Call Now
1800-102-2727We will be looking through the relationships in this chapter. The salient features of this chapter are the basic concepts that it holds to understand the chapters taught further in the class. We covered different operations on sets in the previous chapter, which lead to discovering more about sets. In this chapter, we'll look at another operation called the cartesian product of sets. Finally, we'll be able to incorporate the idea of a relationship. Let us consider a pair of elements that are arranged in a certain order. A subset of the cartesian product A x B is obtained by defining a relationship between the first element x and the second element y from the ordered pairs in A x B, a relation R from set A to set B. The set of all second elements from the ordered pairs in a relation R is the range of that relation R. The chapter covers types of relations relation, graphical representation of a relation, domain, codomain, and range of a relation, the elements that are there in the Cartesian product of two sets, how to represent a relation, how to represent a Cartesian product of two sets in a diagram. Students will learn to integrate pairs of things from two separate sets before introducing connections between the two sets. This is known as the ordered pair. A function's selection is the set of all images. The domain and spectrum of a real function are also real numbers or one of its subsets.
Students who have questions about the topics should consult RD Sharma Solutions Class 11 Maths. The solutions are developed by subject matter specialists with extensive research expertise.
Let A be the set of all human beings in a town at a particular time. Determine whether each of the following relations are reflexive, symmetric and transitive:
R = {(x, y) : x and y work at the same place}
We have been given that,
A is the set of all human beings in a town at a particular time. Here, R is the binary relation on set A.
So, recall that
R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz. Using these criteria, we can solve these.
We have,
R = {(x, y): x and y work at the same place}
Since x & x are the same people then, x & x works at the same place.
Take yourself, for example, if you work at Bloomingdale then you work at Bloomingdale. Since you can’t work in two places at a particular time,
So, ∀ x ∈ A, then (x, x) ∈ R.
If x & y works at the same place, then, y and x also work at the same place.
If you & your friend, Chris was working in Bloomindale, then Chris and you are working in Bloomingdale only.
The only difference is in the way of writing, either you write your name and your friend’s name or your friend’s name and your name, it’s the same.
So, if (x, y) ∈ R, then (y, x) ∈ R
∀ x, y ∈ A
If x & y works at the same place and y & z works at the same place. Then, x & z also works at the same place.
Say, if she & I was working in Bloomingdale and she & you were also working in Bloomingdale. Then, you and I are working in the same company.
So, if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∈ R.
∀ x, y, z ∈ A
Hence, R is reflexive, symmetric and transitive. 1 B. Question
Let A be the set of all human beings in a town at a particular time. Determine whether each of the following relations are reflexive, symmetric and transitive:
R = {(x, y) : x and y live in the same locality}
We have been given that,
A is the set of all human beings in a town at a particular time. Here, R is the binary relation on set A.
So, recall that
R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz. Using these criteria, we can solve these.
We have,
R = {(x, y): x and y live in the same locality}
Since x & x are the same people, then, x & x live in the same locality.
Take yourself, for example, if you lived in colony x then you live in colony x. Since you can’t live in two places at a particular time.
So, ∀ x ∈ A, then (x, x) ∈ R.
If x & y live in the same locality, then, y & x also lives in the the same locality.
If you & your friend, Chris are neighbors, then you and Chris are neighbors only.
The only difference is in the way of writing, either you write your name and your friend’s name or your friend’s name and your name, it’s the same.
So, if (x, y) ∈ R, then (y, x) ∈ R.
∀ x, y ∈ R
If x & y lives in the same locality and y & z lives in the same locality. Then, x & z also lives in the same locality.
Say, if she and I were living in colony x and she & you were also working in colony x. Then, you and I are living in the same colony.
So, if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∈ R.
∀ x, y, z ∈ A
Hence, R is reflexive, symmetric and transitive. 1 C. Question 1. B
Let A be the set of all human beings in a town at a particular time. Determine whether each of the following relations are reflexive, symmetric and transitive:
R = {(x, y) : x is wife of y}
We have been given that,
A is the set of all human beings in a town at a particular time. Here, R is the binary relation on set A.
So, recall that
R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz. Using these criteria, we can solve these.
We have,
Question 1.C
R = {(x, y): x is wife of y}
Since x and x are the same people. Then, x cannot be the wife of itself. A person cannot be a wife of itself.
Wendy is the wife of Sam; Wendy can’t be the wife of herself. So, ∀ x ∈ A, then (x, x) ∉ R.
If x is the wife of y.Then, y cannot be the wife of x.
If Wendy is the wife of Sam, then Sam is the husband of Wendy. Sam cannot be the wife of Wendy.
So, if (x, y) ∈ R, then (y, x) ∉ R.
∀ x, y ∈ A
If x is the wife of y and y is the wife of z, which is not logically possible. Then, x is not the wife of z.
It’s easy, take Wendy, Sam, and Mac.
If Wendy is the wife of Sam, Sam can’t be the wife of Mac.
Thus, the possibility of Wendy being the wife of Mac also eliminates. So, if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∉ R.
∀ x, y, z ∈ A
Hence, R is neither reflexive, nor symmetric, nor transitive. 1 D. Question 1. D
Let A be the set of all human beings in a town at a particular time. Determine whether each of the following relations are reflexive, symmetric and transitive:
R = {(x, y) : x is father of y}
We have been given that,
A is the set of all human beings in a town at a particular time. Here, R is the binary relation on set A.
So, recall that
R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz. Using these criteria, we can solve these.
We have,
R = {(x, y): x is father of y}
Since x and x are the same people. Then, x cannot be the father of itself. A person cannot be a father of itself. Leo is the father of Thiago
So, ∀ x ∈ A, then (x, x) ∉ R.
If x is the father of y.
Then, y cannot be the father of x.
If Sam is the father of Mac, then Mac is the son of Sam. Mac cannot be the father of Sam.
So, if (x, y) ∈ R, then (y, x) ∉ R.
∀ x, y ∈ A
If x is the father of y and y is the father of z,then, x is not the father of z. Take Mickey, Sam, and Mac.
If Mickey is the father of Sam, and Sam is the father of Mac. Thus, Mickey is not the father of Mac, but the grandfather of Mac. So, if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∉ R.
∀ x, y, z ∈ A
Hence, R is neither reflexive, nor symmetric, nor transitive.
