NCERT EXERCISE 1.1

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NCERT Solutions for Class 12 Maths Chapter – 1 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}  
Sol.
A = {1, 2, 3 … 13, 14}
R = {(x, y): 3x − y = 0}
∴ 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. [3(3) − 1 ≠ 0]
Also, R is not transitive as (1, 3), (3, 9) ∈ R, but (1, 9) ∉ R. [3(1) − 9 ≠ 0] 
Hence, 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} 
Sol.
R = {(x, y): y = x + 5 and x < 4} = {(1, 6), (2, 7), (3, 8)} It is clear that (1,1) ∉ R.
R is not reflexive.
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.
Hence, 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} 
Sol.
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.
So, (x, x) ∈ R
[as 4 is divisible by 2]
[as 2 is not divisible by 4]
Let (x, y), (y, z) ∈ R. Then, y is divisible by x and z is divisible by y.
z is divisible by x. ⇒ (x, z) ∈ R
R is transitive.
Hence, R is reflexive and transitive but not symmetric.
(iv) Relation R in the set Z of all integers defined as
R = {(x, y): x – y is as integer)
Sol.
R = {(x, y): x − y is an integer}
Now, for every x ∈ Z, (x, x) ∈ R as x − x = 0 is an integer.
∴ R is reflexive.
Now, for every x, y ∈ Z, if (x, y) ∈ R, then x − y is an integer. ⇒ −(x − y) is also an integer.
⇒ (y − x) is an integer.
∴ (y, x) ∈ R
∴ R is symmetric.
Now, Let (x, y) and (y, z) ∈ R, where x, y, z ∈ Z.
⇒ (x − y) and (y − z) are integers.
⇒ x − z = (x − y) + (y − z) is an integer.
∴ (x, z) ∈ R
∴ R is transitive.
Hence, 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}
Sol.
R = {(x, y): x and y work at the same place}
⇒ (x, x) ∈ R [as x and x work at the same place] ∴ R is reflexive.
If (x, y) ∈ R, then x and y work at the same place.
⇒ 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. ⇒ 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}
Sol.
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.
⇒ y and x live in the same locality.
⇒ (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}
Sol.
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. [Since, y is 7 cm smaller than x]
∴ (y, x) ∉ R
Indeed if x is exactly 7 cm taller than y, then y is exactly 7 cm shorter than x. ∴ R is not symmetric.
Now,
Let (x, y), (y, z) ∈ 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.
Hence, R is neither reflexive, nor symmetric, nor transitive.
(d) R = {(x, y) : x is wife of y}
Sol.
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
Indeed, if x is the wife of y, then y is the husband of x.
∴ R is not transitive.
Let (x, y), (y, z) ∈ R
⇒ 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
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
(e) R = {(x, y) : x is father of y}
Sol.
R = {(x, y): x is the father of y} (x, x) ∉ R
As x cannot be the father of himself. ∴ R is not reflexive.
Now, let (x, y) ∉ R.
⇒ x is the father of y.
⇒ y cannot be the father of y.
Indeed, y is the son or the daughter of y.
∴ (y, x) ∉ R
∴ 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. ⇒ x is not the father of z.
Indeed, x is the grandfather of z.
∴ (x, z) ∉ R
∴ 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.Sol.
(i) R = {(a, b): a ≤ b2}
It can be observed that ( 1/2 , 1/2 ) ∉ R,
since, 1/2 > (1/2 )²
∴ R is not reflexive.
Now, (1, 4) ∈ R as 1 < 4² But, 4 is not less than 1².
∴ (4, 1) ∉ R
∴ 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 3 :
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.

Solution 4 :
R = {(a, b): a ≤ b}
Clearly (a, a) ∈ R
∴ 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 5 :
R = {(a, b): a ≤ b³}
It is observed that ( ½ , ½) ∉ R, [since, ½ > (½)³]
∴ 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, since 3 < (3/2)³ and 3/2 < (6/5)³
But (3, 6/5 ) ∉ R as 3 > (6/5)³
∴ 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 6 :
Let A = {1, 2, 3}.
A relation R on A is defined as R = {(1, 2), (2, 1)}.
It is clear that (1, 1), (2, 2), (3, 3) ∉ R.
∴ 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 7 :
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 8 :
A = {1, 2, 3, 4, 5} and R = {(a, b): |a-b| is even}
It is clear that for any element a ∈ A, we have |a-a| = 0 (which is even). ∴ R is reflexive.
Let (a, b) ∈ R.
⇒|𝑎 − 𝑏| 𝑖𝑠 𝑒𝑣𝑒𝑛
⇒|-(a-b)|= |b – a| is also even
⇒ (b, a) ∈ R
∴ 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 − 𝑏| 𝑖𝑠 𝑒𝑣𝑒𝑛.
⇒ (a, c) ∈ R
∴ R is transitive.
Hence, R is an equivalence relation.
Now, 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. [as 1 – 2, 1 – 4, 3 – 2, 3 – 4, 5 – 2 and 5 – 4 all are odd]

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