Relations and Functions EXERCISE 1.1

Question 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}
(ii) Relation R in the set N of natural numbers defined as
R = {(x, y): y = x + 5 and x < 4}
(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as
R = {(x, y): y is divisible by x}
(iv) Relation R in the set Z of all integers defined as
R = {(x, y): x − y is as integer}
(v) Relation R in the set A of human beings in a town at a time given by
(a) R = {(x, y): x and y work at the same place}
(b) R = {(x, y): x and y live in the same locality}
(c) R = {(x, y): x is exactly 7 cm taller than y}
(d) R = {(x, y): x is wife of y}
(e) R = {(x, y): x is father of y}
Solution
(i) A = {1, 2, 3 ... 13, 14}
R = {(x, y): 3x − y = 0}
This can be written in another form as below
R = {(1, 3), (2, 6), (3, 9), (4, 12)}
R is not reflexive since (1, 1), (2, 2) ... (14, 14) ∉ R.
Also, R is not symmetric as (1, 3) ∈R, but (3, 1) ∉ R.
Also, R is not transitive as (1, 3), (3, 9) ∈R, but (1, 9) ∉ R.
So, R is neither reflexive, nor symmetric, nor transitive.
(ii) A = {1, 2, 3 ... 13, 14}
R = {(x, y): y = x + 5 and x < 4}
This can be written in another form as below
 {(1, 6), (2, 7), (3, 8)}
It is seen that (1, 1) ∉ R.
So, R is not reflexive.
(1, 6) ∈ R But, (1, 6) ∉ R.
So, 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.
So, R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
(iii) A = {1, 2, 3, 4, 5, 6}
R = {(x, y): y is divisible by x}
We know that any number (x) is divisible by itself.
(x, x) ∈R
So, R is reflexive.
Now,
(2, 4) ∈R [as 4 is divisible by 2]
But,
(4, 2) ∉ R. [as 2 is not divisible by 4]
So, R is not symmetric.
Let (x, y), (y, z) ∈ R. Then, y is divisible by x and z is divisible by y.
Therefore, z is divisible by x.
So, R is transitive.
Hence, R is reflexive and transitive but not symmetric.
(iv) R = {(x, y): x − y is an integer}
Now, for every x ∈ Z, (x, x) ∈R as x − x = 0 is an integer.
So, R is reflexive.
Now, for every x, y ∈ Z if (x, y) ∈ R, then x − y is an integer.
Now − (x − y) is also an integer.
So, (y − x) is an integer.
Therefore (y, x) ∈ R
∴R is symmetric.
Now,
Let (x, y) and (y, z) ∈R, where x, y, z ∈ Z.
Now (x − y) and (y − z) are integers.
Then x − z = (x − y) + (y − z) is an integer.
So (x, z) ∈R
Therefore, R is transitive.
Hence, R is reflexive, symmetric, and transitive.
(v) (a) R = {(x, y): x and y work at the same place}
(x, x) ∈ R
∴ R is reflexive.
If (x, y) ∈ R, then x and y work at the same place.
 or y and x work at the same place.
 (y, x) ∈ R.
∴R is symmetric.
Now, let (x, y), (y, z) ∈ R
 x and y work at the same place and y and z work at the same place.
or  x and z work at the same place.
 (x, z) ∈R
∴ R is transitive.
Hence, R is reflexive, symmetric, and transitive.
(b) R = {(x, y): x and y live in the same locality}
Clearly (x, x) ∈ R as x and x is the same human being.
∴ R is reflexive.
If (x, y) ∈R, then x and y live in the same locality.
or y and x live in the same locality.
or (y, x) ∈ R
∴R is symmetric.
Now, let (x, y) ∈ R and (y, z) ∈ 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) ∈ R
∴ R is transitive.
Hence, R is reflexive, symmetric, and transitive.
(c) R = {(x, y): x is exactly 7 cm taller than y}
Now,
(x, x) ∉ R
Since human being x cannot be taller than himself.
R is not reflexive.
Now, let (x, y) ∈R.
 x is exactly 7 cm taller than y.
Then, y is not taller than x.