Relations R1, R2, R3 and R4 are defined on a set A = {a, b, c} as follows : R1 = {(a, a) (a, b) (a, c) (b, b) (b, c), (c, a) (c, b) (c, c)}
R2 = {(a, a)}
R3 = {(b, a)}
R4 = {(a, b) (b, c) (c, a)}
Find whether or not each of the relations R1, R2, R3, R4 on A is (i) reflexive (iii) symmetric (iii) transitive.
We have set, A = {a, b, c}
Here, R1, R2, R3, and R4 are the binary relations on set A. So, recall that for any binary relation R on set A. We have, R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.
So, using these results let us start determining given relations. We have
R1 = {(a, a) (a, b) (a, c) (b, b) (b, c) (c, a) (c, b) (c, c)} (i). Reflexive:
For all a, b, c ∈ A. [∵ A = {a, b, c}]
Then, (a, a) ∈ R1
(b, b) ∈ A
(c, c) ∈ A
[∵ R1 = {(a, a) (a, b) (a, c) (b, b) (b, c) (c, a) (c, b) (c, c)}] So, ∀ a, b, c ∈ A, then (a, a), (b, b), (c, c) ∈ R.
If (a, a), (b, b), (c, c), (a, c), (b, c) ∈ R1
Then, clearly (a, a), (b, b), (c, c), (c, a), (c, b) ∈ R1
∀ a, b, c ∈ A
But, we need to try to show a contradiction to be able to determine the symmetry. So, we know (a, b) ∈ R1
But, (b, a) ∉ R1
So, if (a, b) ∈ R1, then (b, a) ∉ R1.
∀ a, b ∈ A
If (b, c) ∈ R1 and (c, a) ∈ R1
But, (b, a) ∉ R1 [Check the Relation R1 that does not contain (b, a)]
∀ a, b ∈ A
[∵ R1 = {(a, a) (a, b) (a, c) (b, b) (b, c) (c, a) (c, b) (c, c)}] So, if (b, c) ∈ R1 and (c, a) ∈ R1, then (b, a) ∉ R1.
∀ a, b, c ∈ A
Now, we have R2 = {(a, a)}
Here, only (a, a) ∈ R2
for a ∈ A. [∵ A = {a, b, c}]
[∵ R2 = {(a, a)}]
So, for a ∈ A, then (a, a) ∈ R2.
For symmetry,
If (x, y) ∈ R, then (y, x) ∈ R
∀ x, y ∈ A.
Notice, in R2 we have R2 = {(a, a)}
So, if (a, a) ∈ R2, then (a, a) ∈ R2. Where a ∈ A.
Here,
(a, a) ∈ R2 and (a, a) ∈ R2 Then, obviously (a, a) ∈ R2 Where a ∈ A.
[∵ R2 = {(a, a)}]
So, if (a, a) ∈ R2 and (a, a) ∈ R2, then (a, a) ∈ R2, where a ∈ A.
Now, we have R3 = {(b, a)}
∀ a, b ∈ A [∵ A = {a, b, c}] But, (a, a) ∉ R3
Also, (b, b) ∉ R3 [∵ R3 = {(b, a)}]
So, ∀ a, b ∈ A, then (a, a), (b, b) ∉ R3
If (b, a) ∈ R3
Then, (a, b) should belong to R3.
∀ a, b ∈ A. [∵ A = {a, b, c}]
But, (a, b) ∉ R3 [∵ R3 = {(b, a)}]
So, if (a, b) ∈ R3, then (b, a) ∉ R3
∀ a, b ∈ A
We have (b, a) ∈ R3 but do not contain any other element in R3. Transitivity can’t be proved in R3.
[∵ R3 = {(b, a)}]
So, if (b, a) ∈ R3 but since there is no other element.
Now, we have
R4 = {(a, b) (b, c) (c, a)} (i). Reflexive:
∀ a, b, c ∈ A [∵ A = {a, b, c}] But, (a, a) ∉ R4
Also, (b, b) ∉ R4 and (c, c) ∉ R4 [∵ R4 = {(a, b) (b, c) (c, a)}]
So, ∀ a, b, c ∈ A, then (a, a), (b, b), (c, c) ∉ R4
If (a, b) ∈ R4, then (b, a) ∈ R4 But (b, a) ∉ R4
[∵ R4 = {(a, b) (b, c) (c, a)}]
So, ∀ a, b ∈ A, if (a, b) ∈ R4, then (b, a) ∉ R4.
⇒ R4 is not symmetric.
It is sufficient to show only one case of ordered pairs violating the definition.
We have,
(a, b) ∈ R4 and (b, c) ∈ R4
⇒ (a, c) ∈ R4 But, is it so? No, (a, c) ∉ R4
So, it is enough to determine that R4 is not transitive.
∀ a, b, c ∈ A, if (a, b) ∈ R4 and (b, c) ∈ R4, then (a, c) ∉ R4.
Test whether the following relations R1, R2 and R3 are (i) reflexive (ii) symmetric and (iii) transitive : R1 on Q0 defined by (a, b) ϵ R1⇔ a = 1/b
Here, R1, R2, R3, and R4 are the binary relations.
So, recall that for any binary relation R on set A. We have, R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.
So, using these results let us start determining given relations. We have
R1 on Q0 defined by (a, b) ∈ R1⇔ _{}
∀ a, b ∈ Q0,
(a, a), (b, b) ∈ R1 needs to be proved for reflexivity. If (a, b) ∈ R1
Then, _{} …(1) So, for (a, a) ∈ R1
Replace b by a in equation (1), we get
But, we know
⇒ (a, a) ∉ R1
So, ∀ a ∈ Q0, then (a, a) ∉ R1
If (a, b) ∈ R1 Then, (b, a) ∈ R1
∀ a, b ∈ Q0 If (a, b) ∈ R1
We have, _{} …(2) Now, for (b, a) ∈ R1
Replace a by b & b by a in equation (2), we get
⇒ (b, a) ∈ R2
So, if (a, b) ∈ R1, then (b, a) ∈ R1
∀ a, b ∈ Q0
If (a, b) ∈ R1 and (b, c) ∈ R1
We need to eliminate b. We have
Putting _{} in _{}, we get
But, _{}
⇒ (a, c) ∉ R1
So, if (a, b) ∈ R1 and (b, c) ∈ R1, then (a, c) ∉ R1
∀ a, b, c ∈ Q0
Test whether the following relations R1, R2 and R3 are (i) reflexive (ii) symmetric and (iii) transitive : R2 on Z defined by (a, b) ϵ R2⇔ |a – b| ≤ 5
Here, R1, R2, R3, and R4 are the binary relations.
So, recall that for any binary relation R on set A. We have, R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.
