Relations & Functions(Ex. 1.1)

 Exercise 1.1

1. Determine whether each of the following relations are reflexive, symmetric, and transitive:

(i) Relation R in the set A={1,2,3,......,13,14} defined as

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

Solun:- Given A={1,2,3,.........13,14}

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

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation.

So, R is not a reflexive relation because (1,1), (2,2) ………(14,14) ∉ R.

R is not a symmetric relation because of (1,3) ∈ R but (3,1) ∉ R.

R is not a transitive relation because of (1,3) ∈ R and (3,9) ∈ R but (1,9) ∉ R.

So, R is neither 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}

Solun:- Given A={1,2,3,................................}=N

R={(1,6), (2,7), (3,8)}

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation.

So, R is not a reflexive relation because of (1,1), (2,2) ∉ R.

R is not a symmetric relation because of (1,6) ∈ R but (6,1) ∉ R.

R is not a transitive relation.

So, 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}

Solun:- Given A={1,2,3,4,5,6}

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation.

So, R is a reflexive relation because (x,x) ∈ R because x is always divisible by x.

R is not a symmetric relation because of (3,6) ∈ R but (6,3) ∉ R because 6 is divisible by 3 but 3 is not divisible by 6.

R is a transitive relation because of (1,3) ∈ R and (3,6) ∈ R because 3 is divisible by 1 and also 6 is also divisible by 3 then (1,6) ∈ R because 6 is divisible by 1.

So, R is reflexive, transitive relation but not a symmetric relation.

(iv) Relation R in the set of Z of all integers defined as

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

Solun:- Given set is Z and set of integers

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation.

So, R is a reflexive relation because (x,x) ∈ R, x-x = 0 that is integer.

R is a symmetric relation because of (x,y) ∈ R then (y,x) ∈ R because y-x is also an integer.

R is a transitive relation because

(x,y) ∈ R  because given x-y is an integer and also (y,z) ∈ R i.e. y-z is also an integer

Then x-z=(x-y)+(y-z) that is integer so (x,z) ∈ R

So, R is reflexive, symmetric, and transitive.

(v) Relation R in the 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}

Solun:- Given set A of human beings in a town at a particular time

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation.

So, R is a reflexive relation because (x,x) ∈ R one person works at the same place at a particular time.

R is a symmetric relation because of (x,y) ∈ R has given x and y work at the same place then y and x also work at the same place (y,x) ∈ R.

R is a transitive relation because of (x,y) ∈ R has given x and y work at the same place and also (y,z) ∈ R y and z work at the same place then x and z also work at the same place (x,z) ∈ R.

So, R is reflexive, symmetric, and transitive.

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

Solun:- Given set A of human beings in a town at a particular time

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation.

So, R is a reflexive relation because (x,x) ∈ R one person lives in the same locality at a particular time.

R is a symmetric relation because of (x,y) ∈ R has given x and y live in the same locality then y and x also live in the same locality (y,x) ∈ R.

R is a transitive relation because of (x,y) ∈ R has given x and y live in the same locality and also (y,z) ∈ R y and z live in the same locality then x and z also live in the same locality (x,z) ∈ R.

So, R is reflexive, symmetric, and transitive.

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

Solun:- Given set A of human beings in a town at a particular time

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation.

So, R is a reflexive relation because (x,x) ∈ R x is cannot be taller than himself at a particular time.

R is not a symmetric relation because (x,y) ∈ R has given x is exactly 7 cm taller than y but it is not necessary that y is also exactly 7 cm taller than x (y,x) ∉ R.

R is not a transitive relation because of (x,y) ∈ R has given x is exactly 7 cm taller than y and also (y,z) ∈ R y is exactly 7 cm taller than z. Then x is 14 cm taller than z (x,z) ∉ R.

So, R is neither reflexive relation, nor symmetric, nor a transitive relation.

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

Solun:- Given set A of human beings in a town at a particular time

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation.

So, R is not a reflexive relation because (x,x) ∉ R, it is not possible that x is the wife of x at a particular time.

