Relations And Functions

It is given that Relation R in the set A = {1, 2, 3, ..., 13, 14} defined as

R = {(x, y) : 3x – y = 0}

Then, R = {(1,3), (2,6), (3,9), (4,12)}

R is not reflexive as (1,1), (2,2) …… (14,14) ∉ R

Also, R is not symmetric as (1,3) ϵ R, but (3,1) ∉ R

A relation is said to be transitive if (x, z) ϵ R whenever (x, y) ϵ R and (y, z) ϵ R
And, also R is not transitive as (1,3), (3,9) ϵ R, but (1,9) ∉ R

Therefore, R is neither reflexive, nor symmetric, nor transitive.

It is given that Relation R in the set N of natural numbers defined as

R = {(x, y) : y = x + 5 and x < 4}

REFLEXIVE
A relation is said to be reflexive if (x, x) ϵ R, where x is from domain.
we can see that (1,1) ∉ R

⇒ R is not reflexive.

Now, (1,6) ϵ r but (6,1) ∉ R.

⇒ R is not symmetric.

Now, since there is no pair in R such that (x, y) and (y, z) ∈R, then (x, z) cannot belong to R.

∴R is not transitive.

Therefore, R is neither reflexive, nor symmetric, nor transitive.

It is given that relation R in the set A = {1, 2, 3, 4, 5, 6} as

R = {(x, y) : y is divisible by x}

We know that any number (x) is divisible by itself.

⇒ (x,x) ϵ R

⇒ R is reflexive.

Now, (2,4) ϵ R but (4,2) ∉ R.

⇒ R is not symmetric.

Let (x,y), (y,z) ϵ R. Then, y is divisible x and z is divisible by y.

⇒ z is divisible by x.

⇒ (x,z) ϵ R

⇒ R is transitive.

Therefore, R is reflexive and transitive but not symmetric.

It is given that Relation R in the set Z of all integers defined as

R = {(x, y) : x – y is an integer}

Now, for every x ϵ Z, (x, x) ϵ R, as x –x = 0 is an integer.

⇒ R is reflexive.

Now, for every x, y ϵ Z if (x, y) ϵ R, then x – y is an integer.

⇒ -(x – y) is also an integer.

⇒ (y – x) is an integer.

⇒ (y, x) ϵ R

⇒ R is symmetric.

Now, for every (x, y) and (y, z) ϵ R where x, y, z ϵ R,

⇒ (x – y) and (y – z) is an integer.

⇒ (x – z) = (x – y) + (y – x) is an integer.

⇒ (x, z) ϵ R

⇒ R is transitive.

Therefore, R is reflexive, symmetric and transitive.

(a) It is given that R = {(x, y) : x and y work at the same place}

⇒ (x, x) ϵ R

⇒ R is reflexive.

Now, if (x, y) ϵ R, then x and y work on the same place.

⇒ y and x work at the same place.

⇒ (y, x) ϵ R

⇒ R is symmetric.

Now, let (x, y), (y, z) ϵ R

⇒ x and y work at the same place and y and z work at the same place.

⇒ x and z work at the same place

⇒ (x, z) ϵ R

⇒ R is transitive.

Therefore, R is reflexive, symmetric and transitive.

(b) It is given that R = {(x, y) : x and y live in the same locality}

⇒ (x,x) ϵ R as x and x live in the same human being.

⇒ R is reflexive.

Now, if (x,y) ϵ R, then x and y live in the same locality.

⇒ y and x live in the same locality.

⇒ (y,x) ϵ R

⇒ R is symmetric.

Now, let (x,y), (y,z) ϵ R

⇒ x and y live in the same locality and y and z live in the same locality.

⇒ x and z live in the same locality

⇒ (x,z) ϵ R

⇒ R is transitive.

Therefore, R is reflexive, symmetric and transitive.

(c) It is given that R = {(x, y) : x is exactly 7 cm taller than y}

⇒ (x,x) ∉ R as human being x cannot be taller than himself.

⇒ R is not reflexive.

Now, if (x,y) ϵ R, then x is exactly 7 cm taller than y.

⇒ But y is not taller than x.

⇒ (y,x) ∉ R

⇒ R is not symmetric.

Now, let (x,y), (y,z) ϵ R

⇒ x is exactly 7 cm taller than y and y is exactly 7 cm taller than z.

⇒ x is exactly 14 cm taller than z.

⇒ (x,z) ∉ R

⇒ R is not transitive.

Therefore, R is neither reflexive, nor symmetric, nor transitive.

(d) It is given that R = {(x, y) : x is wife of y}

⇒ (x,x) ∉ R as x cannot be the wife of herself.

⇒ R is not reflexive.

Now, if (x,y) ϵ R, then x is the wife of y.

⇒ But y is not wife of x.

⇒ (y,x) ∉ R

⇒ R is not symmetric.

Now, let (x,y), (y,z) ϵ R

⇒ x is the wife of y and y is the wife of z.

⇒ This cannot be possible.

⇒ (x,z) ∉ R

⇒ R is not transitive.

Therefore, R is neither reflexive, nor symmetric, nor transitive.

(e) It is given that R = {(x, y) : x is father of y}

⇒ (x,x) ∉ R as x cannot be the father of himself.

⇒ R is not reflexive.

Now, if (x,y) ϵ R, then x is the father of y.

⇒ But y is not father of x.

⇒ (y,x) ∉ R

⇒ R is not symmetric.

Now, let (x,y), (y,z) ϵ R

⇒ x is the father of y and y is the father of z.

⇒ x is not the father of z.

⇒ Indeed x is the grandfather of z.

⇒ (x,z) ∉ R

⇒ R is not transitive.

Therefore, R is neither reflexive, nor symmetric, nor transitive.

It is given that R = {(a, b) : a ≤ b²}

Check for reflexive:

We can see that

Since,

Therefore, R is not reflexive.

Now, (1,4) ϵ R as 1 < 4²

But 4 is not less than 1².

Then, (4,1) ∉ R

Therefore, R is not symmetric.

Now, (3, 2), (2, 1.5) ϵ R

But, 3 > (1.5)² = 2.25.

Then, (3, 1.5) ∉ R

Therefore, R is not transitive.

Therefore, R is neither reflexive, nor symmetric, nor transitive.

Let us take A = {1, 2, 3, 4, 5, 6}

A relation R is defined on set A as:

R = {(a, b): b = a + 1}

Then, R = {(1,2), (2,3), (3,4), (4,5), (5,6)}

Now, we will find (a, a) ∉ R, where a ϵ A

For instance,

(1,1), (2,2), (3,3), (4,4), (5,5), (6,6) ∉ R

Therefore, R is not reflexive.

We can see that (1,2) ϵ R, but (2,1) ∉ R.

Therefore, R is not symmetric.

And now, (1,2), (2,3) ϵ R

But, (1,3) ∉ R

Therefore, R is not transitive.

Therefore, R is neither reflexive, nor symmetric, nor transitive.

It is given that R = {(a, b): a ≤ b},

It is clear that (a, a) ϵ r as a = a.

Therefore, R is reflexive.

Now let us take (2,4) ϵ R (2 < 4)

But, (4,2) ∉ Ras 4 is greater than 2.

Therefore, R is not symmetric.

Now, let (a, b), (b, c) ϵ R

Then, a ≤ b and b ≤ c

⇒ a ≤ c

⇒ (a, c) ϵ R

Therefore, R is a transitive.

Therefore, R is reflexive and transitive but not symmetric.

It is given that R = {(a, b) : a ≤ b³}

Now, It can observed that as

Therefore, R is not reflexive.

Now, (1,3) ϵ R(as 1 < 3³ = 27)

But, (3,1) ∉ R(as 3³ > 1)

Therefore, R is not symmetric.

Now, if we have

Therefore, R is not transitive.

Therefore, R is neither reflexive, nor symmetric, nor transitive.

Let us take A = {1, 2, 3}

A relation R is defined on set A as R = {(1,2), (2,1)}

It is seen that (1,1), (2,2), (3,3) ∉ R.