So, using these results let us start determining given relations. We have
R2 on Z defined by (a, b) ∈ R2⇔ |a – b| ≤ 5
∀ a ∈ Z,
(a, a) ∈ R2 needs to be proved for reflexivity. If (a, b) ∈ R2
Then, |a – b| ≤ 5 …(1) So, for (a, a) ∈ R1
Replace b by a in equation (1), we get
|a – a| ≤ 5
⇒ 0 ≤ 5
⇒ (a, a) ∈ R2
So, ∀ a ∈ Z, then (a, a) ∈ R2
∀ a, b ∈ Z
If (a, b) ∈ R2
We have, |a – b| ≤ 5 …(2)
Replace a by b & b by a in equation (2), we get
|b – a| ≤ 5
Since, the value is in mod, |b – a| = |a – b|
⇒ The statement |b – a| ≤ 5 is true.
⇒ (b, a) ∈ R2
So, if (a, b) ∈ R2, then (b, a) ∈ R2
∀ a, b ∈ Q0
∀ a, b, c ∈ Z
If (a, b) ∈ R2 and (b, c) ∈ R2
⇒ |a – b| ≤ 5 and |b – c| ≤ 5
Since, inequalities cannot be added or subtract. We need to take example to check for,
|a – c| ≤ 5
Take values a = 18, b = 14 and c = 10 Check: |a – b| ≤ 5
⇒ |18 – 14| ≤ 5
⇒ |4| ≤ 5 is true. Check: |b – c| ≤ 5
⇒ |14 – 10| ≤ 5
⇒ |4| ≤ 5
Check: |a – c| ≤ 5
⇒ |18 – 10| ≤ 5
⇒ |8| ≤ 5 is not true.
⇒ (a, c) ∉ R2
So, if (a, b) ∈ R2 and (b, c) ∈ R2, then (a, c) ∉ R1
∀ a, b, c ∈ Z
Test whether the following relations R1, R2 and R3 are (i) reflexive (ii) symmetric and (iii) transitive : R3 on R defined by (a, b) ϵ R3⇔ a^{2 }– 4 ab + 3b^{2 }= 0.
Here, R1, R2, R3, and R4 are the binary relations.
So, recall that for any binary relation R on set A. We have, R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.
So, using these results let us start determining given relations. We have
R3 on R defined by (a, b) ∈ R3⇔ a^{2 }– 4ab + 3b^{2 }= 0
∀ a ∈ R,
(a, a) ∈ R3 needs to be proved for reflexivity.
If (a, b) ∈ R3, then we have a^{2 }– 4ab + 3b^{2 }= 0 Replace b by a, we get
a^{2 }– 4aa + 3a^{2 }= 0
⇒ a^{2 }– 4a^{2 }+ 3a^{2 }= 0
⇒ –3a^{2 }+ 3a^{2 }= 0
⇒ 0 = 0, which is true.
⇒ (a, a) ∈ R3
So, ∀ a ∈ R, (a, a) ∈ R3
∀ a, b ∈ R
If (a, b) ∈ R3, then we have a^{2 }– 4ab + 3b^{2 }= 0
⇒ a^{2 }– 3ab – ab + 3b^{2 }= 0
⇒ a (a – 3b) – b (a – 3b) = 0
⇒ (a – b) (a – 3b) = 0
⇒ (a – b) = 0 or (a – 3b) = 0
⇒ a = b or a = 3b …(1)
Replace a by b and b by a in equation (1), we get
⇒ b = a or b = 3a …(2)
Results (1) and (2) does not match.
⇒ (b, a) ∉ R3
∀ a, b, c ∈ R
If (a, b) ∈ R3 and (b, c) ∈ R3
⇒ a^{2 }– 4ab + 3b^{2 }= 0 and b^{2 }– 4bc + 3c^{2 }= 0
⇒ a^{2 }– 3ab – ab + 3b^{2 }= 0 and b^{2 }– 3bc – bc + 3c^{2}
⇒ a (a – 3b) – b (a – 3b) = 0 and b (b – 3c) – c (b – 3c) = 0
⇒ (a – b) (a – 3b) = 0 and (b – c) (b – 3c) = 0
⇒ (a – b) = 0 or (a – 3b) = 0 And (b – c) = 0 or (b – 3c) = 0
⇒ a = b or a = 3b And b = c or b = 3c
What we need to prove here is that, a = c or a = 3c Take a = b and b = c
Clearly implies that a = c.
[∵ if a = b, just substitute a in place of b in b = c. We get, a = c] Now, take a = 3b and b = 3c
If a = 3b
Substitute _{} in b = 3c. We get
⇒ a = 9c, which is not the desired result.
⇒ (a, c) ∉ R3
4. Question
Let A = {1, 2, 3}, and let R1 = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3), (2, 2), (2, 1), (3, 3)}, R2={(2,2),
(3,1), (1, 3)}, R3 = {(1, 3),(3, 3)}. Find whether or not each of the relations R1, R2, R3 on A is (i) reflexive (ii) symmetric (iii) transitive.
We have been given, A = {1, 2, 3}
Here, R1, R2, and R3 are the binary relations on A.
So, recall that for any binary relation R on set A. We have, R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.
So, using these results let us start determining given relations. Let us take R1.
R1 = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}
∀ 1, 2, 3 ∈ A [∵ A = {1, 2, 3}]
(1, 1) ∈ R1
(2, 2) ∈ R2
(3, 3) ∈ R3
So, for a ∈ A, (a, a) ∈ R1
∀ 1, 2, 3 ∈ A
If (1, 3) ∈ R1, then (3, 1) ∈ R1
[∵ R1 = {(1, 1), (1, 3), (3, 1) , (2, 2), (2, 1), (3, 3)}]
But if (2, 1) ∈ R1, then (1, 2) ∉ R1
[∵ R1 = {(1, 1), (1, 3), (3, 1), (2, 2),(2, 1), (3, 3)}]
So, if (a, b) ∈ R1, then (b, a) ∉ R1
∀ a, b ∈ A
∀ 1, 2, 3 ∈ A
If (1, 3) ∈ R1 and (3, 3) ∈ R1
Then, (1, 3) ∈ R1
[∵ R1 = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}]
But, if (2, 1) ∈ R1 and (1, 3) ∈ R1
Then, (2, 3) ∉ R1
So, if (a, b) ∈ R1 and (b, c) ∈ R1, then (a, c) ∉ R1
∀ a, b, c ∈ A
Now, take R2.