So, (y, x) ∉R
Therefore, R is not symmetric.
Now,
Let (x, y), (y, z) ∈ R.
Then x is exactly 7 cm taller than y and y is exactly 7 cm taller than z.
So, x is exactly 14 cm taller than z.
(x, z) ∉R
Therefore, R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
(d) R = {(x, y): x is the wife of y}
Now,
(x, x) ∉ R
Since x cannot be the wife of herself.
∴R is not reflexive.
Now, let (x, y) ∈ R
x is the wife of y.
Clearly y is not the wife of x.
(y, x) ∉ R
Therefore, R is not transitive.
Let (x, y), (y, z) ∈ R
So  x is the wife of y and y is the wife of z.
This case is not possible. Also, this does not imply that x is the wife of z.
(x, z) ∉ R
Therefore, R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
(e) R = {(x, y): x is the father of y}
(x, x) ∉ R
As x cannot be the father of himself.
Therefore, R is not reflexive.
Now, let (x, y) ∈R.
So, x is the father of y.
Now y cannot be the father of x.
 (y, x) ∉ R
Therefore, R is not symmetric.
Now, let (x, y) ∈ R and (y, z) ∈ R.
 x is the father of y and y is the father of z.
Then clearly x is not the father of z. Infact x is grandfather of z
 (x, z) ∉ R
Therefore, R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor 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
R = {(a, b): a ≤ b²}
It can be observed that
(1/3,1/3) does not belong to this, as 1/ 3 > (1/3)²
Therefore, R is not reflexive.
Now, (1, 4) ∈ R as 1 < 4²
But, 4 is not less than 1².
So (4, 1) ∉ R
Therefore, R is not symmetric.
Now,
(3, 2), (2, 1.5) ∈ R
(as 3 < 2² = 4 and 2 < (1.5)² = 2.25)
But, 3 > (1.5)² = 2.25
∴(3, 1.5) ∉ R
∴ 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 A = {1, 2, 3, 4, 5, 6}.
A relation R is defined on set A as:
R = {(a, b): b = a + 1}
∴R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
We can find (a, a) ∉ R, where a ∈ A.
For instance,
(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) ∉ R
∴R is not reflexive.
It can be observed that (1, 2) ∈ R, but (2, 1) ∉ R.
∴R is not symmetric.
Now, (1, 2), (2, 3) ∈ R
But,
(1, 3) ∉ R
∴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.
Answer:
R = {(a, b); a ≤ b}
Clearly (a, a) ∈ R as a = a.
∴R is reflexive.
Now,
(2, 4) ∈ R (as 2 < 4)
But, (4, 2) ∉ R as 4 is greater than 2.
∴ R is not symmetric.
Now, let (a, b), (b, c) ∈ R.
Then,
a ≤ b and b ≤ c
⇒ a ≤ c
⇒ (a, c) ∈ R
∴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
R = {(a, b): a ≤ b³ }
It is observed that
(1/3,1/3) does not belong to this, as 1/ 3 > (1/3)³
Therefore, R is not reflexive.
Now,
(1, 2) ∈ R (as 1 < 2³ = 8)
But,
(2, 1) ∉ R (as 2³ > 1)
∴ R is not symmetric.
We have (3, 3/2) , (3/2,6/5)  ∈ R
But
(3,6/5) ∉ R [as 3 > (6/5)³]
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 A = {1, 2, 3}.
A relation R on A is defined as R = {(1, 2), (2, 1)}.
It is seen that (1, 1), (2, 2), (3, 3) ∉ R.
Therefore, R is not reflexive.
Now, as (1, 2) ∈ R and (2, 1) ∈ R, then R is symmetric.
Now, (1, 2) and (2, 1) ∈ R
However,
(1, 1) ∉ R
∴ 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 same number of pages} is an 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}
Now, R is reflexive since (x, x) ∈ R as x and x has 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
∴R is symmetric.
Now, let (x, y) ∈R and (y, z) ∈ R.
⇒ x and y and 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
∴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 element 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},
For any element, a ∈A, we have |a -a| =0 (which is even).
∴R is reflexive.
Let (a, b) ∈ R.
