Mathematics

Session

Set, Relation & FunctionSession - 2

Session Objectives

Class Test

Class Exercise - 1

If A = {1, 2, 3} and B = {3, 8}, then is A B A B

(a) {(3, 1), (3, 2), (3, 3), (3, 8)}

(b) {(1, 3), (2, 3), (3, 3), (8, 3)}

(c) {(1, 2), (2, 2), (3, 3), (8, 8)}

(d) {(8, 3), (8, 2), (8, 1), (8, 8)}

Solution

A B 1, 2, 3, 8

A B 3

A B A B 1,3 , 2, 3 , 3, 3 , 8, 3

Hence, answer is (b).

Class Exercise - 2

Let A and B be two non-empty setssuch that . Then provethat A × B and B × A have n2

elements in common.

O A B n

Solution

As A B C D A C B D

A B B A A B B A

A B A B

O A B B A O A B A B

O A B .O A B

2 2O A B n

A B and B A have n2 elements is common.

Class Exercise - 3

R be a relation on set of natural numbers N defined as . Find the following.

(i) R, R–1 as sets of ordered pairs

(ii) Domain of R and R–1

(iii) Range of R and R–1

(iv) R–1 oR (v) R–1 in set-builder form

R a, b | 3a b 12, a, b N

Solution

(i) R a, b |3a b 12, a, b N= {(1, 9), (2, 6), (3, 3)}

R–1 = {(9, 1), (6, 2), (3, 3)}

(ii) Domain (R) = {1, 2, 3}Domain (R–1) = {9, 6, 3}

(iii)Range (R) = {9, 6, 3}Domain (R–1) = {1, 2, 3}

(iv){(9, 1), (6, 2), (3, 3)} o {(1, 9), (2, 6), (3, 3)}= {(1, 1), (2, 2), (3, 3)}

(v) R–1 = {(a, b) | a + 3b = 12, a, }b N

Class Exercise - 4

Let R be a relation from A = {2, 3, 4, 5, 6}to B = {3, 6, 8, 9, 12} defined as Express R asset of ordered pairs, find the domain andthe range of R, and also find R–1 inset-builder form (where x | y means xdivides y).

R x, y | x A, y B, x | y .

Solution

R x, y | x A, y B, x | y

= {(2, 6), (2, 8), (2, 12), (3, 3), (3, 6), (3, 9), (3, 12),

(4, 8), (4, 12), (6, 6), (6, 12)}

Domain (R) = {2, 3, 4, 6}

Range (R) = {6, 8, 12, 3, 9}

R–1 = {(6, 2), (8, 2), (12, 2), (3, 3), (6, 3), (9, 3), (12, 3), (8, 4), (12, 4), (6, 6), (12, 6)}

and R–1 x is divisible by y} x, y |x B, y A,

Class Exercise - 5

Let R be a relation on Z defined as ExpressR and R–1 as set of ordered pairs.Hence, find the domain of R and R–1.

2 2R x, y | x, y Z, x y 25 .

Solution

R = {(0, 5), (0, –5), (3, 4), (3, –4), (4, 3), (4, –3), (5, 0), (–3, 4), (–3, –4), (–4, 3), (–4, –3), (–5, 0)}

R–1 = {(5, 0), (–5, 0), (4, 3), (–4, 3), (3, 4), (–3, 4), (0, 5), (4, –3), (–4, –3), (3, –4), (–3, –4), (0, –5)} = R

Domain (R) = Domain (R–1) = {0, 3, 4, 5, –3, –4, –5}

Class Exercise - 6

Let S be the set of all the straight lines on a plane, R be a relation on S defined as . Then check R for reflexivity, symmetry and transitivity.

1 2 1 2 1 2R l , l |l l , l , l S

Solution

Reflexive:

as a line cannot be perpendicular to itself.

1 1 1l , l R for any l S

Symmetric: Let 1 2 1 2l , l R i.e. l l

2 1l l

2 1l , l R

R is symmetric

Transitive: Let , i.e. 1 2 2 3l , l , l , l R 1 2 2 3l l and l l

1 3 1 3l , l R as l || l .

Not transitive

Class Exercise - 7

Let f = ‘n/m’ means that n is factorof m or n divides m, where .Then the relation ‘f’ is

(a) reflexive and symmetric(b) transitive and symmetric(c) reflexive, transitive and symmetric(d) reflexive, transitive and not symmetric

n, m N

Solution

Reflexive: a/a a N

As a is factor of a a 1.a

Reflexive

Symmetric: Let i.e. a/b or a is a factor of b

a, b f b ka for some k N

1

a bk

b is not a factor of a until and unless a = b

not symmetric (but antisymmetric)

Solution contd..

Transitive: Let a, b , b, c f

a /b and b / c

1 2 1 2b k a and c k b for some k , k N

2 1c k k a a / c , i.e. a is factor of c

a, c f

Hence, answer is (d).

Class Exercise - 8

Let where R is set of realsdefined as Check S for reflexive, symmetric andtransitive.

S R R, 2 2x, y S or xSy x y 1.

Solution

Reflexive: Let , i.e. a, a S

2 2 2 1a a 1 2a 1 a

2

Hence, only for two values of R not

Not reflexive

a, a S R

Symmetric: If , i.e. a2 + b2 = 1 a, b S

2 2b a 1

b, a S

Symmetric

Solution contd..

Transitive: If a, b , b, c S

i.e. a2 + b2 = 1 and b2 + c2 = 1

2 2a c

2 2a c may not be 1

Not transitive

Class Exercise - 9

Let A be a set of all the points in space. Let R be a relation on A such that a1 Ra2 if distance between the points a1 and a2 is less than one unit. Then which of the following is false?

(a) R is reflexive(b) R is symmetric(c) R is transitive(d) R is not an equivalence relation

Solution

Reflexive: Distance between a1 and a1

is 0 less than one unit.

Hence, 1 1 1a Ra a A

Reflexive

Symmetric: If Distance between isless than 1 unit.

1 2a Ra 1 2a and a

Distance between a2 and a1 is less than 1 unit.

a2Ra1 Symmetric

Transitive: If 1 2 2 3a R a and a R a

Solution contd..

Distance between a1 and a2 is lessthan 1 unit and distance between a2 anda3 is less than 1 unit

Distance between a1 and a3 is less than 1 unit.

For example, 0.9 0.9

1 2 3a a a

Distance between a1 and a3 = 1.8 > 1

Not transitive

Hence, R is not an equivalence relation.

Hence, answer is (c).

Class Exercise - 10

Let N denote the set of all naturalnumbers and R be the relation onN × N defined by

Show that R is an equivalence relation.

a, b R c, d ad b c bc a d .

Solution

Reflexive: (a, b) R (a, b)

As ab(b + a) = ba(a + b)

Reflexive

Symmetric: If (a, b) R (c, d)

ad(b + c) = bc(a + d)

cb(d + a) = da(c + b)

(c, d) R (a, b)

R is symmetric

Solution contd..

Transitive: If (a, b) R (c, d) and (c, d) R (e, f)

ad(b + c) = bc(a + d) and cf(d + e) = de(c + f)

1 1 1 1 1 1 1 1

andb c a d d e c f

1 1 1 1 1 1 1 1

andc d a b c d e f

1 1 1 1 1 1 1 1a b e f a f b e

af(b + e) = be(a + f) (a, b) R (e, f)

Transitive

Hence, R is an equivalence relation.

Thank you