# Class 12 NCERT Solutions- Mathematics Part I – Chapter 1 Relations And Functions – Exercise 1.1 | Set 1

• Last Updated : 05 Apr, 2021

### (i) Relation R in the set A={ 1, 2,3, . . ., 13, 14} defined as R={ ( x , y):3x-y=0}

Solution:

A={1 ,2 ,3,…,13, 14}

R={(x,y): 3x-y=0}

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

R is not reflexive since (1,1),(2,2),(3,3),…,(14,14)∉R

Also, R is not reflexive since (1,3)∈R, but (3,1)∉R.[since 3(3)-1≠0]

Also, R is not transitive as (1,3), (3,9)∈R, but (1,9)∉R.[since 3(1)-9≠0]

Hence, R is not reflexive, nor symmetric nor transitive.

### (ii) Relation R in the set N of natural numbers defined as R={ ( x, y) : y=x+5 and x<4}

Solution:

R={(x, y): y=x+5 and x<4}={(1,6), (2, 7), (3,8)}

It is seen that (1, 1)∉R. Therefore, R is not reflexive.

(1, 6)∈R. But, (6, 1)∉R so, R is not symmetric.

Now, since their is no pair in R such that (x, y) and (y, z)∈R, so (x, z) can not belong to R. Therefore, R is not transitive.

So, we can conclude that R is neither reflexive, nor symmetric, nor transitive.

### (iii) Relation R in the set A= {1, 2, 3, 4, 5, 6} as R={(x, y): y is divisible by x}

Solution:

A={1, 2, 3, 4, 5, 6}

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

We know that any number is always divisible by itself.⇒ (x, x)∈ R. Therefore, R is reflexive.

We see, (2, 4)∈R [4 is divisible by 2]. But, (4, 2)∉R [2 is not divisible by 4]. Therefore, R is not symmetric.

Let’s assume (x, y), (y, z) ∈ R. Then, y is divisible by x and z is divisible by y. Therefore, z is divisible by x.⇒ (x, z) ∈ R. Therefore, R is transitive.

### (iv) Relation R in the set Z of all integers defined as R={(x, y): x-y is an integer}

Solution:

R={(x, y): x-y is an integer}

For every x ∈ Z, (x, x) ∈ R [x-x=0 which is an integer]. Therefore, R is reflexive.

For every x, y ∈ Z if (x, y) ∈ R, then x-y is an integer. ⇒-(x-y) is also an integer. ⇒ (y-x) is also an integer. Therefore, R is symmetric.

Let’s assume, (x, y) and (y, z) ∈ R, where x, y and z ∈ Z. ⇒ (x-y) and (y-z) are integers. ⇒ x-z=(x-y)+(y-z) is an integer. ⇒ (x, z) ∈ R. Therefore, R is transitive.

### (v) Relation R in a set A of human beings in a town at a particular time, given by:

(a)R={(x,y) : x and y work at the same place.

#### Solution:

We can see (x,x) ∈ R .Therefore, R is reflexive.

If (x,y) ∈ R, then x and y work at the same place. So, (y,x) ∈ R. Therefore, R is symmetric.

Let, (x,y), (y,z) ∈ R.

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

⇒ x and z work at the same place.

Therefore, R is transitive.

(b) R={(x,y): x and y live in the same locality}.

Solution:

We can see (x,x) ∈ R. Therefore, R is reflexive.

If (x,y) ∈ R, then x and y live in same locality. So, (y,x) ∈ R. Therefore, R is symmetric.

Let (x,y) ∈ R and (y,z) ∈ R. So, x, y and z live in the same locality. So, (x,z) ∈ R. Therefore, R is transitive.

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

Solution:

(x,x)∉R since, human being can not be taller than himself. So, R is not reflexive.

Let (x,y) ∈ R ,then x is exactly 7 cm taller than y. Then, y is not taller than x. Therefore, R is not symmetric.

Let (x,y), (y,z) ∈ R, then x is exactly 7 cm taller than y and y is exactly 7 cm taller than z which means x is 14 cm taller than z. So, (x,z)∉R . Therefore, R is not transitive.

(d) R={(x,y): x is wife of y}.

Solution:

(x,x) ∉ R. Since, x can not be the wife of herself. Therefore, R is not reflexive.

Let (x,y) ∈ R, then x is the wife of y. So, y is not the wife of x ,i.e., (y,x) ∉ R. Therefore, R is not symmetric.

Let (x,y), (y,z) ∈ R, then x is the wife of y and y is the wife of z which is not possible. So, R can not be transitive.