Therefore, R is not reflexive.

Now, we can see that (1,2) ϵ R and (2,1) ϵ R

Therefore, R is symmetric.

And now, (1,2), (2,1) ϵ R

But, (1,1) ∉ R

Therefore, R is not transitive.

Therefore, R is symmetric but neither reflexive, nor transitive.

It is given that the set A of all the books in a library of a college. Then,

R = {(x, y): x and y have same number of pages}

Now, R is reflexive

Since, (x,x) ϵ R as x and x have same number of pages.

Let (x,x) ϵ R

⇒ x and y have the same number of pages

⇒ y and x have the same number of pages.

⇒ (y,x) ϵ R

Therefore, R is symmetric.

Now, let (x,y) ϵ R and (y,z) ϵ R

⇒ x and y have the same number of pages and y and z have the same number of pages.

⇒ x and z have the same number of pages.

⇒ (x,z) ϵ R

Therefore, R is transitive.

Therefore, R is an equivalence relation.

It is given that the relation R in the set A = {1, 2, 3, 4, 5}

given by R = {(a, b) : |a – b| is even},

It is clear that for any element a ϵ A, we have |a – a|= 0(which is even)

Therefore, R is reflexive.

Now, Let (a,b) ϵ R

⇒ |a – b| is even

⇒ |-(a – b)| = |b - a| is also even.

⇒ (b,a) ϵ R

Therefore, R is symmetric.

Now, Let (a,b) ϵ R and (b,c) ϵ R .

⇒ |a – b| is even and |b – c| is even

⇒ (a – b) is even and (b – c) is even

⇒ (a – c) = (a – b) + (b – c) is even

⇒ |a – c| is even

⇒ (a,c) ϵ R

Therefore, R is transitive.

Therefore, R is an equivalence relation.

Now, all elements of the set {1,3,5} are related to each other as all the elements of this subset are odd.

Therefore, the modules of the difference between any two elements will be even.

Similarly, all elements of the sets {2,4} are related each other as all the elements of this subset are even.

Also, no element of the subset {1, 3, 5} can be related to any element of {2, 4} as all elements of {1, 3, 5} are odd and all elements of {2,4} are even.

Therefore, the modulus of the difference between the two elements will not be even.

It is given that the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by

R = {(a, b) : |a – b| is a multiple of 4}

For any element a ϵ A, we have (a,a) ϵ R as |a-a|=0 is a multiple of 4.

Therefore, R is reflexive.

Now, Let (a,a) ϵ R

⇒ |a – b| is a multiple of 4

⇒ |b – a| = |a – b| is a multiple of 4

⇒ (b,a) ϵ R

Therefore, R is symmetric.

Now, Let (a,b), (b,c) ϵ R

⇒ |a – b| is a multiple of 4 and |b - c| is a multiple of 4

⇒ |a – c| = |(a – b) + (b - c)| is a multiple of 4

⇒ (a,c) ϵ R

Therefore, R is transitive.

Therefore, R is an equivalence relation.

The set of elements related to 1 is {1,5,9}

|1-1| = 0 is multiple of 4

|5-1| = 4 is multiple of 4

|9-1| = 8 is multiple of 4.

It is given that the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by

R = {(a, b) : a = b}

For any element a ϵ A, we have (a,a) ϵ R as a = a.

Therefore, R is reflexive.

Now, Let (a,a) ϵ R

⇒ a = b

⇒ b = a

⇒ (b,a) ϵ R

Therefore, R is symmetric.

Now, Let (a,b), (b,c) ϵ R

⇒ a = b and b = c

⇒ a = c

⇒ (a,c) ϵ R

Therefore, R is transitive.

Therefore, R is an equivalence relation.

The set of elements related to 1 will be those elements from set A which are equal to 1.

Therefore, the set of elements related to 1 is {1}.

Let A = {3,4,5}

Define a relation R on A as R = {(3,4), (4,3)}

Relation R is not reflexive as (3,3), (4,4) and (5,5) ∉ R.

Now, as (3,4) ϵ R and also (4,3) ϵ R,

R is symmetric.

⇒ (3,4), (4,3) ϵ R, but (3,3) ∉ R

⇒ R is not transitive.

Therefore, relation R is symmetric but not reflexive or transitive.

Let a relation R in R defined as:

R = {(a,b): a<b}

For any a ϵ R, we have (a,a)) ∉ R as a cannot be strictly less than a itself.

In fact, a = a,

Therefore, R is not reflexive.

Now, (1,2) ϵ R but 2 > 1

⇒ (2,1)) ∉ R.

⇒ R is not symmetric.

Now, let (a,b), (b,c) ϵ R

⇒ a < b and b < c

⇒ a < c

⇒ (a,c) ϵ R

⇒ R is transitive.

Therefore, relation R is transitive but not reflexive and a symmetric.

Let us take A = {2,4,6}

Define a relation R on A as:

A = {(2,2), (4,4), (6,6), (2,4), (4,2), (4,6), (6,4)}

Relation of R is reflexive as for every a ϵ A,

(a,a) ϵ R

⇒ (2,2), (4,4), (6,6) ϵ R,

Relation R is symmetric as (a,b) ϵ R

⇒ (b,a) ϵ R for all a ,b ϵ R

And Relation R is not transitive as (2,4), (4,6) ϵ R,

but (2,6) ∉ R

Therefore, relation R is reflexive and symmetric but not transitive.

| Let us define a relation R in R as

R = {(a,b) : a³ ≥ b³}

It is clear that (a,a) ϵ R as a³ = a³

⇒ R is reflexive.

Now, (2,1) ϵ R

But (1,2) ∉ R

⇒ R is not symmetric.

Now, let (a,b) (b,c) ϵ R

⇒ a³≥ b³ and b³≥ c³

⇒ a³≥ c³

⇒ (a,c) ϵ R

⇒ R is transitive.

Therefore, relation R is reflexive and transitive but not symmetric.

Let A = {-7, -8}

Define a relation R on A as:

R = {(-7, -8), (-8, -7), (-7, -7)}

Relation R is not reflexive as (-8, -8) ∉ R

Relation R is symmetric as (-7, -8) ϵ R and (-8, -7) ϵ R

But it is seen that (-7, -8), (-8, -7) ϵ R.

Also, (-7, -7) ϵ R.

⇒ R is transitive.

Therefore, relation R is symmetric and transitive but not reflexive.

It is given that

R = {(P, Q) : distance of the point P from the origin is same as the distance of the point Q from the origin},

Now, it is clear that

(P,P) ϵ R since the distance of point P from origin is always the same as the distance of the same point P from the origin.

Therefore, R is reflexive.

Now, Let us take (P,Q) ϵ R,

⇒ The distance of point P from origin is always the same as the distance of the same point Q from the origin.

⇒ The distance of point Q from origin is always the same as the distance of the same point P from the origin.

⇒ (Q,P) ϵ R

Therefore, R is symmetric.

Now, Let (P,Q), (Q,S) ϵ R

⇒ The distance of point P and Q from origin is always the same as the distance of the same point Q and S from the origin.

⇒ The distance of points P and S from the origin is the same.

⇒ (P,S) ϵ R

Therefore, R is transitive.

Therefore, R is equivalence relation.

The set of all points related to P ≠ (0,0) will be those points whose distance from the origin is the same as the distance of point P from the origin.

So, if O(0,0) is the origin and OP = k, then the set of all points related to P is at a distance of k from the origin.

Therefore, this set of points forms a circle with the centre as the origin and this circle passes through point P.

It is given that the relation R defined in the set A of all triangles as

R = {(T₁, T₂): T₁ is similar to T₂},

Now, R is reflexive as every triangle is similar to itself.

Now, if (T₁, T₂) ϵ R, then T₁ is similar to T₂.

⇒ T₂ is similar to T₁.

⇒ (T₁, T₂) ϵ R

Therefore, R is symmetric.

