Relations and Functions Test - 3

R is not reflexive as (1, 1)^∉ R. Moreover, (1, 2) ^∈ R but (2, 1)^∉ R. Thus, R is not symmetric. Now, for (a, b)^∈ R and (b, c) ^∈ R, (a, c) ^∈ R . So, R is transitive.

For commutativity:

a * b = a + b – ab and b * a = b + a – ba = a + b – ab

So, a * b = b * a. So, * is commutative.

For associativity:

a * (b * c) = a * (b + c – bc) = a + b + c – bc – ab – ac +abc, and

(a * b) * c = (a + b – ab) * c = a + b – ab + c – ac – bc + abc.

So, a * (b * c) = (a * b) * c.

The given set R= {(1,x) , ( 2,z), (1,y), (3,x)}

 is a subset of the cross product A x B  and any subset of the cross product A x B  is a relation from A to B , so
R= {(1,x) , ( 2,z), (1,y), (3,x)} is a relation from A to B.

 f(x) = [x] , the greatest integer £ x and g(x ) = |x | , x ^∈ R .

g o f(-1/3) = g{f(-1/3)} = g(-1) = | -1 |= 1

f o g (-1/3) = f{| -1/3 |= f{1/3} = [1/3]= 0

^⇒g o f(-1/3) - f o g (-1/3 )= 1 - 0 = 1

(3* 4) *x = - 1

(3* 4) *x = (3 + 4 +3 .4)*x = (7 +12)*x =19*x= -1 ( given a*b = a + b +ab )

^⇒19 +x + 19(x) = 19 + 20x = -1

^⇒20 x = -20

^⇒x =-1

0 is the identity element of this table or this operation.

as, 0*a=a*0 = 0 ^∀ a ^∈ S={0,1,2,3,4,5}

From the 6 ^th row and 6 ^th column ,we observe 5*1 = 1*5= 0,the identity . So 5 ^-1 = 1

From the table, the second row and second column are the same as the original set.

0*0 = 0, 1*0 = 0*1 = 1 , 2*0= 0*2 = 2 , 3*0= 0*3 = 3 , 4*0 = 0*4 =4, 0*5 = 5*0 = 5

^⇒ ’0’is the identity element of the operation ‘*’