Let R be a relation in N defined by a R b if 5 divides a − b , ( a , b ∈ N ) . Prove that a R b ⇒ a 2 R b 2 .

It is given that a R b if 5 divides a - b , a , b ∈ N . Now, a R b ⇒ a - b = 5 k , for some k . Multiplying both sides by a + b , we get: a - b a + b = 5 k a + b ⇒ a - b a + b = 5 k a + b ⇒ a 2 - b 2 = 5 k a + b ⇒ a 2 - b 2 is divisible by 5 . Hence, a 2 R b 2 . Thus, a R b ⇒ a 2 R b 2 Hence proved.

Find the inverse of the function f : R → R defined by f ( x ) = 3 x + 1 2 ∀ x ∈ R .

In order to prove that f is invertible we will show that f is a bijection. f is an injection: Let x , y ∈ R . Then, f x = f y ⇒ 3 x + 1 2 = 3 y + 1 2 ⇒ 3 x + 1 = 3 y + 1 ⇒ 3 x = 3 y ⇒ x = y Thus, f x = f y ⇒ x = y ∀ x , y ∈ R . So, f is an injection. f is a surjection: Let y be an arbitrary element of R . Then, f x = y ⇒ 3 x + 1 2 = y ⇒ x = 2 y - 1 3 Clearly, 2 y - 1 3 ∈ R ∀ y ∈ R . Thus, for all y ∈ R , ∃ x = 2 y - 1 3 ∈ R such that f x = f 2 y - 1 3 = 3 2 y - 1 3 + 1 2 = y Thus, f is a surjection. Hence, f : R → R is a bijection and so it is invertible. Now, f ( x ) = 3 x + 1 2 Let f x = y . Then, ⇒ y = 3 x + 1 2 ⇒ 2 y = 3 x + 1 ⇒ 2 y - 1 = 3 x ⇒ x = 2 y - 1 3 ⇒ f - 1 y = 2 y - 1 3 Thus, f - 1 : R → R is given by f - 1 x = 2 x - 1 3

Show that the relation R on N × N defined by ( a , b ) R ( c , d ) ⇔ a + d = b + c , ∀ ( a , b ) , ( c , d ) ∈ N × N is an equivalence relation.

Given that, R be the relation in N × N defined by ( a , b ) R ( c , d ) ⇔ a + d = b + c for a , b , c , d in N × N R is reflexive if ( a , b ) R ( a , b ) for a , b in N × N Let ( a , b ) R ( a , b ) ⇒ a + b = b + a which is true since addition is commutative on N . ⇒ ( a , b ) R ( a , b ) ⇒ R is reflexive. R is symmetric if ( a , b ) R ( c , d ) ⇒ ( c , d ) R ( a , b ) for a , b , c , d in N × N Let ( a , b ) R ( c , d ) ⇒ a + d = b + c ⇒ b + c = a + d ⇒ c + b = d + a [since addition is commutative on N ] ⇒ c , d R ( a , b ) ⇒ ( c , d ) R ( a , b ) ⇒ R is symmetric. R is transitive if a , b R c , d and c , d R e , f ⇔ ( a , b ) R e , f for ( a , b ) , ( c , d ) , e , f in N × N Let a , b R c , d and c , d R e , f ⇒ a + d = b + c and c + f = d + e ⇒ ( a + d ) + ( c + f ) = ( b + c ) + ( d + e ) ⇒ a + f = b + e ⇒ ( a , b ) R e , f ⇒ R is transitive. Since, R is reflexive, symmetric and transitive and hence it is an equivalence relation. Hence, proved

Find the inverse of the function f : R → R defined by f ( x ) = 3 x + 1 2 ∀ x ∈ R .