Now, if (T₁, T₂), (T₂, T₃) ϵ R,

⇒ T₁ is similar to T₂ and T₂ is similar to T₃.

⇒ T₁ is similar to T₃.

⇒ (T₁, T₃) ϵ R

Therefore, R is transitive.

Therefore, R is equivalence relation.

Now, we can see that,

Therefore, the corresponding sides of triangles T₁ and T₃ are in the same ratio.

Thus, triangle T₁ is similar to triangle T₃.

Therefore, T₁ is related to T₃.

It is given that the relation R defined in the set A of all polygons as

R = {(P₁, P₂): P₁ and P₂ have same number of sides},

Then, R is reflexive since (P₁, P₂) ϵ R as the same polygon has the same number of sides with itself.

Let (P₁, P₂) ϵ R

⇒ P₁ and P₂ have the same number of sides.

⇒ P₂ and P₁ have the same number of sides.

⇒ (P₂, P₁) ϵ R

Therefore, R is symmetric.

Now, let (P₁, P₂), (P₂, P₃) ϵ R

⇒ P₁ and P₂ have the same number of sides. Also, P₂ and P₃ have the same number of sides.

⇒ P₁ and P₃ have the same number of sides.

⇒ (P₁, P₃) ϵ R

Therefore, R is transitive.

Thus, R is an equivalence relation.

The elements in A related to the right-angled triangle (T) with sides 3, 4 and 5 are those polygons which have 3 sides.

Therefore, the set of all elements in A related to triangle T is the set of all triangels.

It is given that the relation in L defined as

R = {(L₁, L₂): L₁ is parallel to L₂}

R is reflexive as any line L₁ is parallel to itself

⇒ (L₁, L₂) ϵ R

Now, Let (L₁, L₂) ϵ R

⇒ L₁ is parallel to L₂.

⇒ L₂ is parallel to L₁.

⇒ (L₂, L₁) ϵ R

Therefore, R is symmetric.

Now, Let (L₁, L₂), (L₂, L₃) ϵ R

⇒ L₁ is parallel to L₂. Also, L₂ is parallel to L₃.

⇒ L₁ is parallel to L₃.

⇒ (L₁, L₃) ϵ R

Therefore, R is transitive.

Therefore, R is an equivalence relation.

The set of all lines related to the line y = 2x + 4 is the set of all lines that are parallel to the line

y = 2x + 4

Slope of line y = 2x + 4 is m = 2

We know that parallel lines have the same slopes.

The line parallel to the given line is of the form y = 2x + c where, c ϵ R.

Therefore, the set of all lines related to the given line by y = 2x + c, where c ϵ R.

It is given that the relation in the set {1, 2, 3, 4} given by

R = {(1, 2), (2, 2), (1, 1), (4,4), (1, 3), (3, 3), (3, 2)}

It is seen that (a, a) ϵ R, for every a ϵ {1,2,3,4}

Therefore, R is reflexive.

We can seen that (1, 2) ϵ R, but (2, 1) ∉R.

Therefore, R is not symmetric.

(a, b), (b, c) ϵ R

⇒ (a, c) ϵ R

here (1,3) ϵ R , (3,2) ϵ R and (1,2) ϵ R

Therefore, R is transitive.

Therefore, R is reflexive and transitive but not symmetric.

It is given that the relation in the set N given by

R = {(a, b): a = b – 2, b > 6}

Now, b > 6, (2, 4) ∉ R

Also, as 3 ≠ 8 – 2, (3, 8) ∉ R

And 8 ≠ 7 – 2

Therefore, (8,7) ∉ R

Now, consider (6,8)

We have 8 > 6 and also 6 = 8 - 2

Therefore, (6, 8) ϵ R.

It is given that f : R* → R* defined by

check for one-one:

For a function to be one-one, if f(x) = f(y) then x = y.

f(x) = f(y)

⇒ x = y

⇒Therefore, f is one – one.

We can see that y ϵ R, there exists , such that

⇒ f is onto.

Therefore, function f is one-one and onto.

Now, Let us consider g: N → R_* defined by

Then, we get,

g(x₁) = g(x₂)

⇒ x₁ = x₂

⇒ g is one–one.

It can be observed that g is not onto as for 1.2 ϵ R there does not exist any x in N such that

Therefore, function g is one –one but not onto.

It is given that f : N → N given by f (x) = x²

We can see that for x, y ϵ N,

f(x) = f(y)

⇒ x² = y²

⇒ x = y

⇒ f is injective.

Now, let 2 ϵ N. But, we can see that there does not exists any x in N such that

f(x) = x² = 2

⇒ f is not surjective.

Therefore, function f is injective but not surjective.

It is given that f : Z → Z given by f (x) = x²

We can see that f(-1) = f(1) = 1, but -1 ≠ 1

⇒ f is not injective.

Now, let -2 ϵ Z. But, we can see that there does not exists any x in Z such that

f(x) = x² = -2

⇒ f is not surjective.

Therefore, function f is neither injective nor surjective.

It is given that f : R → R given by f (x) = x²

We can see that f(-1) = f(1) = 1, but -1 ≠ 1

⇒ f is not injective.

Now, let -2 ϵ R. But, we can see that there does not exists any x in R such that

f(x) = x² = -2

⇒ f is not surjective.

Therefore, function f is neither injective nor surjective.

It is given that f : N → N given by f (x) = x³

We can see that for x, y ϵ N,

f(x) = f(y)

⇒ x³ = y³

⇒ x = y

⇒ f is injective.

Now, let 2 ϵ N. But, we can see that there does not exists any x in N such that

f(x) = x³ = 2

⇒ f is not surjective.

Therefore, function f is injective but not surjective.

It is given that f : Z → Z given by f (x) = x³

We can see that for x, y ϵ N,

f(x) = f(y)

⇒ x³ = y³

⇒ x = y

⇒ f is injective.

Now, let 2 ϵ Z. But, we can see that there does not exists any x in Z such that

f(x) = x³ = 2

⇒ f is not surjective.

Therefore, function f is injective but not surjective.

It is given f : R → R, given by f (x) = [x]

We can see that f(1.2) = [1.2] = 1

f(1.9) = [1.9] = 1

⇒ f(1.2) = f(1.9), but 1.2 ≠ 1.9.

⇒ f is not one- one.

Now, let us consider 0.6 ϵ R.

We know that f(x) = [x] is always an integer.

⇒ there does not exist any element x ϵ R such that f(x) = 0.6

⇒ f is not onto.

Therefore, the greatest integer function is neither one-one nor onto.

It is given that f : R → R, given by f (x) = | x|

We can see that f(-1) = |-1| = 1, f(1) = |1| = 1

⇒ f(-1) = f(1), but -1 ≠ 1.

⇒ f is not one-one.

Now, we consider -1 ϵ R.

We know that f(x) = |x| is always positive

Therefore, there doesn't exist any element x in domain R such that f(x) = |x| = -1

⇒ f is not onto.

Therefore, modulus function is neither one-one nor onto.

It is given that f : R → R, given by

We can observed that f(1) = f(2) = but 1 ≠ 2.

Thus, f is not one – one.

Now, as f(x) takes only 3 values (1, 0, -1) for the element -2 in co-domain R, there exists any x in domain R such that f(x) = -2

Thus, f is not onto.

Therefore, the signum function is neither one-one nor onto.

It is given that A = {1, 2, 3}, B = {4, 5, 6, 7}

f: A → B is defined as f = {(1, 4), (2, 5), (3, 6)}

Therefore, f(1) = 4, f(2) = 5, f(3) = 6

We can see that the images of distinct elements of A under f are distinct.

Therefore, function f is one- one.

It is given that f : R → R defined by f (x) = 3 – 4x

Let x₁, x₂ϵ R such that f(x₁) = f(x₂)

⇒ 3 – 4x₁ = 3 – 4x₂

⇒ -4x₁ = -4x₂

⇒ x₁ = x₂

⇒ f is one- one

For any real number (y) in R, there exist in R such that

⇒ f is onto.

Therefore, function f is bijective.