Now that means |a-b| is even
Or |-(b-a) | is even
Or |b-a| is even
∴R is symmetric.
Now, let (a, b) ∈ R and (b, c) ∈ R.
Now that means |a-b| and |b-c| are even
Now |a-c| = |(a-b) – (b-c) |
So (a, c) ∈ R
Therefore, R is transitive.
Hence, 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. 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 Î 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 Î Z : 0 ≤ x ≤ 12} = {0,1,2,3,4…..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| is a multiple of 4.
∴R is reflexive.
Now, let (a, b) ∈ R ⇒ |a-b| is a multiple of 4.
Now |b-a| =|-(a-b)|  which is multiple of 4
⇒ (b, a) ∈ R
∴R is symmetric.
Now, let (a, b), (b, c) ∈ R.
|a-b| is multiple of 4
|b-c| is multiple of 4
|a-c | = |(a-b) – (b-c) |
So (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
 
(ii) R = {(a, b): a = b}
For any element, a ∈A, we have (a, a) ∈ R, since a = a.
∴R is reflexive.
Now, let (a, b) ∈ R.
⇒ a = b
⇒ b = a
⇒ (b, a) ∈ R
∴R is symmetric.
Now, let (a, b) ∈ R and (b, c) ∈ R.
⇒ a = b and b = c
⇒ a = c
⇒ (a, c) ∈ R
∴ 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:
Given 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 = {1, 2, 3}.
Define a relation R on A as R = {(1, 2), (2, 1)}.
Relation R is not reflexive as (1, 1), (2, 2), (3, 3) ∉ R.
Now, as (1, 2) ∈ R and (2, 1) ∈ R, R is symmetric.
(1, 2), (2, 1) ∈ R, but (1,1) ∉ 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 a itself. In fact, a = a.
Therefore, R is not reflexive.
Now,
(2, 3) ∈ R (as 2 < 3)
But, 3 is not less than 2.
(3, 2) ∉ R
Therefore, R is not symmetric.
Now, let (a, b), (b, c) ∈ R.
a < b and b < c
or a < c
(a, c) ∈ R
Therefore, R is transitive.
Hence, relation R is transitive but not reflexive and symmetric.
(iii)Let A = {1, 2, 3}.
Define a relation R on A as:
A = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}
Relation R is reflexive since for every a ∈ A, (a, a) ∈R i.e., (1, 1), (2,2), (3, 3)} ∈ R.
Relation R is symmetric since (a, b) ∈ R ⇒ (b, a) ∈ R for all a, b ∈ R.
Relation R is not transitive as  (1,2) and (2,3) ∈ R but (1,3) ∉ R
Hence, relation R is reflexive and symmetric but not transitive
iv)
(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 = a.
Therefore, R is  reflexive.
Now,
(3, 2) ∈ R (as 3 >   2)
But, 3 is not less than 2.
(2, 3) ∉ R
Therefore, R is not symmetric.
Now, let (a, b), (b, c) ∈ R.
a ≥ b and b ≥ c
or a ≥ c
(a, c) ∈ R
Therefore, R is transitive.
Hence, relation R is transitive and reflexive but not symmetric.
(v)Let A = {1, 2}.
Define a relation R on A as:
A = {(1, 1), (1, 2), (2, 1)}
Relation R is not reflexive since for every a ∈ A, (a, a) ∉ R
Relation R is symmetric since (a, b) ∈ R ⇒ (b, a) ∈ R for all a, b ∈ R.
Relation R is transitive as  (1,2) and (2,1) ∈ R  and  (1,1) ∈ R
Hence, relation R is symmetric and transitive but not reflexive
 
 
Question 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 point related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.
Solution
R = {(P, Q): distance of point P from the origin is the same as the distance of point Q from the origin}
Clearly, (P, P) ∈ R since the distance of point P from the origin is always the same as the distance of the same point P from the origin.
Therefore, R is reflexive.
Now,
Let (P, Q) ∈ R.
That means the distance of point P from the origin is the same as the distance of point Q from the origin.
Clearly The distance of point Q from the origin is the same as the distance of point P from the origin.