(e) R={(x,y): x is father of y}.

Solution:

(x,x) ∉ R. Since, x can not be the father of himself. Therefore, R is not reflexive.

Let (x,y) ∈ R, then x is the father of y. So, y can not be the father of x. So, (y,x) ∉ R. Therefore, R is not symmetric.

Let (x,y), (y,z) ∈ R, then x is the father of y and y is the father of z which means x is the grandfather of z. So, So, (x, z) ∉ R. Therefore, R is not transitive.

### Question 2. Show that the relation R in the set R of real numbers, defined as R= {(a, b): a ≤ b2} is neither reflexive nor symmetric nor transitive.

Solution:

It can be observed that (½, ½) ∉ R, since ½>(½)2 =¼. Therefore, R is not reflexive.

(1,4) ∈ R as 1<42 .But, (4,1) ∉  R. Therefore, R is not symmetric.

(3,2), (2,1.5) ∈ R. But, 3> (1.5)2 =2.25 . So, (3,1.5) ∉  R. Therefore, R is not transitive.

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

### Question 3. 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}.

Solution:

Let the set {1, 2, 3, 4, 5, 6} be named A.

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

We can see (x, x) ∉  R. Since, x ≠ x+1. Therefore, R is not reflexive.

It is observed that (1,2) ∈ R but, (2,1) ∉  R. Therefore, R is not symmetric.

We can see, (1,2), (2, 3) ∈ R, but (1,3) ∉  R. Therefore, R is not transitive.

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

### Question 4. Show that the relation R in R defined as R={ (a, b): a≤ b}, is reflexive and transitive but not symmetric.

Solution:

Clearly, (a,a) ∈ R as a=a. Therefore, R is reflexive.

(2,4) ∈ R (as 2<4) but (4,2) ∉  R as 4 is greater than 2. Therefore, R is not symmetric.

Let (a,b), (b,c) ∈ R. Then, a≤ b and b≤ c.

⇒a ≤ c.

(a, c) ∈ R. Therefore, R is transitive.

Hence, R is reflexive and transitive but not symmetric.

### Question 5. Check whether the relation R in R  defined as R ={ (a, b): a ≤ b3 } is reflexive, symmetric or transitive.

Solution:

It is observed that (½, ½) ∉  R as ½ > (½)3 =(1/8). Therefore, R is not reflexive.

(1,2) ∈ R(as 1<8) but, (2,1) ∉  R. Therefore, R is not symmetric.

We have, (3, 3/2), (3/2, 6/5) ∈ R but, (3, 6/5) ∉  R. Therefore, R is not transitive.

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

### Question 6. Show that the relation R in the set {1, 2, 3} given by R={(1,2), (2,1)} is symmetric but neither reflexive nor transitive.

Solution:

Let the set {1, 2, 3} be named A.

It is seen that, (1, 1), (2,2), (3,3)∉  R. Therefore, R is not reflexive.

As (1, 2) ∈ R and (2, 1) ∈ R. Therefore, R is symmetric.

However, (1, 1)∉  R. Therefore, R is not transitive.

Hence, R is symmetric but neither reflexive nor transitive.

### Question 7. Show that the relation R in the set A of all the books in a library of a college , given by R={(x,y): x and y have the same number of pages} is a equivalence relation.

Solution:

Set A is the set of all books in the library of a college.

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

R is reflexive since (x,x) ∈ R as x and x have the same number of pages.

Let (x,y) ∈ R

⇒x and  y have the same number of pages

⇒y and x have the same number of pages.

⇒(y,x)∈ R

Therefore , R is symmetric.

Let (x,y) ∈ R and (y,z)∈ 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) ∈ R

Therefore, R is transitive.

Hence, R is an equivalence relation.

### Question 8. Show that the relation R in the set A ={1,2,3,4,5} given by R={(a,b):|a-b| is even} , is an equivalence relation . Show that all the elements of { 1,3,5} are related to each other and all the elements of {2, 4} are related to each other . But no elements of {1,3,5} is related to any element of {2,4}.

Solution:

A={ 1,2,3,4,5}

R={(a,b):|a-b| is even }

It is clear that for any element a∈ A, we have |a-a|=0 (which is even).

Therefore, R is reflexive.

Let (a,b) ∈ R.

⇒|a-b| is even.

⇒|-(a-b)|=|b-a| is also even.

⇒(b,a)∈ R

Therefore, R is symmetric.

Now , let (a,b)∈ R and (b,c)∈ 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.   [Sum of two even integers is even]

⇒|a-c| is even.