It is given that f : R → R defined by f (x) = 1 + x²

Let x₁, x₂ϵ R such that f(x₁) = f(x₂)

Now, f(1) = f(-1) = 2

⇒ f(x₁) = f(x₂) which does means that x₁ = x₂

⇒ f is not one – one

Now consider an element -2 in co- domain R.

We can see that f(x) = 1 + x² is always positive.

⇒ there does not exist any x in domain R such that f(x) = -2

⇒ F is not onto.

Therefore, function f is neither one-one nor onto.

It is given that f : A × B → B × A is defined as f (a, b) = (b, a)

Now let us consider (a₁, b₁), (a₂, b₂) ϵ A × B

Such that f(a₁, b₁) = f(a₂, b₂)

⇒ (b₁, a₁) = (b₂, a₂)

⇒ b₁ = b₂ and a₁ = a₂

⇒ (a₁, b₁) = (a₂, b₂)

⇒ f is one-one.

Now, let (b, a) ϵ B × A be any element.

Then, there exists (a, b) ϵ A × B such that f(a, b) = (b, a)

⇒ f is onto.

Therefore, f is bijective.

It is given that

f : N → N be defined by

We can observed that:

(by using the definition of f)

Thus, f(1) = f(2), where 1 ≠ 2.

Therefore, f is not one-one.

Now, let us consider a natural number (n) in co domain N.

Case I: When n is odd.

Then, n = 2r +1 for some r ϵ N.

⇒ there exist 4r + 1 ϵ N such that f(4r+1) =

Case II: When n is even.

Then, n = 2r for some r ϵ N.

⇒ there exist 4r ϵ N such that f(4r) =

Therefore, f is onto.

⇒ Function f is not one-one but it is onto.

Thus, Function f is not bijective function.

It is given that A = R – {3} and B = R – {1}

f: A → B defined by

Now, let x, y ϵ A such that f(x) = f(y)

⇒ (x-2)(y-3) = (y-2)(x-3)

⇒ xy – 3x -2y + 6 = xy -3y -2x +6

⇒ -3x -2y = -3y -2x

⇒ x = y

⇒ f is one –one.

Let y ϵ B = R – {1}

Then, y ≠ 1.

The function f is onto if there exist x ϵ A such that f(x) = y

Now, f(x) = y

⇒ x – 2 = xy -3y

⇒ x(1-y) = -3y + 2

⇒ y ϵ B, there exists ϵ A such that

⇒ f is onto.

Therefore, function f is one- one and onto.

f: R → R be defined as f(x) = x⁴.

Let x, y ϵ R such that f(x) = f(y)

⇒ x⁴ = y⁴

⇒ x = y

Therefore, f(x₁) = f(x₂) which does not implies x₁ = x₂.

For instance, f(1) = f(-1) = 1

Therefore, f is not one-one.

Now, an element 2 in co-domain R.

We can see that there does not exist any x in domain R such that

f(x) = 2

Therefore, f is not onto.

Therefore, function f is neither one-one nor onto.

It is given that f: R → R be defined as f (x) = 3x.

Let x, y ϵ R such that f(x) = f(y).

⇒ 3x = 3y

⇒ x = y

⇒ f is one–one.

Also, for any real number (y) in co–domain R, there exists in R such that

Therefore, f is onto.

Therefore, function f is one-one and onto.

It is given that f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} be given by

f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Then,

gof(1) = g (f(1)) = g (2) = 3

gof(3) = g (f(3)) = g (5) = 1

gof(4) = g (f(4)) = g (1) = 3

Therefore, gof = {(1,3), (3,1), (4,3)}

(i) (f + g)oh = foh + goh

Let us consider ((f + g)oh)(x) = (f + g)(h(x))

= f(h(x)) + g(h(x))

= (foh)(x) + (goh)(x)

= {(foh) + (goh)}(x)

Then, ((f + g)oh)(x) = {(foh) +(goh)}(x) ∀ x ϵ R

Therefore, (f + g)oh = foh + goh.

(ii) (f.g)oh = (foh).(goh)

Let us consider ((f.g)oh)(x) = (f.g)(h(x))

= f(h(x)).g(h(x))

= f (h(x)).g(h(x))

= (fog)(x).(goh)(x)

= {(fog).(goh)}(x)

Then, ((f.g)oh)(x) = {(fog).(goh)}(x) ∀ x ϵ R

Therefore, (f.g)oh = (fog).(goh)

(i) f(x) = |x| and g(x) = |5x -2|

Then, (gof)(x) = g (f(x)) = g(|x|) = |5|x|-2|

(fog)(x) = f(g(x)) = f (|5x -2 |) = ||5x -2 || = |5x -2 |

(ii)

Then, (gof)(x) = g(f(x)) = g(8x³) = = 2x

(fog)(x) = f(g(x)) =

It is given that

Therefore, fof(x) = x, for all x ≠

= > fof = 1

Therefore, the given function f is invertible and the inverse of f itself.

It is given that f : {1, 2, 3, 4} → {10} with

f = {(1, 10), (2, 10), (3, 10), (4, 10)}

From the given definition of f,

We can see that f is a many one function as:

f(1) = f(2) = f(3) = f(4) = 10

therefore, f is not one-one.

Therefore, function f does not have an inverse.

It is given that g : {5, 6, 7, 8} → {1, 2, 3, 4} with

g = {(5, 4), (6, 3), (7, 4), (8, 2)}

From the given definition,

We can see that f is a many one function as:

g(5) = g(7) = 4

[As at two points function have same values, the function is not one-one]

i.e. g is not one- one.

Therefore, function g does not have an inverse.

It is given that h: {2, 3, 4, 5} → {7, 9, 11, 13} with

h = {(2, 7), (3, 9), (4, 11), (5, 13)}

We can see that all distinct elements of the set (2, 3, 4, 5) have distinct images under h.

⇒ h is one–one.

Also, h is onto since for every element of the set {7, 9, 11, 13},

there exists an element x in the set {2, 3, 4, 5} such that h(x) = y.

Therefore, h is a one–one and onto function.

Therefore, h has an inverse.

It is given that f : [–1, 1] → R, given by

Now, Let f(x) = f(y)

⇒ xy + 2x = xy +2y

⇒ 2x = 2y

⇒ x = y

⇒ f is a one- one function.

Now, Let y = , xy = x + 2y so x =

So, for every y in the range there exists x in the domain such that f(x) = y

⇒ f is onto function.

⇒ f: [-1,1] → Range f is one-one and onto

⇒ the inverse of the function : f: [-1, 1] → Range f exists.

Let g: Range f → [-1, 1]be the inverse of range f.

Let y be an arbitrary element of range f.

Since, f :[-1, 1] → Range f is onto, we get:

y = f(x) for same x ϵ [-1, 1]

⇒ xy +2y = x

⇒ x(1 - y) = 2y

⇒ x = , y ≠ 1

Now, Let us define g: Range f → [-1, 1]

g(y) = , y ≠ 1

Now, (gof)(x) = g(f(x)) =

(fog)(y) = f(g(y)) =

Thus, gof = I_[-1,1] and fog = I_{Range f}

⇒ f^-1 = g

Therefore, f^-1(y) = , y ≠ 1

It is given that f : R → R given by f (x) = 4x + 3

Let f(x) = f(y)

⇒ 4x +3 = 4y +3

⇒ 4x = 4y

⇒ x = y

⇒ f is one- one function.

Now, for y ϵ R, Let y = 4x +3

⇒ for any y ϵ R, there exists x = ϵ R

such that, f(x) =

⇒ F is onto function.

Since, f is one –one and onto

⇒ f^-1 exists.

Let us define g: R → R by g(x) =

Now, (gof)(x) = g(f(x)) = g(4x + 3) =

(fog)(y) = f(g(xy)) =

Therefore, gof = fog = I_R

Therefore, f is invertible and the inverse of f is given by

f^-1 (y) = g(y) = .

It is given that f : R₊→ [4, ∞) given by f (x) = x² + 4.