R is not a symmetric relation because (x,y) ∈ R has given x is the wife of y but it is not possible that y is the wife of x at a particular time (y,x) ∉ R.

R is not a transitive relation because (x,y) ∈ R has given x is the wife of y and also (y,z) ∈ R y is the wife of z but it is not necessary that x is also the wife of z (x,z) ∉ R.

So, R is neither reflexive relation nor symmetric, and nor transitive relation.

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

Solun:- Given set A of human beings in a town at a particular time

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation.

So, R is not a reflexive relation because (x,x) ∉ R, it is not possible that x is the father of x at a particular time.

R is not a symmetric relation because (x,y) ∈ R has given x is the father of y but it is not possible that y is also the father of x at a particular time (y,x) ∉ R.

R is not a transitive relation because (x,y) ∈ R has given x is the father of y and also (y,z) ∈ R y is the father of z but it is not necessary that x is the father of z (x,z) ∉ R.

So, R is neither reflexive relation nor symmetric, and nor transitive relation.

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.

Solun:- Given set R is a set of real numbers.

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation.

So, R is not a reflexive relation because (x,x) ∉ R, it is not always possible that x is always less than or equal to x2. 

For example:- (1/2) ≤ (1/4) {False}

R is not a symmetric relation because (x,y) ∈ R has given x is always less than or equal to y2 but it is not possible that y is always less than or equal to x2 (y,x) ∉ R.

For example- (1,4)    1<16

But (4,1)  4 is not less than or equal to 1.

R is not a transitive relation because 

(x,y) ∈ R i.e. (3,2)   3<4

And (y,z) ∈ R i.e. (2,1.5)  2<2.25

But (x,z) ∉ R i.e. (3,1.5)   3<2.25 {False}

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

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.

Solun:- Given set A={1, 2, 3, 4, 5, 6}

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

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation.

So, R is not a reflexive relation because (x,x) ∉ R, (1,1) ∉ R, (2,2) ∉ R, (3,3) ∉ R, (4,4) ∉ R, (5,5) ∉ R, (6,6) ∉ R. 

R is not a symmetric relation because of (x,y) ∈ R but (y,x) ∉ R, (1,2) ∈ R but (2,1) ∉ R

R is not a transitive relation because 

(x,y) ∈ R and (y,z) ∈ R But (x,z) ∉ R i.e. (1,2) ∈ R and (2,3) ∈ R But (1,3) ∉ R

So, R is neither reflexive relation nor symmetric, and nor transitive relation.

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

Solun:- Given set R is a set of real numbers

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation.

So, R is a reflexive relation because (x,x) ∈ R, x is always equal to x.

R is not a symmetric relation because of (x,y) ∈ R but (y,x) ∉ R, (1,2) ∈ R but (2,1) ∉ R

2<1 {False}

R is a transitive relation because 

(x,y) ∈ R and (y,z) ∈ R, Also (x,z) ∈ R i.e. (1,2) ∈ R and (2,3) ∈ R then (1,3) ∈ R

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

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

Solun:- Given set R is a set of real numbers

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation.

So, R is not a reflexive relation because (x,x) ∉ R, it is not always possible that x is always less than or equal to x3. 

For example:- (1/2) ≤ (1/8) {False}

R is not a symmetric relation because (x,y) ∈ R has given x is less than or equal to y3 but it is not possible that y is always less than or equal to x3 (y,x) ∉ R.

For example- (1,4)    1<64

But (4,1)  4 is not less than or equal to 1.

R is not a transitive relation because 

(x,y) ∈ R i.e. (3,3/2)   3<27/8

And (y,z) ∈ R i.e. (3/2,6/5)  3/2<216/125

But (x,z) ∉ R i.e. (3,6/5)   3<216/125 {False}

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

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.

Solun:- Given set R={1, 2, 3}

Relation R={(1,2), (2,1)}

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is a symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation.

So, R is not a reflexive relation because of (x,x) ∉ R, (1,1) ∉ R.