In order to prove that f is invertible we will show that f is a bijection. f is an injection: Let x , y ∈ R . Then, f x = f y ⇒ 3 x + 1 2 = 3 y + 1 2 ⇒ 3 x + 1 = 3 y + 1 ⇒ 3 x = 3 y ⇒ x = y Thus, f x = f y ⇒ x = y ∀ x , y ∈ R . So, f is an injection. f is a surjection: Let y be an arbitrary element of R . Then, f x = y ⇒ 3 x + 1 2 = y ⇒ x = 2 y - 1 3 Clearly, 2 y - 1 3 ∈ R ∀ y ∈ R . Thus, for all y ∈ R , ∃ x = 2 y - 1 3 ∈ R such that f x = f 2 y - 1 3 = 3 2 y - 1 3 + 1 2 = y Thus, f is a surjection. Hence, f : R → R is a bijection and so it is invertible. Now, f ( x ) = 3 x + 1 2 Let f x = y . Then, ⇒ y = 3 x + 1 2 ⇒ 2 y = 3 x + 1 ⇒ 2 y - 1 = 3 x ⇒ x = 2 y - 1 3 ⇒ f - 1 y = 2 y - 1 3 Thus, f - 1 : R → R is given by f - 1 x = 2 x - 1 3

Let Z be the set of integers. Show that the relation R = { ( a , b ) : a , b ∈ Z and a + b is even } is an equivalence relation.

Given that, ∀ a , b ∈ Z , R = { ( a , b ) : ( a + b ) is even } R is reflexive if ( a , a ) ∈ R ∀ a ∈ Z For any a ∈ A , we have a + a = 2 a , which is even. ⇒ ( a , a ) ∈ R Thus, R is reflexive. R is symmetric if ( a , b ) ∈ R ⇒ ( b , a ) ∈ R ∀ a , b ∈ Z ( a , b ) ∈ R ⇒ a + b is even. ⇒ b + a is even. ⇒ ( b , a ) ∈ R Thus, R is symmetric . R is transitive if ( a , b ) ∈ R and ( b , c ) ∈ R ⇒ ( a , c ) ∈ R ∀ a , b , c ∈ Z Let ( a , b ) ∈ R and ( b , c ) ∈ R ∀ a , b , c ∈ Z ⇒ a + b = 2 P and b + c = 2 Q where P , Q are any two integers Adding both, we get a + c + 2 b = 2 ( P + Q ) ⇒ a + c = 2 ( P + Q ) - 2 b ⇒ a + c is an even number ⇒ ( a , c ) ∈ R Thus, R is transitive on Z . Since R is reflexive, symmetric and transitive it is an equivalence relation on Z . Hence proved

Show that the operation * on R + (set of all positive real numbers) defined by a * b = a b 3 , ∀ a , b ∈ R + is a binary operation on R + .

Given * on R + , defined by a * b = a b 3 , ∀ a , b ∈ R + . Let a , b ∈ R + then a b ∈ R + and a b 3 ∈ R + . ⇒ a * b ∈ R + Thus, * is a binary operation on R +

Let R be a relation in N defined by a R b if 5 divides a − b , ( a , b ∈ N ) . Prove that a R b ⇒ a 2 R b 2 .

It is given that a R b if 5 divides a - b , a , b ∈ N . Now, a R b ⇒ a - b = 5 k , for some k . Multiplying both sides by a + b , we get: a - b a + b = 5 k a + b ⇒ a - b a + b = 5 k a + b ⇒ a 2 - b 2 = 5 k a + b ⇒ a 2 - b 2 is divisible by 5 . Hence, a 2 R b 2 . Thus, a R b ⇒ a 2 R b 2 Hence proved.

Define identity element for a binary operation defined on a set.

Let * be a binary operation on a set A . An element e is called an identity element in A with respect to * if and only if a * e = e * a = a , ∀ a ∈ A Consider an example: Show that zero is the identity for addition on R . Now, e is the identity of ∗ if a ∗ e = e ∗ a = a i.e. a + e = e + a = a This is possible only if e = 0 . Since, a + 0 = 0 + a = a ∀ a ∈ R . Thus, 0 is the identity element for addition on R .