Now, Let f(x) = f(y)

⇒ x² + 4 = y² + 4

⇒ x² = y²

⇒ x = y

⇒ f is one-one function.

Now, for y ϵ [4, ∞), let y = x² + 4.

⇒ x² = y -4 ≥ 0

⇒ for any y ϵ R, there exists x = ϵ R such that

= y -4 + 4 = y.

⇒ f is onto function.

Therefore, f is one–one and onto function, so f-1 exists.

Now, let us define g: [4, ∞) → R+ by,

g(y) =

Now, gof(x) = g(f(x)) = g(x² + 4) =

And, fog(y) = f(g(y)) = =

Therefore, gof = gof = I_R.

Therefore, f is invertible and the inverse of f is given by

f-1(y) = g(y) =

It is given that f: R+ → [–5, ∞) given by f(x) = 9x² + 6x – 5.

Let y be any element of [-5, ∞)

Now, let y = 9x² + 6x – 5

⇒ y = (3x+1)² -1-5 = (3x+1)2 – 6

⇒ 3x + 1 =

⇒ f is onto and it’s range is f = [-5, ∞)

Now, Let us define g: [-5, ∞) → R₊ as g(y) =

Now, we have:

(gof)(x) = g(f(x)) = g(9x² + 6x – 5)

= g((3x+1)² – 6)

And, (fog)(y) = f(g(y)) =

Thus, gof = I_R anf fog = I(-5, ∞)

Therefore, f is invertible and the inverse of f is given by

It is given that f: X → Y be an invertible function.

Also, suppose f has two inverse

Then, for all y ϵ Y, we get:

fog₁(y) = I₁ (y) = fog₂(y)

= > f(g₁(y)) = f(g₂(y))

= > g₁(y) = g₂(y)

= > g₁ = g₂

Therefore, f has a unique inverse.

It is given that f: {1, 2, 3} → {a, b, c} given by

f(1) = a, f(2) = b and f(3) = c

So, if we define g:

{a,b,c} → {1, 2, 3} as

g(a) = 1, g(b) = 2, g(c) = 3, then we get:

(fog)(a) = f(g(a)) = f(1) = a

(fog)(b) = f(g(b)) = f(2) = b

(fog)(c) = f(g(c)) = f(3) = c

And

(gof)(1) = g(f(1)) = g(a) = 1

(gof)(2) = g(f(2)) = g(b) = 2

(gof)(3) = g(f(3)) = g(c) = 3

Therefore, gof = I_X and fog = I_Y, where X = {1, 2, 3} and Y = {a, b, c}

Thus, the inverse of f exists and f^-1 = g.

Then, f^-1: {a, b, c} → {1, 2, 3} is given by

f^-1(a) = 1, f^-1(b) = 2, f^-1(c) = 3

Let us now find the inverse of f-1,

So, if we define h: {1, 2, 3} → {a, b, c} as

h(1) = a, h(2) = b, h(3) = c, then we get:

(goh)(1) = g(h(1)) = g(a) = 1

(goh)(2) = g(h(2)) = g(b) = 2

(goh)(3) = g(h(3)) = g(c) = 3

And,

(hog)(a) = h(g(a)) = h(1) = a

(hog)(b) = h(g(b)) = h(2) = b

(hog)(c) = h(g(c)) = h(3) = c

⇒ goh = I_X and hog = I_Y, where X = {1, 2, 3} and Y = {a, b, c}.

⇒ The inverse of g exists and g^-1 = h

⇒ (f^-1)^-1 = h

⇒ h = f

⇒ (f^-1)^-1 = f

It is given that f: X → Y be an invertible function.

Then, there exists a function g: Y → X such that gof = I_x and fog = I_y.

Then, f^-1 = g.

Now, gof = I_x and fog = I_y

⇒ f^-1of = I_x and fof^-1 = I_y

Thus, f^-1: Y→X is invertible and f is the inverse of f^-1.

Therefore, (f^-1)^-1 = f.

It is given that f : R → R be given by ,

Then, fof(x) = f(f(x)) =

⇒ fof(x) = x

It is given that R be a function defined as f (x) = .

Let y be any element of Range f.

Then, there exists x ϵ R - such that y = f(x)

⟹ 3xy + 4y = 4x

⟹ x(4 – 3y) = 4y

⟹ x =

Let us define g: Range f → R - as g(y) =

Now, (gof)(x) = g(f(x)) =

And, (fog)(y) = f(g(y)) =

Therefore, gof = and fog = I_{Range f}

Thus, g is the inverse of f

Therefore, The inverse of f is the map

: Range f → R - , which is given by g(y) =

It is given that On Z⁺, define ∗ by a ∗ b = a – b

It is not a binary operation as the image of (1, 2) under * is

1 * 2 = 1 – 2 = -1 Z⁺.

It is given that on Z⁺, define ∗ by a ∗ b = ab

We can see that for each a, b ϵ Z⁺, there is a unique element ab in Z⁺.

⇒ * carries each pair (a, b) to a unique element a * b = ab in Z⁺.

Therefore, * is a binary operation.

It is given that On R, define ∗ by a ∗ b = ab²

We can see that for each a, b ϵ R, there is a unique element ab² in R.

⇒ * carries each pair (a, b) to a unique element a * b = ab² in R.

Therefore, * is a binary operation.

It is given On Z⁺, define ∗ by a ∗ b = | a – b|

We can see that for each a, b ϵ Z⁺, there is a unique element |a – b| in Z⁺.

⇒ * carries each pair (a, b) to a unique element a * b = |a – b| in Z⁺.

Therefore, * is a binary operation.

It is given that On Z⁺ define ∗ by a ∗ b = a

We can see that for each a, b ϵ Z⁺, there is a unique element a in Z⁺.

⇒ * carries each pair (a, b) to a unique element a * b = a in Z⁺.

Therefore, * is a binary operation.

It is given that On Z, define a ∗ b = a – b

a –b ϵ Z. so the operation * is binary.

We can see that 1 * 2 = 1-2 = -1 and 2 * 1 = 2-1 = 1.

⇒ 1 *2 ≠ 2 * 1, where 1,2 ϵ Z.

⇒ the operation * is not commutative.

Also, we get,

(1 * 2) * 3 = (1 – 2) * 3 = -1 * 3 = -1 -3 = -4

1 * (2 * 3) = 1 * (2 – 3) = 1 * -1 = 1 – (-1) = 2

⇒ the operation * is not associative.

It is given that On Q, define a ∗ b = ab + 1

ab + 1 ϵ Q, so operation * is binary

We know that ab = ba for a,b ϵ Q

⇒ ab + 1 = ba + 1for a,b ϵ Q

⇒ a * b = a * b for a,b ϵ Q

⇒ 1 *2 ≠ 2 * 1, where 1,2 ϵ Z.

⇒ The operation * is commutative.

Also, we get,

(1 * 2) * 3 = (1 × 2) * 3 = 3 * 3 = 3 × 3 + 1 = 10

1 * (2 * 3) = 1 * (2 × 3) = 1 * 7 = 1 × 7 + 1 = 8

⇒ (1 * 2) * 3 ≠ 1 * (2 * 3)

⇒ the operation * is not associative.

It is given that On Q, define a ∗ b =

Now, ϵ Q, so the operation * is binary.

We know that ab = ba for a,b ϵ Q

⇒ = for a, b ϵ Q

⇒ a * b = a * b for a,b ϵ Q

⇒ The operation * is commutative.

Also for all a,b,c ϵ Q, we get,

(a * b) * c = () * c =

a * (b * c) = a * () =

⇒ (a * b) * c = a * (b * c)

⇒ the operation * is associative.

It is given that on Z⁺, define a ∗ b = 2^ab

2^abϵ Z⁺, so operation * is binary

We know that ab = ba for a,b ϵ Z⁺

⇒ 2^ab = 2^ba for a,b ϵ Z⁺

⇒ a * b = a * b for a, b ϵ Z⁺

⇒ The operation * is commutative.

