Relations and Functions Miscellaneous NCERT Solutions

Ex. 1: Let f:\mathbb{N}\to Y be a function defined as

f(x)=4x+3

where

Y=\{y \in \mathbb{N} : y=4x+3 \text{ for some } x \in \mathbb{N}\}

Show that f is invertible and find f^{-1} .

Step 1: Show that f is one-one

Assume

f(x_1)=f(x_2)

Then 4x_1+3=4x_2+3

\Rightarrow 4x_1=4x_2

\Rightarrow x_1=x_2

Hence, f is one-one.

Step 2: Show that f is onto

Let y \in Y . By definition of Y, there exists x \in \mathbb{N} such that

y=4x+3

So every element of Y has a pre-image in N.

Hence, f is onto.

Therefore, f is bijective and hence invertible.

Step 3: Find the inverse

Let

y=4x+3

Solving for x:

x=\frac{y-3}{4}

Thus, the inverse function is

f^{-1}(y)=\frac{y-3}{4}, \quad y \in Y

Final Answer: The function is invertible and its inverse is f^{-1}(y)=\frac{y-3}{4} .

Solution (Using composition method)

Let us define a function g:Y \to \mathbb{N} as

g(y)=\frac{y-3}{4}

Now we verify the compositions.

Step 1: Compute g ∘ f

(g \circ f)(x)=g(f(x))

= g(4x+3)

= \frac{(4x+3)-3}{4}

i.e.

= \frac{4x}{4}=x

Step 2: Compute f ∘ g

(f \circ g)(y)=f(g(y))

= f\left(\frac{y-3}{4}\right)

= 4\left(\frac{y-3}{4}\right)+3

i.e.

= y-3+3=y

Conclusion

Since

g \circ f = I_{\mathbb{N}} and f \circ g = I_{Y} ,

the function f is bijective and hence invertible.

Therefore,

f^{-1}(y)=\frac{y-3}{4}, \quad y \in Y

Ex. 2: If R_1 and R_2 are equivalence relations on a set A , show that R_1 \cap R_2 is also an equivalence relation on A .

Solution

To prove that R_1 \cap R_2 is an equivalence relation, we must show that it is:

• Reflexive
• Symmetric
• Transitive

1. Reflexive

Since R_1 is an equivalence relation, for every a \in A ,

(a,a) \in R_1

Also, since R_2 is an equivalence relation,

(a,a) \in R_2

Therefore,

(a,a) \in R_1 \cap R_2

Hence, R_1 \cap R_2 is reflexive.

2. Symmetric

Let (a,b) \in R_1 \cap R_2 .

Then (a,b) \in R_1 and (a,b) \in R_2 .

Since both are symmetric relations,

(b,a) \in R_1 and (b,a) \in R_2

Therefore,

(b,a) \in R_1 \cap R_2

Hence, it is symmetric.

3. Transitive

Let (a,b) \in R_1 \cap R_2 and (b,c) \in R_1 \cap R_2 .

Then:

(a,b), (b,c) \in R_1 and (a,b), (b,c) \in R_2

Since both are transitive relations,

(a,c) \in R_1 and (a,c) \in R_2

Therefore,

(a,c) \in R_1 \cap R_2

Hence, it is transitive.

Conclusion

Since R_1 \cap R_2 is reflexive, symmetric, and transitive, it is an equivalence relation on A .

Ex. 3: Let R be a relation on the set A of ordered pairs of positive integers defined by

(x,y)\,R\,(u,v) \iff xv = yu

Show that R is an equivalence relation.

Solution

To prove that R is an equivalence relation, we verify:

• Reflexive
• Symmetric
• Transitive

1. Reflexive

Let (x,y) \in A . Then

(x,y) R (x,y) \iff x\cdot y = y \cdot x

which is true for all positive integers.

Hence, R is reflexive.

2. Symmetric

Assume (x,y) R (u,v) , so

xv = yu

Rewriting, we get

uy = vx

Thus,

(u,v) R (x,y)

Hence, R is symmetric.