(Q, P) ∈ R
Therefore, R is symmetric.
Now,
Let (P, Q), (Q, S) ∈ R.
The distance of points P and Q from the origin is the same and, the distance of points Q and S from the origin is the same.
So, the distance of points P and S from the origin is the same.
(P, S) ∈ R
Therefore, R is transitive.
Therefore, R is an 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.
In other words, if O (0, 0) is the origin and OP = k, then the set of all points related to P is at distance k from the origin.
Hence, this set of points forms a circle with the centre as the origin and this circle passes through point P.
Question 12:
Show that the relation R defined in the set A of all triangles as R = {(T₁, T₂): T₁ is similar to T₂}, is 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?
Solution
R = {(T₁, T₂): T₁ is similar to T₂},
Now R is clearly reflexive since every triangle is similar to itself.
Now,
if (T₁, T₂) ∈ R, then
 T₁ is similar to T₂.
 T₂ is similar to T₁.
So, (T₂, T₁)  ∈ R
Therefore, R is symmetric.
Now,
Let (T₁, T₂), (T₂, T₃) ∈ R.
 T₁ is similar to T₂ and T₂ is similar to T₃.
 T₁ is similar to  T₃.
Therefore, R is transitive.
Thus, R is an equivalence relation.
Now, we can observe that:
The corresponding sides of triangles T₁ and T₃ are in the same ratio.
3/6=4/8=5/10= 1/2
Then, triangle T₁ is similar to triangle T₃.
So These two belongs to this relation
Question 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?
Solution
R = {(P₁, P₂):  P₁ and P₂ have same the number of sides}
R is reflexive since (P₁, P₁) ∈ R as the same polygon has the same number of sides with itself.
Let (P₁, P₂) ∈ R.
That means P₁ and P₂ have the same number of sides.
Or P₂ and P₁ have the same number of sides.
(P₂, P₁) ∈ R
Therefore, R is symmetric.
Now,
Let (P₁, P_­2), (P₂, P₃) ∈ R.
 P₁ and P₂ have the same number of sides.
P₂ and P₃ have the same number of sides.
So P₁ and P₃ have the same number of sides.
(P₁, P₃) ∈ R
Therefore, R is transitive.
Hence, 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 (since T is a polygon with 3 sides).
Hence, the set of all elements in A related to triangle T is the set of all triangles.
Question 14:
Let L be the set of all lines in 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.
Solution
R = {(L₁, L₂): L₁ is parallel to L₂}
Now R is reflexive as any line L₁ is parallel to itself i.e., (L₁, L₁) ∈ R.
Now,
Let (L₁, L₂) ∈ R.
Now that means L₁ is parallel to L₂.
So L₂ is parallel to L₁.
 (L₂, L₁) ∈ R
Therefore, R is symmetric.
Now,
Let (L₁, L₂), (L₁, L₃) ∈R.
⇒ L₁ is parallel to L₂. Also, L₂ is parallel to L₃.
⇒ L₁ is parallel to L₃.
Therefore, R is transitive.
Hence, 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
It is known that parallel lines have the same slopes.
The line parallel to the given line is of the form y = 2x + c, where c ∈ R.
Hence, the set of all lines related to the given line is given by y = 2x+ c, where c ∈ R.
Question 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.
Solution:
R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}
It is seen that (a, a) ∈ R, for every a ∈ {1, 2, 3, 4}.
Therefore, R is reflexive.
It is seen that (1, 2) ∈ R, but (2, 1) ∉ R.
Therefore, R is not symmetric.
Also, it is observed that (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈
{1, 2, 3, 4}.
Therefore, R is transitive.
So, R is reflexive and transitive but not symmetric.
The correct answer is B.
Question 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
Solution
R = {(a, b): a = b − 2, b > 6}
Now, since b > 6, (2, 4) ∉ R
Also, as 3 ≠ 8 − 2, (3, 8) ∉ R
And, as 8 ≠ 7 − 2
(8, 7) ∉ R
Now, consider (6, 8).
We have 8 > 6 and  6 = 8 − 2.
Therefore,  (6, 8) ∈ R
The correct answer is C.

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