Also, we get,

(1 * 2) * 3 = 2^(1×2) *3 = 4 * 3 = 2^(4×3) = 2¹²

1 * (2 * 3) = 1 * 2^{(2 × 3)} = 1 * 2⁶ = 1 × 64 =2⁶⁴

⇒ (1 * 2) * 3 ≠ 1 * (2 * 3)

⇒ The operation * is not associative.

It is given that On Z⁺, define a ∗ b = a^b

a^bϵ Z⁺, so operation * is binary

We know that ab = ba for a, b ϵ Z⁺

⇒ 1 * 2 =1² and 2 * 1 = 2¹ =2

⇒ 1 * 2 ≠ 2 *1, where 1,2 ϵ Z⁺

⇒ The operation * is not commutative.

Also, we get,

(1 * 2) * 3 = 2³ *3 = 8 * 3 = 24

1 * (2 * 3) = 1 * 3² = 1 * 9 = 9

⇒ (1 * 2) * 3 ≠ 1 * (2 * 3), where 1,2,3 ϵ Z⁺

⇒ The operation * is not associative.

It is given that On R, * – {– 1}, define a ∗ b =

ϵ R for b ≠ -1, so the operation * is binary.

We can see that 1 * 2 = and 2 * 1=

⇒ 1 * 2 ≠ 2 * 1; where 1,2 ϵ R – {-1}

⇒ the operation * is not commutative.

Now, we can observed that

(1 * 2) * 3 =

1 * (2 * 3) = 1 *

⇒ (1 * 2) * 3 ≠ 1 * (2 * 3), where 1,2,3 ϵ R * – {– 1}

⇒ The operation * is not associative.

The binary operation ^ on the set {1, 2, 3, 4, 5} is defined as a^b = min {a, b}

for a, b ϵ {1, 2, 3, 4, 5}

Therefore, the operation table of the given operation ^ can be given as:

(i) (2 ∗ 3) ∗ 4 = 1 * 4 = 1

2 ∗ (3 ∗ 4) = 2 * 1 = 1

(ii) For every a, b ϵ {1, 2, 3, 4, 5},

We have, a * b = b * a

⇒ the operation * is commutative.

(iii) (2 ∗ 3) = 1

⇒ (2 ∗ 3) ∗ (4 * 5) = 1 * 1 = 1

The binary operation *’ on the set {1, 2, 3, 4, 5} is defined as

a *’ b = HCF of a and b.

The operation table for the operation *’ is given by:

From the above table we can see that the operation tables for the operation * and *’ are the same.

Therefore, the operation *’ is same as the operation *.

(i) It is given that the binary operation on N given by a ∗ b = L.C.M. of a and b.

Then, 5 * 7 = LCM of 5 and 7 = 35

20 * 16 = LCM of 20 and 16 = 80.

(ii) It is given that the binary operation on N given by a ∗ b = L.C.M. of a and b.

We know that LCM of a and b = LCM of b and a, a,b ϵ N.

⇒ a * b = b * a

Therefore, the operation * is commutative.

(iii) It is given that the binary operation on N given by a ∗ b = L.C.M. of a and b.

For a, b, c ϵ N

(a * b) * c = (LCM of a and b ) * c = LCM of a, b and c

a * (b * c) = a * (LCM of b and c) =LCM of a, b and c

⇒ (a * b)* c = a * (b * c)

Therefore, the operation * is associative.

(iv) It is given that the binary operation on N given by a ∗ b = L.C.M. of a and b.

We know that LCM of a and 1 = a = LCM of 1 and 1, a ϵ N

⇒ a * 1 = a = 1 * a, a ϵ N

Therefore, 1 is the identity of * in N.

(v) It is given that the binary operation on N given by a ∗ b = L.C.M. of a and b.

An element a in N is invertible w.r.t. the operation * if there exists an element b in N,

Such that a * b = e =b * a

Now, if e = 1

⇒ LCM of a and b = 1= LCM of b and a

⇒ This is only possible when a = b = 1

Therefore, 1 is the only invertible element of N w.r.t. the operation *.

The operation * on the set A = {1, 2, 3, 4, 5} is defined as

a * b = LCM of a and b.

2 * 3 = LCM of 2 and 3 = 6.

But 6 does not belong to the given set.

Therefore, the given operation * is not a binary operation.

It is given that the binary operation on N defined by a ∗ b = H.C.F. of a and b.

We know that HCF of a and b = HCF of b and a, a, b ϵ N.

⇒ a * b = b * a

⇒ The operation * is commutative.

For a, b c ϵ N, we get,

(a * b) * c = (HCF of a and b) * c = HCF of a, b and c

a * (b * c) = a * (HCF of b and c) = HCF of a, b and c

⇒ (a * b) * c = a * (b * c)

⇒ The operation * is associative.

Now, an element e ϵ N will be the identity for the operation.

Now, if a * e = a = e * a, a ϵ N.

But, this is not true for any a ϵ N.

Therefore, the operation * does not have any identity in N.

It is given that ∗ be a binary operation on the set Q of rational numbers as

a ∗ b = a – b

We can observed that for 2,3,4 ϵ Q

2 * 3 = 2-3 = -1 and 3 * 2 = 3-2 =1

2 * 3 ≠ 3 * 2

⇒ the operation * is not commutative.

Also, (2 * 3) * 4 = (-1) * 4 = -1 -4 = -5 and

2 * (3 * 4) = 2 * (-1) = 2 – (-1) = 3.

⇒ (2 * 3) * 4 ≠ 2 * (3 * 4)

Therefore, the operation * is not associative.

It is given that ∗ be a binary operation on the set Q of rational numbers is defined as

a ∗ b = a² + b²

For a, b ϵ Q, we get,

a * b = a² + b² = b² + a² = b * a

⇒ the operation is commutative.

We can see that (1 * 2) * 3 ≠ 1 *(2 * 3), where 1,2,3 ϵ Q

Therefore, the operation * is not associative.

It is given that ∗ be a binary operation on the set Q of rational numbers is defined as

a ∗ b = a + ab

We can see that 1 * 2 = 1 + 1 × 2 = 1 + 2 = 3

and 2 * 1 = 2 + 2 × 1 = 2+2 = 4

⇒ 1 *2 ≠ 2 * 1: where 1,2 ϵ Q.

⇒ the operation * is not commutative.

Also, We can see that (1 * 2) * 3 ≠ 1 *(2 * 3), where 1,2,3 ϵ Q

Therefore, the operation * is not associative.

It is given that ∗ be a binary operation on the set Q of rational numbers is defined as

a ∗ b = (a – b)²

For a, b ϵ Q, we have,

a ∗ b = (a – b)²

b ∗ a = (b – a)² = [-(a – b)]² = (a –b)²

⇒ a * b = b * a

⇒ the operation * is commutative.

Also, We can see that (1 * 2) * 3 ≠ 1 *(2 * 3), where 1,2,3 ϵ Q

Therefore, the operation * is not associative.

It is given that ∗ be a binary operation on the set Q of rational numbers is defined as

a ∗ b =

For a, b ϵ Q, we get,

a * b = = b * a

⇒ the operation is commutative.

For a, b, c ϵ Q, we get,

(a * b) * c =

a * (b * c) = a *

⇒ (a * b) * c = a * (b * c)

Therefore, the operation * is associative.

It is given that ∗ be a binary operation on the set Q of rational numbers is defined as

a ∗ b = ab²

We can see that for 2, 3 ϵ Q

2 * 3 = 2. 3² = 18 and 3 * 2 = 3. 2² = 12

⇒ 2 * 3 ≠ 3 * 2

Therefore, the operation * is not commutative.

Also, We can see that (1 * 2) * 3 ≠ 1 *(2 * 3), where 1,2,3 ϵ Q

Therefore, the operation * is not associative.

(i) An element e ϵ Q will be the identity element for the operation * if

a * e = a = e * a, a ϵ Q.

a * b = a – b

This operation is not commutative,

Therefore, it does not have identity element.