3. Transitive

Let

(x,y) R (u,v) \Rightarrow xv = yu

and

(u,v) R (p,q) \Rightarrow uq = vp

We need to show:

(x,y) R (p,q) \Rightarrow xq = yp

From xv = yu , multiply both sides by q :

xvq = yuq

From uq = vp , multiply both sides by y :

yuq = yvp

Thus,

xvq = yvp

Since v \ne 0 , cancel v :

xq = yp

Hence,

(x,y) R (p,q)

Conclusion

Since the relation is reflexive, symmetric, and transitive, R is an equivalence relation.

Ex. 4: Find the number of all one-one functions from set A=\{1,2,3\} to itself.

Solution

We are given:

A = \{1,2,3\}, \quad n(A)=3

Step 1: Understand the condition

A one-one function means:

Different elements of the domain must have different images.

Since domain and codomain both have 3 elements, this becomes a permutation problem.

Step 2: Use formula for one-one functions

Number of one-one functions from a set with n elements to itself is:

n!

Step 3: Apply the formula

Here n = 3 , so

\text{Number of one-one functions} = 3!

= 3 \times 2 \times 1 = 6

Alternative Explanation (Conceptual View)

Each one-one function is just a rearrangement of elements of the set A.

So we are counting the number of permutations of 3 distinct elements.

Final Answer \boxed{6}

Ex. 5: Let A = \{1, 2, 3\} . Then show that the number of relations containing (1, 2) and (2, 3) which are reflexive and transitive but not symmetric is three.

Solution

We are given the set

A=\{1,2,3\}

Since the relation must be reflexive, the following pairs must always be included:

(1,1),(2,2),(3,3)

Also, the relation must contain:

(1,2),(2,3)

Step 1: Apply transitivity

Since (1,2) and (2,3) belong to the relation, transitivity forces:

(1,3)

Thus the smallest required relation is:

R_1=\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}

This relation is reflexive, transitive, but not symmetric.

Step 2: Form other valid relations

If we add the pair (2,1) to R_1 , we obtain:

R_2=R_1\cup\{(2,1)\}

or R_2=\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3),(2,1)\}

This relation remains reflexive and transitive, but not symmetric.

Similarly, if we add the pair (3,2) to R_1 , we obtain:

R_3=R_1\cup\{(3,2)\}

or R_3=\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3),(3,2)\}

This relation is also reflexive and transitive, but not symmetric.

Step 3: Why other additions are not allowed

We cannot add both pairs (2,1) and (3,2) together, or the single pair (3,1) , because transitivity would then force the remaining missing pairs to be included.

As a result, the relation would become symmetric, which is not allowed.

Final Answer:

Therefore, the total number of such relations is \boxed{3}

Ex. 6: Show that the number of equivalence relations in the set \{1, 2, 3\} containing (1, 2) and (2, 1) is two.

Solution

We are given the set

A=\{1,2,3\}

Since the relation must be an equivalence relation, it must always be reflexive, symmetric, and transitive.

Also, the relation contains:

(1,2),(2,1)

This means 1 and 2 are connected in both directions.

Step 1: Start with the smallest required relation

Since reflexivity must hold, we include:

(1,1),(2,2),(3,3)

Together with the given pairs:

(1,2),(2,1)

So the smallest required relation is:

R_1=\{(1,1),(2,2),(3,3),(1,2),(2,1)\}

Step 2: Remaining possible pairs

The remaining pairs are:

(1,3),(3,1),(2,3),(3,2)

Now we check what happens if we connect 3 with the other elements.

Step 3: If we add any connection involving 3

Suppose we add:

(2,3)

Then symmetry forces:

(3,2)

Since 1 is already related to 2, transitivity now forces:

(1,3),(3,1)

Thus all ordered pairs of A\times A are included, giving the universal relation.

Step 4: Final possibilities

• Case 1: Keep 3 separate → relation is R_1
• Case 2: Connect all elements → universal relation

Hence only two equivalence relations are possible.

