For the two sets A and B , given n ( A × B ) = 6 and three of the elements of A × B are 2 , 5 , ( 4 , 6 ) , ( 8 , 6 ) . Then the remaining elements are

( 2 , 6 ) , ( 4 , 5 ) , ( 8 , 5 )

( 2 , 4 ) , ( 4 , 8 ) , ( 5 , 6 )

( 2 , 6 ) , ( 4 , 5 ) , ( 4 , 8 )

( 2 , 4 ) , ( 4 , 5 ) , ( 6 , 8 )

If A = { 1 , 2 } , B = { 1 , 2 , 3 } , then ( A × B ) ∩ ( B × A ) is

{ ( 1 , 1 ) , ( 1 , 2 ) , ( 2 , 1 ) , ( 2 , 2 ) }

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

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

Show that the function f : N → N defined by f n = n + 1 is not onto, by drawing arrow diagrams.

Given, the function f : N → N defined by f n = n + 1 . N = 1 , 2 , 3 , 4 , 5 , 6 , . . . . . . . f 1 = 1 + 1 = 2 f 2 = 2 + 1 = 3 f 3 = 3 + 1 = 4 f 4 = 4 + 1 = 5 f 5 = 5 + 1 = 6 Thus, f = 1 , 2 , 2 , 3 , 3 , 4 , 4 , 5 , 5 , 6 , . . . . . . . . . . . . Different elements have different images. Here, the range is not equal to the codomain. Therefore, the given function is not an onto function.

If A = 1 , 2 , 3 , B = 4 , 5 and C = { 5 , 6 } , then find A × ( B - C ) .

Given sets, A = { 1 , 2 , 3 } , B = { 4 , 5 } and C = { 5 , 6 } . We need to find, A × ( B - C ) Now, B - C = 4 ∴ A × ( B - C ) = { 1 , 2 , 3 } × 4 ∴ A × ( B - C ) = { ( 1 , 4 ) , ( 2 , 4 ) , ( 3 , 4 ) }

If A = 1 , 2 , 3 , B = 4 , 5 and C = { 5 , 6 } , then verify that A × ( B - C ) = ( A × B ) - ( A × C ) .

Given sets, A = { 1 , 2 , 3 } , B = { 4 , 5 } and C = { 5 , 6 } . We need to prove that, A × ( B - C ) = ( A × B ) - ( A × C ) From LHS, we have A × ( B - C ) Now, B - C = 4 ∴ A × ( B - C ) = { 1 , 2 , 3 } × 4 ∴ A × ( B - C ) = { ( 1 , 4 ) , ( 2 , 4 ) , ( 3 , 4 ) } → 1 From RHS, we have ( A × B ) - ( A × C ) A × B = { ( 1 , 4 ) , ( 1 , 5 ) , ( 2 , 4 ) , ( 2 , 5 ) , ( 3 , 4 ) , ( 3 , 5 ) } And A × C = 1 , 5 , 1 , 6 , 2 , 5 , 2 , 6 , 3 , 5 , 3 , 6 ∴ ( A × B ) - ( A × C ) = { ( 1 , 4 ) , ( 2 , 4 ) , ( 3 , 4 ) } → 2 Hence, from 1 and 2 A × ( B - C ) = ( A × B ) - ( A × C ) Hence proved.

Let A = 1 , 2 and B = 3 , 4 . Find A × B .

Given sets, A = 1 , 2 and B = 3 , 4 We know that the Cartesian product of two sets A and B is the set of all possible ordered pairs where the elements of A are first and the elements of B are second. It is given by A × B = a , b : a ∈ A , b ∈ B ∴ A × B = 1 , 2 × 3 , 4 ⇒ A × B = ( 1 , 3 ) , ( 1 , 4 ) , ( 2 , 3 ) , ( 2 , 4 ) .

Show that the function f : N → N defined by f n = n + 1 is not onto, by drawing arrow diagrams.

Given, the function f : N → N defined by f n = n + 1 . N = 1 , 2 , 3 , 4 , 5 , 6 , . . . . . . . f 1 = 1 + 1 = 2 f 2 = 2 + 1 = 3 f 3 = 3 + 1 = 4 f 4 = 4 + 1 = 5 f 5 = 5 + 1 = 6 Thus, f = 1 , 2 , 2 , 3 , 3 , 4 , 4 , 5 , 5 , 6 , . . . . . . . . . . . . Different elements have different images. Here, the range is not equal to the codomain. Therefore, the given function is not an onto function.