(ii) An element e ϵ Q will be the identity element for the operation * if

a * e = a = e * a, a ϵ Q.

a * b = a² + b²

If a * e = a, then a² + e² = a.

For a = -2, (-2)⁴ + e² ≠ -2.

Therefore, there is no identity element.

(iii) An element e ϵ Q will be the identity element for the operation * if

a * e = a = e * a, a ϵ Q.

Now, a * b = a + ab

This is not commutative.

Therefore, there is no identity element.

(iv) An element e ϵ Q will be the identity element for the operation * if

a * e = a = e * a, a ϵ Q.

a * b = (a – b)²

If a * e = a, then (a – e)² = a.

A square is always positive, thus for a = -2, (-2 –e)² ≠ -2.

Therefore, there is no identity element.

(v) An element e ϵ Q will be the identity element for the operation * if

a * e = a = e * a, a ϵ Q.

a * b =

If a * e = a, then

Therefore, e =4 is the identity element.

a * 4 =4 * a = .

(vi) An element e ϵ Q will be the identity element for the operation * if

a * e = a = e * a, a ϵ Q.

Now, a * b = ab2

This operation is not commutative,

Therefore, there is not have identity element.

It is given that A = N × N and ∗ be the binary operation on A defined by

(a, b) ∗ (c, d) = (a + c, b + d)

Let (a, b), (c, d) ϵ A

Then, a, b, c, d ϵ N

Now, we have,

(a, b) * (c, d) = (a + c, b + d)

(c, d) * (a, d) = (c + a, d + b) = (a + c, b + d)

⇒ (a, b) * (c, d) = (c, d) * (a, b)

⇒ The operation * is commutative.

Now, (a, b), (c, d), (e, f) ϵ A

Then, a, b, c, d, e, f ϵ N

Now, we have,

((a, b) * (c, d)) * (e, f) = (a + c, b + d) * (e, f) = (a + c + e, b + d + f)

(a, b)* ((c, d) * (e, f)) = (a, b) * (c + e, d + f) = ( a + c + e, b + d + f)

Then, ((a, b) * (c, d)) * (e, f) = (a, b)* ((c, d) * (e, f))

Therefore, the operation * is associative.

Now, an element e = (e₁, e₂) ϵ A will be an identity for the operation *

if a * e = a = e * a a = (a₁, a₂) ϵ A,

(a₁ + e₁, a₂ + e₂) = (a₁, a₂) = (e₁ + a₁, e₂ + a₂)

which is not true for any element in A.

Therefore, the operation * does not have any identity element.

For an arbitrary binary operation ∗ on a set N, a ∗ a = a ∀a ∈ N.

The above statement is false.

Explanation: It is given that an operation ∗ on a set N, a ∗ a = a ∀a ∈ N

Then, in particular, for b = a = 3, we get,

3 * 3 = 3 + 3 = 6 ≠ 3

If ∗ is a commutative binary operation on N, then a ∗ (b ∗ c) = (c ∗ b) ∗ a

The above statement if true.

Explanation: RHS = (c * b) * a

= (b * c) * a (* is commutative)

= a * (b * c) (as * is commutative)

= LHS.

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

Hence Proved

On N, the operation * is defined as a * b = a3 + b3

For, a, b ϵ N, we get,

a * b = a³ + b³ = b³ + a³ = b * a [Addition is commutative in N]

⇒ the operation * is commutative.

We can observed that (1*2)*3 = (1³+2³)*3 = 9 * 3 = 93 + 33 = 729 + 27 = 756

Also, 1*(2*3) = 1*(2³ +3³) = 1*(8 +27) = 1 × 35 = 1³ +35³ = 1 + (35)³ = 1 + 42875 = 42876.

Therefore, (1 * 2)*3 ≠ 1*(2*3); where 1,2,3 ϵ N

Therefore, the operation * is not associative.

Therefore, the operation * is commutative, but not associative.

It is given that f: R → R be defined as f (x) = 10x + 7

Let f(x) = f(y), where x, y ϵ R.

⇒ 10x + 7 = 10y + 7

⇒ x = y

⇒ f is a one – one function.

For y ϵ R, let y = 10x + 7.

⇒ x =

Therefore, for any y ϵ R, there exists x = such that

⇒ f is onto.

⇒ f is an invertible function.

Let us define g : R → R as

Now, we get:

gof(x) = g(f(x)) = g(10x + 7)

And,

⇒ gof = IR and gof = IR

Therefore, the required function g : R → R is defined as .

It is given that f : W → W be defined as

f(n) =

Let f(n) = f(m)

We can see that if n is odd and m is even, then we will have n -1 = m +1.

⇒ n – m = 2

⇒ this is impossible

Similarly, the possibility of n being even and m being odd can also be ignored under a similar argument.

Therefore, both n and m must be either odd or even.

Now, if both n and m are odd, then we get:

f(n) = f(m)

⇒ n -1 = m -1

⇒ n = m

Again, if both n and m are even, the we get:

f(n) = f(m)

⇒ n +1 = m + 1

⇒ n = m

⇒ f is one – one.

Now, it is clear that any odd number 2r + 1 in co-domain N is the image of 2r in domain N and any even number 2r in co – domain N is the image of 2r +1 in domain N.

⇒ f is onto.

⇒ f is an invertible function.

Now, let us define g : W → W be defined as

Now, when n is odd:

gof(n) = g(f(n)) = g (n-1) = n -1 +1 = n (when n is odd, then n-1 is even)

And when n is even:

gof(n) = g(f(n)) = g (n+1) = n +1 -1 = n (when n is even, then n+1 is odd)

Similarly, when m is odd:

fog(m) = f(g(m)) = f (m-1) = m -1 +1 = m

And when n is even:

fog(m) = f(g(m)) = f (m+1) = m +1 -1 = m

Therefore, gof = I_W and fog = I_W

Therefore, f is invertible and the inverse of f is given by f^-1 = g, which is the same as f.

Thus, the inverse of f is f itself.

It is given that f : R → R is defined by f(x) = x² – 3x + 2.

f(f(x)) = f (x² – 3x + 2)

= (x² – 3x + 2)² - 3(x² – 3x + 2) + 2

= x⁴ + 9x² + 4 – 6x³ -12x + 4x² – 3x² + 9x – 6 + 2

= x⁴ - 6x³ + 10x² - 3x

It is given that f: R → {x ∈ R : – 1 < x < 1} defined by , x ∈ R

Now, suppose that f(x) = f(y), where x,y ϵ R.

We can see that if x is positive and y is negative, then we get:

Since, x is positive, and y is negative.

Then, 2xy ≠ x –y.

Thus, the case of x being positive and y being negative can be ruled out.

Similarly, x being negative and y being positive can also be ruled out.

Therefore, x and y have to be either positive or negative.

When x and y are both positive, we get:

f(x) = f(y)

And when x and y are both negative, we get:

f(x) = f(y)

⇒ f is one- one.

Now, let y ϵ R such that -1 < y < 1.

It y is negative, then there exists such that

⇒ f is onto.

Therefore, f is one – one and onto.

Let f : R → R given by f (x) = x³.

Suppose f(x) = f(y), where x, y ϵ R.

⇒ x³ = y³ …(1)

Now, we need to show that x = y.

Suppose x ≠ y, their cubes will also not be equal.

⇒ x³ ≠ y³

However, this will be contraction to (1).

Thus, x = y

Therefore, f is injective.

Define f: N → Z as f(x) = x and g: Z → Z as g(x) = |x|

Now, we can see that

g(-1) = |-1| = 1

g(1) = |1| = 1

⇒ g(-1) = g(1) , but -1 ≠ 1

⇒ g is not injective.

Now, gof: N → Z is defined as gof(x) = g(f(x)) = g(x) = |x|

Let x, y ϵ N such that gof(x) = gof(y).

⇒ |x| = |y|

⇒ x = y

Therefore, gof is injective.

It is given that f: N → N by, f(x) = x + 1

And, g : N → N by,

Now, consider element 1 in co-domain N. So, it is clear that this element is not an image of any of the elements in domain N.