Therefore, R is transitive.

Hence, R is an equivalence relation.

All elements of the set {1,2,3} are related to each other as all the elements of this subset are odd. Thus, the modulus of the difference between any two elements will be even.

Similarly, all elements of the set {2,4} are related to 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 . Thus, the modulus of the difference between the two elements (from each of these two subsets) will not be even.

### Question 9: Show that each of the relation R in the set A={x∈ Z:0<=x<=12} , given by

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

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

is an equivalence relation. Find the set of all elements related to 1 in each case.

#### Solution:

A={x∈ Z:0<=x<=12}={0,1,2,3,4,5,6,7,8,9,10,11,12}

#### (i) R={(a,b):|a-b| is a multiple of 4}

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

Therefore, R is reflexive.

Now , let (a,b)∈R ⇒|a-b| is a multiple of 4.

⇒|-(a-b)|=|b-a| is a multiple of 4.

⇒(b,a)∈R

Therefore, R is symmetric.

Let (a,b) ,(b,c) ∈ R.

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

⇒(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| is a multiple of 4.

⇒(a,c)∈R

Therefore, R is transitive.

Hence, R is an equivalence relation .

The set of elements related to 1 is {1,5,9} since |1-1|=0 is a multiple of 4,

|5-1|=4 is a multiple of 4,and

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

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

For any element a∈A, we have (a,a) ∈ R , since a=a.

Therefore, R is reflexive.

Now , let (a,b)∈R.

⇒a=b

⇒b=a

⇒(b,a)∈R

Therefore, R is symmetric.

Now, let (a,b)∈R and (b,c)∈R.

⇒a=b and b=c

⇒a=c

⇒(a,c)∈R

Therefore, R is transitive.

Hence, R is an equivalence relation.

The elements in R that are related to 1 will be those elements from set A which are equal to 1.

Hence, the set of elements related to 1 is {1].

### Question 10: Give an example of a relation. Which is

(i) Symmetric but neither reflexive nor transitive.

(ii) Transitive but neither reflexive nor symmetric.

(iii)Reflexive and symmetric but not transitive.

(iv) Reflexive and transitive but not symmetric.

(v) Symmetric and transitive but not reflexive.

Solution:

(i) Let A ={5,6,7}.

Define a relation R on A as R ={(5,6),(6,5)}.

Relation R is not reflexive as (5,5) , (6,6),(7,7)∉R.

Now, as (5,6)∈R and also (6,5)∈R, R is symmetric.

⇒(5,6),(6,5)∈R , but (5,5)∉R

Therefore , R is not transitive.

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

(ii) Consider a relation R in R defined as:

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

For any a∈R, we have (a,a)∉R since a cannot be strictly less than itself . In fact a=a.

Therefore, R is not reflexive. Now,

(1,2)∈R (as 1<2)

But, 2 is not less than 1.

Therefore, (2,1)∉R

Therefore, R is not symmetric.

Now, let (a,b),(b,c)∈R.

⇒a<b and b<c

⇒a<c

⇒ (a,c)∈R

Therefore, R is transitive.

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

(iii) Let A={4,6,8}.

Define a relation R on A as:

A={(4,4),(6,6),(8,8),(4,6),(6,4),(6,8),(8,6)}

Relation R is reflexive since for every a∈A , (a,a)∈R i.e.,(4,4),(6,6),(8,8)∈R.

Relation R is symmetric since (a,b)∈R⇒(b,a)∈R for all a,b∈R.

Relation R is not transitive since (4,6),(6,8)∈R , but (4,8)∉R.

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

(iv) Define a relation R in R as:

R={ (a,b): a3 ≥ b3 }

Clearly (a,a)∈R as a^3=a^3

Therefore, R is reflexive.

Now, (2,1)∈R(as 23>=13)

But, (1,2)∉ R (as 13< 23)

Therefore, R is not symmetric

Let (a,b), (b,c) ∈ R.

⇒a3 >= b3 and b3 >=c3

⇒a3>=c3

⇒(a, c)∈R

Therefore, R is transitive.

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

(v) Let A={-5,-6},

Define a relation R on A as:

R={(-5,-6), (-6,-5), (-5,-5)}

Relation R  is not reflexive as (-6,-6)∉ R.

Relation R is symmetric as (-5,-6)∈ R and (-6,-5)∈ R.

It is seen that (-5,-6),(-6,-5)∈R . Also , (-5,-5)∈R.

Therefore, the relation R is transitive .

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

### Chapter 1 Relations And Functions – Exercise 1.1 | Set 2

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