R is a symmetric relation because of (x,y) ∈ R and also (y,x) ∈ R, (1,2) ∈ R then (2,1) ∈ R

R is not a transitive relation because of (x,y) ∈ R and (y,z) ∈ R, then (x,z) ∉ R.

Hence, R is a symmetric but neither reflexive nor transitive relation.

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

Solun:- Given set A of all books in a library of a college

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

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is a symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation.

So, R is a reflexive relation i.e. (x,x) ∈ R, because x and x have the same number of pages.

R is a symmetric relation i.e. (x,y) ∈ R and also (y,x) ∈ R because x and y have the same number of pages then y and x have also the same number of pages

R is a transitive relation i.e. (x,y) ∈ R and (y,z) ∈ R, then (x,z) ∈ R because x and y have the same number of pages and also y and z have the same number of pages then x and z also have the same number of pages

Hence, R is a reflexive, symmetric, and transitive relation so it is an equivalence relation.

8. Show that the relation R in the set A {1, 2, 3, 4, 5, 6} 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 element of {1,3,5} is related to any element of {2,4}.

Solun:- Given set A={1, 2, 3, 4, 5, 6}

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

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is a symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation.

So, R is a reflexive relation i.e. (a,a) ∈ R, because |a-a|=0 (even) 

R is a symmetric relation i.e. (a,b) ∈ R and also (b,a) ∈ R because |a-b|=|b-a| that is even

R is a transitive relation i.e. (a,b) ∈ R and (b, a) ∈ R, then (a,c) ∈ R because 

|a-b| is even and |b-c| is even

|a-c|=|a-b|+|b-c| So |a-c| is also a even number.

Hence, R is a reflexive, symmetric, and transitive relation so it is an equivalence relation.

Now, elements {1,3,5} are related to each other because the difference between these elements is even.

And, elements {2,4} are related to each other because the difference between these elements is even.

And, {1,3,5} is not related to the element {2,4} because the difference between these elements is odd.

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 elemets related to 1 in each case. 

Solun:- Given set A={x ∈ Z:0≤x≤12}

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

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is a symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation.

So, R is a reflexive relation i.e. (a,a) ∈ R, because |a-a|=0 (multiple by 4) 

R is a symmetric relation i.e. (a,b) ∈ R and also (b,a) ∈ R because |a-b| is multiple by 4 and also |b-a|=|-(a-b)|=|a-b| that is multiple by 4.

R is a transitive relation i.e. (a,b) ∈ R and (b, a) ∈ R, then (a,c) ∈ R because 

|a-b| and |b-c| is multiple by 4

|a-c|=|a-b|+|b-c| So |a-c| is also multiple by 4.

Hence, R is a reflexive, symmetric, and transitive relation so it is an equivalence relation.

Elements related to 1 are:

R={(1,1), (5,1), (9,1)}

Related elements are {1,5,9}

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

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is a symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation.

So, R is a reflexive relation i.e. (a, a) ∈ R, because a is always equal to a 

R is a symmetric relation i.e. (a,b) ∈ R and also (b,a) ∈ R because given a=b and also b=a.

R is a transitive relation i.e. (a,b) ∈ R and (b, a) ∈ R, then (a,c) ∈ R because given a=b and b=a then

a=b=c and (a,c) ∈ R

Hence, R is a reflexive, symmetric, and transitive relation so it is an equivalence relation.

Elements related to 1 are:

R={(1,1)}

Related elements are {1}

10. Give an example of a relation. Which is 

(i) Symmetric but neither reflexive nor transitive

Solun:- Let A={1, 2, 3}

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

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is a symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation.

And according to the relation R 

(a, a) ∉ R i.e. (1,1) So it is not a reflexive relation.

(a, b) ∈ R i.e. (1,2) and also (b, a) ∈ R i.e. (2,1) So it is a symmetric relation.

(a, b) ∈ R and (b, c) ∈ R but (a, c) ∉ R So it is not a transitive relation.

(ii) Transitive but neither reflexive nor symmetric

Solun:- Let A={1, 2, 3}

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

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is a symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation.