R2 = {(2, 2), (3, 1), (1, 3)}
∀ 1, 2, 3 ∈ A [∵ A = {1, 2, 3}]
(1, 1) ∉ R2
(2, 2) ∈ R2
(3, 3) ∉ R2
So, for a ∈ A, (a, a) ∉ R2
∀ 1, 2, 3 ∈ A
If (1, 3) ∈ R2, then (3, 1) ∈ R2
[∵ R2 = {(2, 2), (3, 1), (1, 3) }]
If (2, 2) ∈ R2, then (2, 2) ∈ R2
[∵ R2 = {(2, 2), (3, 1), (1, 3)}]
So, if (a, b) ∈ R2, then (b, a) ∈ R2
∀ a, b ∈ A
∀ 1, 2, 3 ∈ A
If (1, 3) ∈ R2 and (3, 1) ∈ R2
Then, (1, 1) ∉ R2
[∵ R2 = {(2, 2), (3, 1), (1, 3) }]
So, if (a, b) ∈ R2 and (b, c) ∈ R2, then (a, c) ∉ R2
∀ a, b, c ∈ A
Now take R3.
R3 = {(1, 3), (3, 3)}
∀ 1, 3 ∈ A [∵ A = {1, 2, 3}]
(1, 1) ∉ R3
(3, 3) ∈ R3
So, for a ∈ A, (a, a) ∉ R3
∀ 1, 3 ∈ A
If (1, 3) ∈ R3, then (3, 1) ∉ R3
[∵ R3 = {(1, 3), (3, 3)}]
So, if (a, b) ∈ R3, then (b, a) ∉ R3
∀ a, b ∈ A
∀ 1, 3 ∈ A
If (1, 3) ∈ R3 and (3, 3) ∈ R3
Then, (1, 3) ∈ R3
[∵ R3 = {(1, 3), (3, 3) }]
So, if (a, b) ∈ R3 and (b, c) ∈ R3, then (a, c) ∈ R3
∀ a, b, c ∈ A
The following relations are defined on the set of real numbers : aRb if a – b > 0
Find whether these relations are reflexive, symmetric or transitive.
Let set of real numbers be ℝ.
So, recall that for any binary relation R on set A. We have, R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.
We have
aRb if a – b > 0
For a ∈ ℝ If aRa,
⇒ a – a > 0
⇒ 0 > 0
But 0 > 0 is not possible. Hence, aRa is not true.
So, ∀ a ∈ ℝ, then aRa is not true.
⇒ R is not reflexive.
∀ a, b ∈ ℝ If aRb,
⇒ a – b > 0
Replace a by b and b by a, we get
⇒ b – a > 0
[Take a = 12 and b = 6. a – b > 0
⇒ 12 – 6 > 0
⇒ 6 > 0, which is a true statement. Now, b – a > 0
⇒ 6 – 12 > 0
⇒ –6 > 0, which is not a true statement as –6 is not greater than 0.]
⇒ bRa is not true.
So, if aRb is true, then bRa is not true.
∀ a, b ∈ ℝ
⇒ R is not symmetric.
∀ a, b, c ∈ ℝ
If aRb and bRc.
⇒ a – b > 0 and b – c > 0
⇒ a – c > 0 or not. Let us check.
a – b > 0 means a > b. b – c > 0 means b > c. a – c > 0 means a > c. If a > b and b > c,
⇒ a > b, b > c
⇒ a > b > c
⇒ a > c
Hence, aRc is true.
So, if aRb is true and bRc is true, then aRc is true.
∀ a, b, c ∈ ℝ
⇒ R is transitive.
The following relations are defined on the set of real numbers :
aRb if 1 + ab > 0
Find whether these relations are reflexive, symmetric or transitive.
Let set of real numbers be ℝ.
So, recall that for any binary relation R on set A. We have, R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.
We have
aRb if 1 + ab > 0
For a ∈ ℝ If aRa,
⇒ 1 + aa > 0
⇒ 1 + a^{2 }> 0
If a is a real number.
[Positive or negative, large or small, whole numbers or decimal numbers are all Real Numbers. Real numbers are so called because they are ‘Real’ not ‘imaginary’.]
This means, even if ‘a’ was negative. a^{2 }= positive.
a^{2 }+ 1 = positive
And any positive number is greater than 0. Hence, 1 + a^{2 }> 0
⇒ aRa is true.
So, ∀ a ∈ ℝ, then aRa is true.
⇒ R is reflexive.
∀ a, b ∈ ℝ If aRb,
⇒ 1 + ab > 0
Replace a by b and b by a, we get
⇒ 1 + ba > 0
Whether we write ab or ba, it is equal. ab = ba
So, 1 + ba > 0
⇒ bRa is true.
So, if aRb is true, then bRa is true.
∀ a, b ∈ ℝ
⇒ R is symmetric.
∀ a, b, c ∈ ℝ
If aRb and bRc.
⇒ 1 + ab > 0 and 1 + bc > 0
⇒ 1 + ac > 0 or not. Let us check.
1 + ab > 0 means ab > –1.
1 + bc > 0 means bc > –1.
1 + ac > 0 means ac > –1. If ab > –1 and bc > –1.
⇒ ac > –1 should be true. Take a = –1, b = 0.9 and c = 1 ab > –1
⇒ (–1)(0.9) > –1
⇒ –0.9 > –1, is true on the number line. bc > –1
⇒ (0.9)(1) > –1
⇒ 0.9 > –1, is true on the number line. ac > –1
⇒ (–1)(1) > –1
⇒ –1 > –1, is not true as –1 cannot be greater than itself.
⇒ ac > –1 is not true.
⇒ 1 + ac > 0 is not true.
⇒ aRc is not true.
So, if aRb is true and bRc is true, then aRc is not true.
∀ a, b, c ∈ ℝ
⇒ R is not transitive.
The following relations are defined on the set of real numbers :
aRb if | a | ≤ b.
Find whether these relations are reflexive, symmetric or transitive.
Let set of real numbers be ℝ.
So, recall that for any binary relation R on set A. We have, R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.
We have aRb if |a| ≤ b
For a ∈ ℝ If aRa,
⇒ |a| ≤ a, is true
If a is a real number.
[Positive or negative, large or small, whole numbers or decimal numbers are all Real Numbers. Real numbers are so called because they are ‘Real’ not ‘imaginary’.]
This means, even if ‘a’ was negative.
|a| = positive. & |a| ≤ a Hence, |a| ≤ a.
⇒ aRa is true.
So, ∀ a ∈ ℝ, then aRa is true.
