Let f , g , h : R → R , given by f x = 2 x + 1 , g x = 1 - 4 x and h x = x 2 + 1 . Find f o g o h and g o f o h .

We have, f x = 2 x + 1 g x = 1 - 4 x h x = x 2 + 1 Now, f o g o h x = f g h x ⇒ f o g o h x = f g x 2 + 1 ⇒ f o g o h x = f 1 - 4 x 2 + 1 ⇒ f o g o h x = f - 3 - 4 x 2 ⇒ f o g o h x = 2 - 3 - 4 x 2 + 1 ⇒ f o g o h x = - 8 x 2 - 5 And, g o f o h x = g f h x ⇒ g o f o h x = g f x 2 + 1 ⇒ g o f o h x = g 2 x 2 + 1 + 1 ⇒ g o f o h x = g 2 x 2 + 3 ⇒ g o f o h x = 1 - 4 2 x 2 + 3 ⇒ g o f o h x = - 11 - 8 x 2

Let f , g , h : R → R , given by f x = 2 x + 1 , g x = 1 - 4 x and h x = x 2 + 1 . Find f o g o h and g o f o h .

We have, f x = 2 x + 1 g x = 1 - 4 x h x = x 2 + 1 Now, f o g o h x = f g h x ⇒ f o g o h x = f g x 2 + 1 ⇒ f o g o h x = f 1 - 4 x 2 + 1 ⇒ f o g o h x = f - 3 - 4 x 2 ⇒ f o g o h x = 2 - 3 - 4 x 2 + 1 ⇒ f o g o h x = - 8 x 2 - 5 And, g o f o h x = g f h x ⇒ g o f o h x = g f x 2 + 1 ⇒ g o f o h x = g 2 x 2 + 1 + 1 ⇒ g o f o h x = g 2 x 2 + 3 ⇒ g o f o h x = 1 - 4 2 x 2 + 3 ⇒ g o f o h x = - 11 - 8 x 2

Let N m be the set of all ordered m - tuples of natural numbers. If x = x 1 , x 2 , x 3 , … … , x m , y = y 1 , y 2 , y 3 , … . , y m where x i , y i ∈ N , i = 1 , 2 , 3 , … . , m and an operation ' + ' is defined by x + y = x 1 + y 1 , x 2 + y 2 , … … x m + y m , then prove that given operation is commutative as well as associative.

We have, x = x 1 , x 2 , x 3 , … … , x m , y = y 1 , y 2 , y 3 , … . , y m Now, x + y = x 1 + y 1 , x 2 + y 2 , … … , x m + y m y + x = y 1 + x 1 , y 2 + x 2 , … … , y m + x m But addition is commutative, hence x 1 + y 1 = y 1 + x 1 , x 2 + y 2 = y 2 + x 2 , . . . . , x m + y m = y m + x m Therefore, x + y = y + x , so " + " is commutative. Also, let z = z 1 , z 2 , z 3 , … . , z m , then x + y + z = x 1 + y 1 + z 1 , x 2 + y 2 + z 2 , x 3 + y 3 + z 3 , . . . . , x m + y m + z m x + y + z = x 1 + y 1 + z 1 , x 2 + y 2 + z 2 , x 3 + y 3 + z 3 , . . . . , x m + y m + z m But addition is associative, hence x + y + z = x + y + z Therefore, " + " is associative.

Prove that f : N → N , defined by f ( m ) = m 2 + m + 1 for all m ∈ N , is one-one but not onto.

Given that, f : N → N defined by f ( x ) = x 2 + x + 1 We know that, a function f : A → B is said to be a one-one function or an injection, if different elements of A have different images in B . An onto function is such that for every element in the codomain there exists an element in domain which maps to it. f ( x ) = f ( y ) , ∀ x , y ∈ N ⇒ x 2 + x + 1 = y 2 + y + 1 ⇒ x 2 + x - y 2 - y = 0 ⇒ ( x - y ) ( x + y ) + x - y = 0 ⇒ ( x - y ) ( x + y + 1 ) = 0 ⇒ x - y = 0 [ x + y + 1 ≠ 0 ∀ x , y ∈ N ] ⇒ x = y Thus f is one-one. Further f ( x ) = x 2 + x + 1 ≥ 3 ∀ x ∈ N Thus f ( x ) = 1 and 2 and have no pre images in N (domain). ∴ f is not onto. Thus f is one-one but not onto.

Let a binary operation' * 'on Q (set of all rational numbers) be defined by a * b = a + 2 b for all a , b ∈ Q . Prove that : Q is closed under the given operation.

We have, a * b = a + 2 b Let x , y ∈ Q , then x * y = x + 2 y Since, x , y ∈ Q ⇔ x + 2 y ∈ Q ∀ x , y ∈ Q , hence x * y ∈ Q , hence Q is closed under given operation.

Let a binary operation' * 'on Q (set of all rational numbers) be defined by a * b = a + 2 b for all a , b ∈ Q . Prove that the given operation is not commutative.

We have, a * b = a + 2 b b * a = b + 2 a Clearly, a * b ≠ b * a , hence given operation is not commutative.

Let a binary operation' * 'on Q (set of all rational numbers) be defined by a * b = a + 2 b for all a , b ∈ Q . Prove that the given operation is not associative.

We have, a * b = a + 2 b Let x , y , z ∈ Q , then x * y * z = x * y + 2 z = x + 2 y + 2 z = x + 2 y + 4 z x * y * z = x + 2 y * z = x + 2 y + 2 z Clearly, x * y * z ≠ x * y * z , hence given operation is not associative.

Check whether the relation R defined in the set 1 , 2 , 3 , 4 , 5 , 6 as R = a , b : b = a + 1 is reflexive, symmetric or transitive.

Suppose A = 1 , 2 , 3 , 4 , 5 , 6 . A relation R is defined on set A as: R = { ( a , b ) : b = a + 1 } ∴ R = { ( 1, 2 ) , 2 , 3 , 3 , 4 , 4 , 5 , 5 , 6 } Reflexive Property : A relation R in a set A is said to be reflexive if x , x ∈ R , for every x ∈ A Now, 1 , 1 , 2 , 2 , 3 , 3 , 4 , 4 , 5 , 5 , 6 , 6 ∉ R ∴ R is not reflexive. Symmetric Property : A relation R in a set A is said to be symmetric if x , y ∈ R implies that y , x ∈ R for all x , y ∈ A . Now it can be observed that 1 , 2 ∈ R , but 2 , 1 ∉ R . Hence , R is not symmetric. Transitive Property : A relation R in a set A is said to be transitive if x , y ∈ R and y , z ∈ R implies that x , z ∈ R for all x , y , z ∈ A . Now, 1 , 2 , 2 , 3 ∈ R but, ( 1 , 3 ) ∉ R ∴ R is not transitive. Hence, R is neither reflexive, nor symmetric, nor transitive.