And according to the relation R 

(a, a) ∉ R i.e. (1,1) So it is not a reflexive relation.

(a, b) ∈ R i.e. (1,2) but (b, a) ∉ R So it is not a symmetric relation.

(a, b) ∈ R i.e. (1,2) and (b, c) ∈ R i.e. (2,3) and also (a, c) ∈ R i.e. So it is a transitive relation.

(iii) Reflexive and symmetric and but not transitive 

Solun:- Let A={1, 2, 3}

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

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is a symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation.

And according to the relation R 

(a, a) ∈ R i.e. (1,1), (2,2), (3,3), So it is a reflexive relation.

(a, b) ∈ R i.e. (1,2) and also (b, a) ∈ R i.e. (2,1) So it is a symmetric relation.

(a, b) ∈ R and (b, c) ∈ R but (a, c) ∉ R So it is not a transitive relation.

(iv) Reflexive and transitive and but not symmetric 

Solun:- Let A={1, 2, 3}

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

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is a symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation.

And according to the relation R 

(a, a) ∈ R i.e. (1,1), (2,2), (3,3) ∈ R, So it is a reflexive relation.

(a, b) ∈ R i.e. (1,2) but (b, a) ∉ R So it is not a symmetric relation.

(a, b) ∈ R i.e. (1,2) and (b, c) ∈ R i.e. (2,3) and also (a, c) ∈ R i.e. (1,3) So it is a transitive relation.

(v) Symmetric and transitive and but not reflexive

Solun:- Let A={1, 2, 3}

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

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is a symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation. According to the relation R 

(a, a) ∉ R i.e. (1,1) So it is not a reflexive relation.

(a, b) ∈ R i.e. (1,2) and also (b, a) ∈ R i.e. (2,1) So it is a symmetric relation.

(a, b) ∈ R i.e. (1,2) and (b, c) ∈ R i.e. (2,3) and also (a, c) ∈ R i.e. (1,3) So it is a transitive relation.

11. Show that the relation R in the set A of points in a plane given by R={(P, Q): distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point P ≠ (0,0) is the circle passing through P with origin as center.

Solun:- Given set A of points in a plane

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

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is a symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation.

So, R is a reflexive relation i.e. (P, P) ∈ R, because the distance of the same point from the origin is also same.

R is a symmetric relation i.e. (P, Q) ∈ R and also (Q, P) ∈ R because given distance of the point P from the origin is same as the distance of the point Q from the origin similarly distance of the point Q from the origin is same as the distance of the point P from the origin

R is a transitive relation i.e. (P, Q) ∈ R and (Q, S) ∈ R, then (P, S) ∈ R because given distance of the point P from the origin is same as the distance of the point Q from the origin and also the distance of the point Q from the origin is same as the distance of the point S from the origin then also the distance of P and S from the origin is same.

Hence, R is a reflexive, symmetric, and transitive relation so it is an equivalence relation.

Let P be a point P ≠ (0,0) i.e. OP=k then all points that are distant k from the origin are related to the circle that passes from P.

12. Show that the relation R defined in the set A of all triangles as R={(T1, T2}: T1 is similar to T2 }, is an equivalence relation. Consider three right angle triangles T1 with sides 3,4,5, T2 with sides 5,12,13, and T3 with sides 6,8,10. Which triangles among T1, T2, and T3 are related?

Solun:- Given set A of all triangles

Relation R={(T1, T2}: T1 is similar to T2 }

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is a symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation.

So, R is a reflexive relation i.e. (T1, T1) ∈ R, because T1 is similar to itself.

R is a symmetric relation i.e. (T1, T2) ∈ R then (T2, T1) ∈ R because given T1 is similar to T2 then T2 is also similar to T1.

R is a transitive relation i.e. (T1, T2) ∈ R and (T2, T3) ∈ R, then (T1, T3) ∈ R because given T1 is similar to T2 and also T2 is similar to T3 then T1 is also similar to T3.