⇒ R is reflexive.
∀ a, b ∈ ℝ If aRb,
⇒ |a| ≤ b
Replace a by b and b by a, we get
⇒ |b| ≤ a, which might be true or not. Let a = 2 and b = 3.
|a| ≤ b
⇒ |2| ≤ 3, is true
|b| ≤ a
⇒ |3| ≤ 2, is not true
⇒ bRa is not true.
So, if aRb is true, then bRa is not true.
∀ a, b ∈ ℝ
⇒ R is not symmetric.
∀ a, b, c ∈ ℝ
If aRb and bRc.
⇒ |a| ≤ b and |b| ≤ c
⇒ |a| ≤ c or not. Let us check.
If |a| ≤ b and |b| ≤ c b ≠ |b|
Say, if b = –2
⇒ –2 ≠ |–2|
⇒ –2 ≠ 2
But, from |a| ≤ b
b ≥ 0 in every case otherwise the statement would not hold true.
⇒ b can only accept positive values including 0.
⇒ b is a whole number.
∴ if |a| ≤ b and |b| ≤ c
⇒ |a| ≤ b, b ≤ c
⇒ |a| ≤ b ≤ c
⇒ |a| ≤ c
⇒ aRc is true.
So, if aRb is true and bRc is true, then aRc is true.
∀ a, b, c ∈ ℝ
⇒ R is transitive.
Check whether the relation R defined on the set A={1,2,3,4,5,6} as R= {(a, b) : b = a + 1} is reflexive, symmetric or transitive.
We have the set A = {1, 2, 3, 4, 5, 6}
So, recall that for any binary relation R on set A. We have, R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz. We have
R = {(a, b): b = a + 1}
∵ Every a, b ∈ A.
And A = {1, 2, 3, 4, 5, 6}
The relation R on set A can be defined as: Put a = 1
⇒ b = a + 1
⇒ b = 1 + 1
⇒ b = 2
⇒ (a, b) ≡ (1, 2) Put a = 2
⇒ b = 2 + 1
⇒ b = 3
⇒ (a, b) ≡ (2, 3) Put a = 3
⇒ b = 3 + 1
⇒ b = 4
⇒ (a, b) ≡ (3, 4) Put a = 4
⇒ b = 4 + 1
⇒ b = 5
⇒ (a, b) ≡ (4, 5) Put a = 5
⇒ b = 5 + 1
⇒ b = 6
⇒ (a, b) ≡ (5, 6) Put a = 6
⇒ b = 6 + 1
⇒ b = 7
⇒ (a, b) ≠ (6, 7) [∵ 7 ∉ A]
Hence, R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
For 1, 2, …, 6 ∈ A [∵ A = {1, 2, 3, 4, 5, 6}]
(1, 1) ∉ R
(2, 2) ∉ R
…
(6, 6) ∉ R
So, ∀ a ∈ A, then (a, a) ∉ R.
⇒ R is not reflexive.
∀ 1, 2 ∈ A [∵ A = {1, 2, 3, 4, 5, 6}]
If (1, 2) ∈ R
Then, (2, 1) ∉ R
[∵ R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}]
So, if (a, b) ∈ R, then (b, a) ∉ R
∀ a, b ∈ A
⇒ R is not symmetric.
∀ 1, 2, 3 ∈ A
If (1, 2) ∈ R and (2, 3) ∈ R
Then, (1, 3) ∉ R
[∵ R = {(1, 2), (2, 3) , (3, 4), (4, 5), (5, 6)}]
So, if (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∉ R.
∀ a, b, c ∈ A
⇒ R is not transitive.
Check whether the relation R on R defined by R = {(a, b) : a ≤ b^{3}} is reflexive, symmetric or transitive.
We have the set of real numbers, R.
So, recall that for any binary relation R on set A. We have, R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz. We have
R = {(a, b): a ≤ b^{3}}
For a ∈ R
If (a, a) ∈ R,
⇒ a ≤ a^{3}, which is not true. Say, if a = – 2.
a ≤ a^{3}
⇒ – 2 ≤ – 8
⇒ –2 ≤ –8, which is not true as – 2 > – 8. Hence, (a, a) ∉ R
So, ∀ a ∈ R, then (a, a) ∉ R.
⇒ R is not reflexive.
∀ a, b ∈ R
If (a, b) ∈ R
⇒ a ≤ b^{3}
Replace a by b and b by a, we get
⇒ b ≤ a^{3}
[Take a = –2 and b = 3. a ≤ b^{3}
⇒ –2 ≤ 3^{3}
⇒ –2 ≤ 27, which is a true statement. Now, b ≤ a^{3}
⇒ 3 ≤ (–2)^{3}
⇒ 3 ≤ –8, which is not a true statement as 3 > –8]
⇒ (b, a) ∉ R
So, if (a, b) ∈ R, then (b, a) ∉ R
∀ a, b ∈ R
⇒ R is not symmetric.
∀ a, b, c ∈ R
If (a, b) ∈ R and (b, c) ∈ R
⇒ a ≤ b^{3 }and b ≤ c^{3}
⇒ a ≤ c^{3 }or not. Let us check.
Take a = 3, _{} and _{}. a ≤ b^{3}
⇒ 3 ≤ 3.37, which is true. b ≤ c^{3}
⇒ 1.5 ≤ 1.728
a ≤ c^{3}
⇒ 3 ≤ 1.728, which is not true as 3 > 1.728. Hence, (a, c) ∉ R.
So, if (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∉ R.
∀ a, b, c ∈ ℝ
⇒ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
Prove that every identity relation on a set is reflexive, but the converse is not necessarily true.
To Prove: Every identity relation on a set is reflexive, but every reflexive relation is not identity relation. Proof:
Let us first understand what ‘Reflexive Relation’ is and what ‘Identity Relation’ is.
Reflexive Relation: A binary relation R over a set A is reflexive if every element of X is related to itself. Formally, this may be written as ∀ x ∈ A: xRx.
Identity Relation: Let A be any set.
Then the relation R= {(x, x): x ∈ A} on A is called the identity relation on A. Thus, in an identity relation, every element is related to itself only.
Let A = {a, b, c} be a set.
Let R be a binary relation defined on A.
Let RA = {(a, a): a ∈ A} is the identity relation on A.
Hence, every identity relation on set A is reflexive by definition. Converse: Let A = {a, b, c} is the set.
Let R = {(a, a), (b, b), (c, c), (a, b), (c, a)} be a relation defined on A. R is reflexive as per definition.