Final Answer: \boxed{2}

1. Show that the function f:\mathbb{R}\to \{x\in\mathbb{R}:-1<x<1\} defined by f(x)=\frac{x}{1+|x|} is one-one and onto.

Solution

We are given the function

f:\mathbb{R}\to \{x\in\mathbb{R}:-1<x<1\}

defined by

f(x)=\frac{x}{1+|x|}

We will show that the function is both one-one and onto.

One-One (Injective):

Let

f(x_1)=f(x_2)

Then

\frac{x_1}{1+|x_1|}=\frac{x_2}{1+|x_2|}

Cross-multiplying, we get

x_1(1+|x_2|)=x_2(1+|x_1|)

x_1+x_1|x_2|=x_2+x_2|x_1|

Rearranging,

x_1-x_2=x_2|x_1|-x_1|x_2|

(x_1-x_2)=|x_1||x_2|\left(\frac{x_2}{|x_2|}-\frac{x_1}{|x_1|}\right)

This simplifies only when

x_1=x_2

Hence, the function is one-one.

Onto (Surjective):

Let

y\in(-1,1)

We must find x\in\mathbb{R} such that

y=\frac{x}{1+|x|}

We consider two cases.

Case 1: x\ge0

Then |x|=x , so

y=\frac{x}{1+x}

Solving for x ,

y(1+x)=x

or y=x-xy=x(1-y)

x=\frac{y}{1-y}

This gives a real value of x whenever 0\le y<1 .

Case 2: x<0

Then |x|=-x , so

y=\frac{x}{1-x}

Solving for x ,

y(1-x)=x

or y=x+xy=x(1+y)

x=\frac{y}{1+y}

This gives a real value of x whenever -1<y<0 .

Thus every element of (-1,1) has a pre-image in \mathbb{R} .

Hence, the function is onto.

Since the function is both one-one and onto, it is bijective.

2. Show that the function f:\mathbb{R}\to\mathbb{R} given by f(x)=x^3 is injective.

Solution

We are given the function

f:\mathbb{R}\to\mathbb{R}

defined by

f(x)=x^3

We need to show that the function is injective (one-one).

Proof:

Let

f(x_1)=f(x_2)

Then

x_1^3 = x_2^3

Taking cube root on both sides, we get

x_1 = x_2

Thus, equal outputs imply equal inputs.

Hence, the function f(x)=x^3 is injective (one-one).

Conclusion: f is one-one.

3. Given a non empty set X , consider P(X) which is the set of all subsets of X . Define the relation R in P(X) as follows:

For subsets A,B\in P(X) , ARB \iff A\subseteq B . Is R an equivalence relation on P(X) ? Justify your answer.

We are given a relation R on P(X) defined by

A R B \iff A \subseteq B

We check whether it is an equivalence relation (reflexive, symmetric and transitive).

1. Reflexive:

For reflexivity, we need A R A for every A \in P(X) .

Since every set is a subset of itself,

A \subseteq A

Therefore, the relation is reflexive.

2. Symmetric:

For symmetry, if A \subseteq B , then we must also have B \subseteq A .

This is not true in general.

For example, let

A=\{1\},\quad B=\{1,2\}

Then

A \subseteq B

but

B \not\subseteq A

Therefore, the relation is not symmetric.

3. Transitive:

If

A \subseteq B \text{ and } B \subseteq C

then clearly

A \subseteq C

Therefore, the relation is transitive.

Conclusion:

The relation is reflexive and transitive, but not symmetric.

Hence, R is not an equivalence relation.

4. Find the number of all onto functions from the set \{1,2,3,\dots,n\} to itself.

Solution

We are given functions from a set having n elements to itself.

So, both the domain and codomain contain n elements.

Step 1: Condition for onto

An onto function means every element of the codomain must have at least one pre-image in the domain.

When the domain and codomain have the same number of elements, every onto function is automatically one-one.

Hence, every onto function in this case is bijective.

Step 2: Count bijective functions

The number of bijective functions from an n -element set to itself is equal to the number of permutations of n distinct elements.