⇒ f is onto.

Now, gof: N → N is defined as:

gof(x) = g(f(x)) = g(x+1) = (x + 1) – 1 [x ϵ N = > (x+1) > 1]

Then we can see that for y ϵ N, there exists x = y ϵ N such that gof(x) = y.

Therefore, gof is onto.

We know that every set is a subset of itself, ARA for all A ϵ P(X).

⇒ R is reflexive.

This cannot be implied to B ⊂ A.

So, if A = {1, 2} and B = {1, 2, 3}, then it cannot be implied that B is related to A.

⇒ R is not symmetric.

So, if ARB and BRC, then A ⊂ B and B ⊂ C.

⇒ A ⊂ C

⇒ R is transitive.

Therefore, R is not an equivalence relation since it is not symmetric.

It is given that ∗: P(X) × P(X) → P(X) given by

A ∗ B = A ∩ B ∀ A, B ϵ P(X).

As we know that,

⇒ A * X = A = x * A A ϵ P(X).

Thus, X is the identity element for the given binary operation *.

Now, an element A ϵ P(X) is invertible if there exists B ϵ P(X) such that

A * B = X = B * A (As X is the identity element)

A ∩ B = X = B ∩ A

This can be possible only when A = X = B.

Therefore, X is the only invertible element in P(X) w.r.t. given operation *.

Hence Proved.

Onto function from the set {1, 2, 3, ..., n} to itself is simply a permutation on n symbols 1, 2, 3, …, n.

Therefore, the total number of onto maps from {1, 2, 3, …, n} to itself is the same as the total number of permutations on n symbols 1, 2, 3, …, n, which is n!

(i) It is given that S = {a, b, c} and T = {1, 2, 3}

F : S → T is defined as:

F = {(a, 3), (b, 2), (c, 1)}

= > F(a) = 3, F(b) = 2, F(c) = 1

Therefore, F^-1: T → S is given by:

F^-1 = {(3, a), (2, b), (1, c)}.

It is given that S = {a, b, c} and T = {1, 2, 3}

F: S → T is defined as:

F = {(a, 2), (b, 1), (c, 1)}

= > F(b) = 1, F(c) = 1, F is not one-one.

Therefore, F is not invertible

⇒ F^-1 does not exists.

It is given that ∗: R × R → R and o: R × R → R defined as

a ∗b = |a – b| and a o b = a, ∀ a, b ∈ R.

For a, b ϵ R, we get:

a*b = |a-b|

b*a = |b-a| = |-(a-b)| = |a-b|

Therefore, a*b = b*a

⇒ the operation * is commutative.

We can see that

(1*2)*3 = (|1-2|)*3 = |1-3| = 2

1*(2*3) = 1*(|2-3|) = 1*1 = 1

Therefore, the operation * is not associative.

Now, consider the operation o:

We can observed that 1o2 = 1 and 2o1 = 2.

⇒ 1o2 ≠ 2o1(where 1, 2 ϵ R)

⇒ the operation o is not commutative.

Let a, b, c ϵ R. Then, we get:

(a o b) o c = a o c = a

a o (b o c) = a o b = a

⇒ (a o b) o c = a o (b o c)

⇒ the operation o is associative.

Now, let a, b, c ϵ R, then we have:

a * (b o c) = a * b = |a –b|

(a * b) o (a * c) = (|a –b|) o (|a –c| = |a –b|

Thus, a * (b o c) = (a * b) o (a * c)

Now,

1 o (2 * 3) = 1 o (|2 –3|) = 1 o 1 = 1

(1 o 2)* (1o3) = 1 * 1 = |1 –1| = 0

Therefore, 1 o (2 * 3) ≠ (1 o 2)* (1o3)( where 1, 2, 3 ϵ R)

Therefore, the operation o does not distribute over *.

It is given that ∗: P(X) × P(X) → P(X) be defined as

A * B = (A – B) ∪ (B – A), A, B ∈ P(X).

Now, let A ϵ P(X). Then, we get,

A * ф = (A – ф) ∪ (ф – A) = A ∪ ф = A

ф * A = (ф - A) ∪ (A - ф) = ф ∪ A = A

⇒ A * ф = A = ф * A, A ϵ P(X)

Therefore, ф is the identity element for the given operation *.

Now, an element A ϵ P(X) will be invertible if there exists B ϵ P(X) such that

A * B = ф = B * A. (as ф is an identity element.)

Now, we can see that A * A = (A –A) ∪ (A – A) = ф ∪ ф = ф A ϵ P(X).

Therefore, all the element A of P(X) are invertible with A^-1 = A.

Let X = {0, 1, 2, 3, 4, 5}

The operation * on X is defined as

An element e ϵ X is the identity element for the operation *,

If a * e = a = e * a a ϵ X.

For a ϵ X, we can see that:

a * 0 = a + 0 = a [a ϵ X = > a + 0 < 6]

0 * A = 0 + a = a [a ϵ X = > a + 0 < 6]

⇒ a * 0 = a = 0 * a a ϵ X.

Therefore, o is the identity element for the given operation *.

An element a ϵ X is invertible if there exists b ϵ X such that

a * b = 0 = b * a.

a = -b or b = 6 – a

But, X = {0, 1, 2, 3, 4, 5} and a, b ϵ X. Then, a ≠ -b.

Therefore, b = 6 – a is the inverse of a ϵ X.

Thus, the inverse of an element a ϵ X, a ≠ 0 is 6 – a, a^-1 = 6 – a.

It is given that A = {–1, 0, 1, 2}, B = {– 4, – 2, 0, 2}

And also, it is given that f, g: A → B be functions defined by f (x) = x² – x, x ∈ A and , x ∈ A

We can see that

f(-1) = (-1)² – (-1) = 1 + 1 = 2

⇒ f(-1) = g(-1)

f(0) = (0)² – 0 = 0 + 0 = 0

⇒ f(0) = g(0)

f(1) = (1)² – 1 = 1- 1 = 0

⇒ f(1) = g(1)

f(2) = (2)² – 2 = 4- 2 = 2

f(2) = g(2)

Thus, f(a) = g(a) . a ϵ A

Therefore, the functions f and g are equal.

This is because relation R is reflexive as (1, 1), (2, 2), (3, 3) ϵ R.

Relation R is symmetric as (1, 2), (2, 1) ϵ R and (1, 3), (3, 1) ϵ R.

But relation R is not transitive as (3, 1), (1, 2) ϵ R but (3, 2) R.

Now, if we add any one of the two pairs (3, 2) and (2, 3) (or both) to relation R,

Then, relation R will become transitive.

Therefore, the total number of desired relations is one.

It is given that A = {1, 2, 3}.

An equivalence relation is reflexive, symmetric and transitive.

The smallest equivalence relations containing (1, 2) is equal to

R₁ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}

Now, only four pairs are left (2, 3), (3, 2), (1, 3) and (3, 1).

So, if we add one pair to R, then for symmetry we must add (3, 2).

Also, for transitivity we required to add (1, 3) and (3, 1).

Thus, the only equivalence relation is the universal relation.

Therefore, the total number of equivalence relations containing (1, 2) is 2.

It is given that

f: R → R be the Signum Function defined as

Also, g: R → R is defined as g(x) = [x], where [x] is the greatest integer less than or equal to x.

Now, let x ϵ (0, 1]

Then, we get,

[x] = 1 if x = 1 and [x] = 0 if 0 < x < 1

Therefore, fog(x) = f(g(x)) = f([x]) =

gof(x) = g(f(x))

= g(1) [x > 0]

= [1] = 1

Then, when x ϵ (0,1), we get fog(x) = 0 and gof(x) = 1

But fog(1) ≠ gof(1)

Therefore, fog and gof do not coincide in (0, 1].

A binary option * on {a, b} is a function from

{a, b} × {a, b} → {a, b}

i.e.y

Therefore, the total number of binary operations on the set {a, b} is 2⁴ that is 16.