[∵ (a, a) ∈ R, (b, b) ∈ R & (c, c) ∈ R]
But, (a, b) ∈ R
(c, a) ∈ R
⇒ R is not identity relation by definition.
9 A. Question
If A = {1, 2, 3, 4}, define relations on A which have properties of being reflexive, transitive but not symmetric.
Recall that for any binary relation R on set A. We have, R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz. Using these properties, we can define R on A.
A = {1, 2, 3, 4}
We need to define a relation (say, R) which is reflexive, transitive but not symmetric. Let us try to form a small relation step by step.
The relation must be defined on A. Reflexive relation:
R = {(1, 1), (2, 2), (3, 3), (4, 4)} …(1)
Transitive relation:
R = {(1, 2), (2, 1), (1, 1)}, is transitive but also symmetric. So, let us define another relation.
R = {(1, 3), (3, 2), (1, 2)}, is transitive and not symmetric. …(2) Let us combine (i) and (ii) relation.
R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 3), (3, 2), (1, 2)} …(A)
(A) can be shortened by eliminating (3, 2) and (1, 2) from R. R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 3)} …(B)
Further (B) can be shortened by eliminating (2, 2) and (4, 4).
R = {(1, 1), (3, 3), (1, 3)} …(C)
All the results (A), (B) and (C) is correct.
Thus, we have got the relation which is reflexive, transitive but not symmetric.
If A = {1, 2, 3, 4}, define relations on A which have properties of being symmetric but neither reflexive nor transitive.
Recall that for any binary relation R on set A. We have, R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz. Using these properties, we can define R on A.
A = {1, 2, 3, 4}
We need to define a relation (say, R) which is symmetric but neither reflexive nor transitive. The relation R must be defined on A.
Symmetric relation:
R = {(1, 2), (2, 1)}
Note that, the relation R here is neither reflexive nor transitive, and it is the shortest relation that can be form.
Similarly, we can also write:
R = {(1, 3), (3, 1)}
Or R = {(3, 4), (4, 3)}
Or R = {(2, 3), (3, 2), (1, 4), (4, 1)}
And so on…
All of these are right answers.
Thus, we have got the relation which is symmetric but neither reflexive nor transitive.
If A = {1, 2, 3, 4}, define relations on A which have properties of being reflexive, symmetric and transitive.
Recall that for any binary relation R on set A. We have, R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz. Using these properties, we can define R on A.
A = {1, 2, 3, 4}
We need to define a relation (say, R) which is reflexive, symmetric and transitive.
The relation must be defined on A. Reflexive Relation:
R = {(1, 1), (2, 2), (3, 3), (4, 4)}
Or simply shorten it and write, R = {(1, 1), (2, 2)} …(1)
Symmetric Relation:
R = {(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3)}
Or simply shorten it and write, R = {(1, 2), (2, 1)} …(2)
Combine results (1) and (2), we get R = {(1, 1), (2, 2), (1, 2), (2, 1)}
It is reflexive, symmetric as well as transitive as per definition. Similarly, we can find other combinations too.
Thus, we have got the relation which is reflexive, symmetric as well as transitive.
Let R be a relation defined on the set of natural numbers N as R = {(x, y) : x, y ∈ N, 2x + y = 41}
Find the domain and range of R. Also, verify whether R is
First let us define what range and domain are.
Range: The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually) after we have substituted the domain. In plain English, the definition means: The range is the resulting y-values we get after substituting all the possible x-values.
Domain: The domain of definition of a function is the set of "input" or argument values for which the function is defined. That is, the function provides an "output" or value for each member of the domain.
We have been given that, R is a relation defined on N. N = set of natural numbers
R = {(x, y): x, y ∈ N, 2x + y = 41} We have the function as
2x + y = 41
⇒ y = 41 – 2x
As y ∈ N (Natural number)
⇒ 41 – 2x ≥ 1
⇒ –2x ≥ 1 – 41
⇒ –2x ≥ –40
⇒ 2x ≤ 40
⇒ x ≤ 20
As, 2x + y = 41
⇒ 2x = 41 – y
As x ∈ N (Natural number)
⇒ 41 – y ≥ 2
⇒ –y ≥ 2 – 41
⇒ –y ≥ –39
⇒ y ≤ 39
We have relation R defined on set N. R = {(x, y): x, y ∈ N, 2x + y = 41} Check for Reflexivity:
∀ x ∈ N
If (x, x) ∈ R
⇒ 2x + x = 41
⇒ 3x = 41
⇒ x = 13.67
But, x ≠ 13.67 as x ∈ N.
⇒ (x, x) ∉ R
So, ∀ x ∈ N, then (x, x) ∉ R
⇒ R is not reflexive.
∀ x, y ∈ N If (x, y) ∈ R
⇒ 2x + y = 41 …(i)
Now, replace x by y and y by x, we get
⇒ 2y + x = 41 …(ii) Take x = 20 and y = 1.
Equation (i) ⇒ 2(20) + 1 = 41
⇒ 40 + 1 = 41
⇒ 41 = 41, holds true.
Equation (ii) ⇒ 2(1) + 20 = 41
⇒ 2 + 20 = 41
⇒ 22 = 41, which is not true as 22 ≠ 41.
⇒ (y, x) ∉ R
So, if (x, y) ∈ R, then (y, x) ∉ R.
∀ x, y ∈ N
⇒ R is not symmetric.
∀ x, y, z ∈ N
If (x, y) ∈ R and (y, z) ∈ R
⇒ 2x – y = 41 and 2y – z = 41
⇒ 2x – z = 41, may be true or not. Let us sole these to find out.
We have
2x – y = 41 …(iii) 2y – z = 41 …(iv)
Multiply 2 by equation (i), we get 4x – 2y = 82 …(v)
Adding equation (v) and (iv), we get (4x – 2y) + (2y – z) = 82 + 41
⇒ 4x – z = 123
⇒ 2x + 2x – z = 123
⇒ 2x – z = 123 – 2x Take x = 40 (as x ∈ N)
⇒ 2x – z = 123 – 2(40)
⇒ 2x – z = 123 – 80
⇒ 2x – z = 43 ≠ 41
⇒ (x, z) ∉ R
So, if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∉ R.
∀ x, y, z ∈ N
⇒ R is not transitive.
Is it true that every relation which is symmetric and transitive is also reflexive? Give reasons.
It is not true that every relation which is symmetric and transitive is also reflexive.