Hence, R is a reflexive, symmetric, and transitive relation so it is an equivalence relation.

According to triangle properties:-

Triangle T1 and T2:-

3/5 ≠ 4/12 ≠ 5/13

T1 and T2 are not related.

Triangle T1 and T3:-

3/6 = 4/8 = 5/10 = (1/2)

T1 and T3 are related.

Triangle T2 and T3:-

5/6 ≠ 12/8 ≠ 13/10

T2 and T3 are not related.

13. Show that the relation R defined in the set A of all polygons as R={(P1, P2}: P1 and P2 have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3,4, and 5?

Solun:- Given set A of all polygons

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

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is a symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation.

So, R is a reflexive relation i.e. (P1, P1) ∈ R, because same polygon has the same number of sides.

R is a symmetric relation i.e. (P1, P2) ∈ R then (P2, P1) ∈ R because given P1 and P2 have the same number of sides then also P2 and P1 have also the same number of sides.

R is a transitive relation i.e. (P1, P2) ∈ R and (P2, P3) ∈ R, then (P1, P3) ∈ R because given P1 and P2 have the same number of sides and P2 and P3 have the same number of sides then also P1 and P3 have also the same number of sides.

Hence, R is a reflexive, symmetric, and transitive relation so it is an equivalence relation.

Right angle triangle T with sides 3, 4, and 5 is related to all polygons which have 3 sides (triangle).

14. Let L be the set of all lines in the XY plane and R be the relation in L defined as R={(L1, L2): L1 is parallel to L2}. Show that R is an equivalence relation. Find the set of all lines related to the line y=2x+4.

Solun:- Given set L be the set of all lines in the XY plane

Relation R={(L1, L2): L1 is parallel to L2}

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is a symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation.

So, R is a reflexive relation i.e. (L1, L1) ∈ R, because L1 is parallel to the L1.

R is a symmetric relation i.e. (L1, L2) ∈ R then (L2, L1) ∈ R because given L1 is parallel to L2 then L2 is also parallel to L1.

R is a transitive relation i.e. (L1, L2) ∈ R and (L2, L3) ∈ R, then (L1, L3) ∈ R because given L1 is parallel to L2 and L2 is parallel to L3 then L1 is also parallel to L3.

Hence, R is a reflexive, symmetric, and transitive relation so it is an equivalence relation.

We know parallel lines have the same slope.

The given line is:- y=2x+4

General equation of a line:- y=mx+c

Then m=2

So all lines y=2x+c is related to the line y=2x+4 with the slope 2. (Here c ∈ R)

15. Let R be 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)}. Choose the correct answer.

(A) R is reflexive and symmetric but not transitive.

(B) R is reflexive and transitive but not symmetric.

(C) R is symmetric and transitive but not reflexive.

(D) R is an equivalence relation.

Solun:- Given set={1, 2, 3, 4} 

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

We know that if (a, a) ∈ R, for every a ∈ A then it is reflexive relation.

If (a1, a2) ∈ R, implies that (a2, a1) ∈ R, for all a1, a2 ∈ A then it is a symmetric relation.

if (a1, a2) ∈ R, and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A then it is transitive relation.

So, R is a reflexive relation because (1,1), (2,2), (3,3), (4,4) ∈ R.

R is not a symmetric relation because of (1,2) ∈ R but (2,1) ∉ R.

R is a transitive relation because of (1,3) ∈ R and (3,2) ∈ R and also (1,2) ∈ R.

So, R is reflexive and transitive but not symmetric.

The correct answer is (B).

16. Let R be the relation in the set N given by R={(a, b): a = b - 2, b > 6}. Choose the correct answer.

(A) (2,4) ∈ R    (B) (3,8) ∈ R    (C) (6,8) ∈ R    (D) (8,7) ∈ R 

Solun:- Given set N is set of natural numbers

Relation R={(a, b): a = b - 2, b > 6}

The correct answer is (C) because (6=8-2) and also 8>6.

Download PDF of Exercise 1.1

See Also:-

Notes of Relations and Functions

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