question 31:

Mark the correct alternative in the following question:

Let L denote the set of all straight lines in a plane. Let a relation R be defined by lRm iff l is perpendicular to m for all l, m L. Then, R is

(a) reflexive                                 (b) symmetric                                  (c) transitive                                  (d) none of these
                                                                                                                                                                                           [NCERT EXEMPLAR]

Answer:

We have,R=l,m:l is perpendicular to m; l,mLAs, l is not perpencular to ll,lRSo, R is not reflexive relationLet l,mRl is perpendicular to mm is also perpendicular to lm,lRSo, R is symmetric relationLet l,mR and m,nRl is perpendicular to m and m is perpendicular to nl is parallel to n                     Lines perpendicular to same line are parallelm,lRSo, R is not transitive relation

Hence, the correct alternative is option (b).

Page No 1.33:

Question 32:

Mark the correct alternative in the following question:

Let T be the set of all triangles in the Euclidean plane, and let a relation R on T be defined as aRb if a is congruent to b for all a, b  T. Then, R is

a) reflexive but not symmetric                                                                               (b) transitive but not symmetric
c) equivalence                                                                                                     (d) none of these

Answer:

We have,R=a,b:a is congruent to b; a,bTAs, aaa,aRSo, R is reflexive relationLet a,bR. Then,abbab,aRSo, R is symmetric relationLet a,bR and b,cR. Then,ab and bcaca,cRSo, R is transitive relation R is an equivalence relation

Hence, the correct alternative is option (c).

Page No 1.33:

Question 33:

Mark the correct alternative in the following question:

Consider a non-empty set consisting of children in a family and a relation R defined as aRb if a is brother of b. Then, R is

(a) symmetric but not transitive                                                                                (b) transitive but not symmetric
(c) neither symmetric nor transitive                                                                       (d) both symmetric and transitive

Answer:

We have,R=a,b:a is brother of bLet a,bR. Then,a is brother of bbut b is not necessary brother of a           As, b can be sister of ab,aRSo, R is not symmetricAlso,Let a,bR and b,cRa is brother of b and b is brother of ca is brother of ca,cRSo, R is transitive

Hence, the correct alternative is option (b).

Page No 1.33:

Question 34:

Mark the correct alternative in the following question:

For real numbers x and y, define xRy iff x-y+2 is an irrational number. Then the relation R is

(a) reflexive                          (b) symmetric                          (c) transitive                          (d) none of these

Answer:

We have,R=x,y:x-y+2 is an irrational number; x,yRAs, x-x+2=2, which is an irrational numberx,xRSo, R is reflexive relationSince, 2,2Ri.e. 2-2+2=22-2, which is an irrational numberbut 2-2+2=2, which is a rational number2,2RSo, R is not symmetric relationAlso, 2,2R and 2,22Ri.e. 2-2+2=22-2, which is an irrational number and 2-22+2=2-2, which is also an irrational numberBut 2-22+2=0, which is a rational number2,22RSo, R is not transitive relation

Hence, the correct alternative is option (a).



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