Take for example:
Take a set A = {1, 2, 3, 4} And define a relation R on A. Symmetric relation:
R = {(1, 2), (2, 1)}, is symmetric on set A. Transitive relation:
R = {(1, 2), (2, 1), (1, 1)}, is the simplest transitive relation on set A.
⇒ R = {(1, 2), (2, 1), (1, 1)} is symmetric as well as transitive relation. But R is not reflexive here.
If only (2, 2) ∈ R, had it been reflexive.
Thus, it is not true that every relation which is symmetric and transitive is also reflexive.
An integer m is said to be related to another integer n if m is a multiple of n. Check if the relation is symmetric, reflexive and transitive.
According to the question,
m is related to n if m is a multiple of n.
∀ m, n ∈ I (I being set of integers) The relation comes out to be:
R = {(m, n): m = kn, k ∈ ℤ}
Recall that for any binary relation R on set A. We have, R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.
∀ m ∈ I
If (m, m) ∈ R
⇒ m = k m, holds.
As an integer is always a multiple of itself, So, ∀ m ∈ I, then (m, m) ∈ R.
⇒ R is reflexive.
∀ m, n ∈ I
If (m, n) ∈ R
⇒ m = k n, holds.
Now, replace m by n and n by m, we get n = k m, which may or not be true.
Let us check:
If 12 is a multiple of 3, but 3 is not a multiple of 12.
⇒ n = km does not hold.
So, if (m, n) ∈ R, then (n, m) ∉ R.
∀ m, n ∈ I
⇒ R is not symmetric.
∀ m, n, o ∈ I
If (m, n) ∈ R and (n, o) ∈ R
⇒ m = kn and n = ko Where k ∈ ℤ
Substitute n = ko in m = kn, we get m = k(ko)
⇒ m = k^{2}o
If k ∈ ℤ, then k^{2}∈ ℤ. Let k^{2 }= r
⇒ m = ro, holds true.
⇒ (m, o) ∈ R
So, if (m, n) ∈ R and (n, o) ∈ R, then (m, o) ∈ R.
∀ m, n ∈ I
⇒ R is transitive.
Show that the relation “≥” on the set R of all real numbers is reflexive and transitive but not symmetric.
We have
The relation “≥” on the set R of all real numbers. Recall that for any binary relation R on set A. We have, R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz. So, let the relation having “≥” be P.
We can write
P = {(a, b): a ≥ b, a, b ∈ R}
∀ a ∈ R
If (a, a) ∈ P
⇒ a ≥ a, which is true.
Since, every real number is equal to itself. So, ∀ a ∈ R, then (a, a) ∈ P.
⇒ P is reflexive.
∀ a, b ∈ R If (a, b) ∈ P
⇒ a ≥ b
Now, replace a by b and b by a. We get b ≥ a, might or might not be true.
Let us check:
Take a = 7 and b = 5. a ≥ b
⇒ 7 ≥ 5, holds. b ≥ a
⇒ 5 ≥ 7, is not true as 5 < 7.
⇒ b ≥ a, is not true.
⇒ (b, a) ∉ P
So, if (a, b) ∈ P, then (b, a) ∉ P
∀ a, b ∈ R
⇒ P is not symmetric.
∀ a, b, c ∈ R
If (a, b) ∈ P and (b, c) ∈ P
⇒ a ≥ b and b ≥ c
⇒ a ≥ b ≥ c
⇒ a ≥ c
⇒ (a, c) ∈ P
So, if (a, b) ∈ P and (b, c) ∈ P, then (a, c) ∈ P
∀ a, b, c ∈ R
⇒ P is transitive.
Thus, shown that the relation “≥” on the set R of all the real numbers are reflexive and transitive but not symmetric.
Give an example of a relation which is
reflexive and symmetric but not transitive.
Recall that for any binary relation R on set A. We have, R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz. Let there be a set A.
A = {1, 2, 3, 4}
We need to define a relation on A which is reflexive and symmetric but not transitive. Let there be a set A.
A = {1, 2, 3, 4}
Reflexive relation:
R = {(1, 1), (2, 2), (3, 3), (4, 4)} …(1)
Symmetric relation:
R = {(3, 4), (4, 3)} …(2)
Combine results (1) and (2), we get
Check for Transitivity:
If (3, 4) ∈ R and (4, 3) ∈ R
Then, (3, 3) ∈ R
∀ 3, 4 ∈ A [∵ A = {1, 2, 3, 4}]
So eliminate (3, 3) from R, we get
Check for Transitivity:
If (4, 3) ∈ R and (3, 4) ∈ R
Then, (4, 4) ∈ R
∀ 3, 4 ∈ A
So, eliminate (4, 4) from R, we get
Thus, the relation which is reflexive and symmetric but not transitive is:
14 B. Question
Give an example of a relation which is reflexive and transitive but not symmetric. Answer
Recall that for any binary relation R on set A. We have, R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz. Let there be a set A.
A = {1, 2, 3, 4}
We need to define a relation on A which is reflexive and transitive but not symmetric. Let there be a set A.
A = {1, 2, 3, 4}
Reflexive relation:
R = {(1, 1), (2, 2), (3, 3), (4, 4)} …(1)
Transitive relation:
R = {(3, 4), (4, 1), (3, 1)} …(2)
Combine results (1) and (2), we get
Check for Symmetry:
If (3, 4) ∈ R
Then, (4, 3) ∉ R
∀ 3, 4 ∈ A [∵ A = {1, 2, 3, 4}]
One example is enough to prove that R is not symmetric.
Thus, the relation which is reflexive and transitive but not symmetric is:
14 C. Question
Give an example of a relation which is symmetric and transitive but not reflexive. Answer
Recall that for any binary relation R on set A. We have, R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz. Let there be a set A.
A = {1, 2, 3, 4}
We need to define a relation on A which is symmetric and transitive but not reflexive.
It is not possible to define such relation which is symmetric and transitive but not reflexive. As every relation which is symmetric and transitive will use identity ordered pair of the form (x, x) to balance the relation (to make the relation symmetric and transitive). Without such identity pair both, symmetry and transitivity will not be possible.
Give an example of a relation which is symmetric but neither reflexive nor transitive. Answer
Recall that for any binary relation R on set A. We have, R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz. Let there be a set A.
A = {1, 2, 3, 4}
We need to define a relation which is symmetric but neither reflexive nor transitive. Let there be a set A.
A = {1, 2, 3, 4}
Symmetric Relation:
{(1, 3), (3, 1)}
This is neither reflexive nor transitive.