Therefore, the number of such functions is

n!

Final Answer: \boxed{n!}

5. Let A=\{-1,0,1,2\} , B=\{-4,-2,0,2\} and f,g:A\to B be functions defined by

f(x)=x^2-x , x\in A and g(x)=2\left|x-\frac12\right|-1 , x\in A . Are f and g equal? Justify your answer.

Solution

Two functions f:A\to B and g:A\to B are equal if

f(a)=g(a)\ \forall a\in A

Step 1: Compute values of f(x)

• f(-1)=(-1)^2-(-1)=1+1=2
• f(0)=0
• f(1)=1-1=0
• f(2)=4-2=2

f=\{(-1,2),(0,0),(1,0),(2,2)\}

Step 2: Compute values of g(x)

• g(-1)=2\left| -1-\frac12 \right|-1=2\cdot\frac32-1=2
• g(0)=2\left| -\frac12 \right|-1=1-1=0
• g(1)=2\left| \frac12 \right|-1=1-1=0
• g(2)=2\left| \frac32 \right|-1=3-1=2

g=\{(-1,2),(0,0),(1,0),(2,2)\}

Step 3: Compare f and g

We see that for every element in A , f(x)=g(x) .

Conclusion:

Since both functions have identical values for all elements of the domain, f=g . Hence, the functions are equal.

6. Let A = \{1, 2, 3\} . Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is

(A) 1    (B) 2    (C) 3    (D) 4

Solution

We are given the set

A=\{1,2,3\}

Since the relation must be reflexive, we must include:

(1,1),(2,2),(3,3)

Also given that the relation contains:

(1,2),(1,3)

Step 1: Apply symmetry

Since the relation is symmetric, we must also include:

(2,1),(3,1)

So the minimal required relation becomes:

R_1=\{(1,1),(2,2),(3,3),(1,2),(2,1),(1,3),(3,1)\}

Step 2: Remaining possible pair

The only remaining pair in A\times A is:

(2,3),(3,2)

These may or may not be included.

Step 3: Check transitivity

We already have:

(2,1) \text{ and } (1,3) \Rightarrow (2,3) is required for transitivity.

So if we do NOT include (2,3) , transitivity fails.

If we include (2,3), symmetry forces (3,2).

Thus both must be added together or not at all.

Step 4: Count valid relations

• Case 1: Do not include (2,3), (3,2) → relation is not transitive
• Case 2: Include both (2,3), (3,2) → relation becomes transitive

We need relations that are not transitive, so only Case 1 is valid.

Hence exactly one relation satisfies all conditions except transitivity.

✅️ Final Answer: (A) \boxed{1}

7. Let A = \{1, 2, 3\} Then number of equivalence relations containing (1, 2) is

(A) 1    (B) 2    (C) 3    (D) 4

Solution

We are given the set

A=\{1,2,3\}

An equivalence relation must be reflexive, symmetric, and transitive.

Also given that the relation contains:

(1,2)

So by symmetry, we must also include:

(2,1)

Step 1: Required pairs

Since reflexivity is mandatory, we must include:

(1,1),(2,2),(3,3)

Thus the smallest possible equivalence relation is:

R_1=\{(1,1),(2,2),(3,3),(1,2),(2,1)\}

Step 2: Remaining pairs

The remaining pairs are:

(1,3),(3,1),(2,3),(3,2)

Step 3: Check possible additions

If we connect 3 with either 1 or 2, then symmetry and transitivity force all remaining pairs to be included.

For example, if we include (2,3) , then symmetry forces:

(3,2)

Since (1,2) and (2,3) are present, transitivity forces:

(1,3)

Similarly, symmetry then gives:

(3,1)

Thus every ordered pair of A\times A gets included, giving the universal relation.

Step 4: Count equivalence relations

• Case 1: R_1=\{(1,1),(2,2),(3,3),(1,2),(2,1)\}
• Case 2: Universal relation A\times A

✅️ Hence, the total number of equivalence relations is: (B) \boxed{2}