Question

A = {1, 2, 3, 4}, B = {-1, 2, 5, 6} and R = {{a, b) / a < b, a E A, bE B} then range of R is (A) {2, 5, 6} (B) A (C) B (D) Ø​

Answer 1

Answer:

Steps:

1) Draw a line segment AB of length 9.3 units.

2) Extend the line by 1 unit more such that BC=1 unit .

3) Find the midpoint of AC.

4) Draw a line BD perpendicular to AB and let it intersect the semicircle at point D.

5) Draw an arc DE such that BE=BD.

Therefore, BE= 9.3units

Answer 2

is an equivalance relation if R is reflexive,symmetric and transitive.

a)checking if it is reflexive;

Given R in A×Aand(a,b)R(c,d)suchthata+d=b+c

For reflexive,consider (a,b)R(a,b)(a,b)∈A

and applying given condition⇒a+b=b+a;which is true for all A

∴Risreflexive.

b)checking if it is symmetric;

given(a,b)R(c,d)suchthata+d=b+c

consider (c,d)R(a,b)onA×A

applying given condition⇒c+b=d+awhichsatisfiesgivencondition

Hence R is symmetric.

c)checking if it is transitive;

Let(a,b)R(c,d)and(c,d)R(e,f)

and(a,b),(c,d),(e,f)∈A×A

applying given condition:⇒a+d=b+c→1andc+f=d+e→2

equation 1⇒a−c=b−d

nowaddequation1and2;

⇒a−c+c+f=b−d+d+e

⇒a+f=b+e

∴(a,b)R(e,f)alsosatisfiesthecondition

Hence R is transitive.

Therfore by above inspection we can say that R is an equivalence relation.

⇒stepsforfindingequivalenceclass[3,4]:

Let (3,4)R(a,b)onA×AwhereA={1,2,3,.......10}

⇒3+b=4+a

leta=1⇒b=2

therfore one pair (a,b)=(1,2)

similarly we can find the pairs (a,b)

Therefore equivalence class of [3,4]={(1,2),(2,3),(3,4),(4,5),(5,6),(6,7),(7,8),(8,9),(9,10)}

Answer By