∵ (1, 1) ∉ R
(3, 3) ∉ R
Hence, R is not reflexive.
∵ (1, 3) ∈ R and (3, 1) ∈ R
Then, (1, 1) ∉ R
Hence, R is not transitive.
Thus, the relation which is symmetric but neither nor transitive is: R = {(1, 3), (3, 1)}
Give an example of a relation which is transitive but neither reflexive nor symmetric. Answer
Recall that for any binary relation R on set A. We have, R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz. Let there be a set A.
A = {1, 2, 3, 4}
We need to define a relation which is transitive but neither reflexive nor symmetric. Let there be a set A.
A = {1, 2, 3}
Transitive Relation:
R = {(2, 4), (4, 1), (2, 1)}
This is neither reflexive nor symmetric.
∵ (1, 1) ∉ R
(2, 2) ∉ R
(4, 4) ∉ R
Hence, R is not reflexive.
∵ if (2, 4) ∈ R
Then, (4, 2) ∉ R
Hence, R is not symmetric.
Thus, the relation which is transitive but neither reflexive nor symmetric is:
R = {(2, 4), (4, 1), (2, 1)}
Given the relation R = {(1, 2), (2, 3)} on the set A = {1, 2, 3}, add a minimum number ordered pairs so that the enlarged relation is symmetric, transitive and reflexive.
Given is:
R = {(1, 2), (2, 3)} on the set A.
A = {1, 2, 3}
Right now, we have R = {(1, 2), (2, 3)}
Symmetric Relation:
We know (1, 2) ∈ R
Then, (2, 1) ∈ R
Also, (2, 3) ∈ R
Then, (3, 2) ∈ R
So, add (2, 1) and (3, 2) in R, so that we get
R’ = {(1, 2), (2, 1), (2, 3), (3, 2)}
Transitive Relation:
We need to make the relation R’ transitive. So, we know (1, 2) ∈ R and (2, 1) ∈ R
Then, (1, 1) ∈ R
Also, (2, 3) ∈ R and (3, 2)
Then, (2, 2) ∈ R
Also, (2, 1) ∈ R and (1, 2) ∈ R
Then, (2, 2) ∈ R
Also, (3, 2) ∈ R and (2, 3) ∈ R
Then, (3, 3) ∈ R
Add (1, 1), (2, 2) and (3, 3) in R’, we get
R’’ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}
Thus, we have got a relation which is reflexive, symmetric and transitive. R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}
The ordered pair added are (1, 1), (2, 2), (3, 3), (3, 2).
Let A = {1, 2, 3} and R = {(1, 2), (1, 1), (2, 3)} be a relation on A. What minimum number of ordered pairs may be added to R so that it may become a transitive relation on A.
We have the relation R such that R = {(1, 2), (1, 1), (2, 3)}
R is defined on set A. A = {1, 2, 3}
Recall that,
A relation R defined on a set A is called transitive if (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R, ∀ a, b, c ∈ A. For transitive relation:
Note in R,
(1, 2) ∈ R and (2, 3) ∈ R
Then, (1, 3) ∈ R
So, add (1, 3) in R.
R = {(1, 2), (1, 1), (2, 3), (1, 3)}
Now, we can see that R is transitive.
Hence, the ordered pair to be added is (1, 3).
Let A = {a, b, c} and the relation R be defined on A as follows R={(a,a), (b, c), (a, b)}. Then, write a minimum number of ordered pairs to be added in R to make it reflexive and transitive.
Recall that,
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz. We have relation R = {(a, a), (b, c), (a, b)} on A.
A = {a, b, c} For Transitive:
If (a, b) ∈ R and (b, c) ∈ R Then, (a, c) ∈ R
∀ a, b, c ∈ A For Reflexive:
∀ a, b, c ∈ R Then, (a, a) ∈ R
(b, b) ∈ R
(c, c) ∈ R
We need to add (b, b), (c, c) and (a, c) in R.
We get
R = {(a, a), (b, b), (c, c), (a, b), (b, c), (a, c)}
Each of the following defines a relation on N :
x > y, x, y ∈ N
Determine which of the above relations are reflexive, symmetric and transitive.
Recall that for any binary relation R on set A. We have, R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.
We have
x > y, x, y ∈ N
This relation is defined on N (set of Natural Numbers) The relation can also be defined as
R = {(x, y): x > y} on N
∀ x ∈ N
We should have, (x, x) ∈ R
⇒ x > x, which is not true. 1 can’t be greater than 1.
2 can’t be greater than 2.
16 can’t be greater than 16. Similarly, x can’t be greater than x. So, ∀ x ∈ N, then (x, x) ∉ R
⇒ R is not reflexive.
∀ x, y ∈ N If (x, y) ∈ R
⇒ x > y
Now, replace x by y and y by x. We get y > x, which may or not be true.
Let us take x = 5 and y = 2. x > y
⇒ 5 > 2, which is true. y > x
⇒ 2 > 5, which is not true.
⇒ y > x, is not true as x > y
⇒ (y, x) ∉ R
So, if (x, y) ∈ R, but (y, x) ∉ R ∀ x, y ∈ N
⇒ R is not symmetric.
∀ x, y, z ∈ N
If (x, y) ∈ R and (y, z) ∈ R
⇒ x > y and y > z
⇒ x > y > z
⇒ x > z
⇒ (x, z) ∈ R
So, if (x, y) ∈ R and (y, z) ∈ R, and then (x, z) ∈ R
∀ x, y, z ∈ N
⇒ R is transitive.
Each of the following defines a relation on N :
x + y = 10, x, y ∈ N
Determine which of the above relations are reflexive, symmetric and transitive.
Recall that for any binary relation R on set A. We have, R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.
We have
x + y = 10, x, y ∈ N
This relation is defined on N (set of Natural Numbers) The relation can also be defined as
R = {(x, y): x + y = 10} on N
∀ x ∈ N
We should have, (x, x) ∈ R
⇒ x + x = 10, which is not true everytime. Take x = 4.
x + x = 10
⇒ 4 + 4 = 10
⇒ 8 = 10, which is not true.
That is 8 ≠ 10.
So, ∀ x ∈ N, then (x, x) ∉ R
⇒ R is not reflexive.
∀ x, y ∈ N If (x, y) ∈ R
⇒ x + y = 10
Now, replace x by y and y by x. We get
y + x = 10, which is as same as x + y = 10.
⇒ y + x = 10
⇒ (y, x) ∈ R
So, if (x, y) ∈ R, and then (y, x) ∈ R ∀ x, y ∈ N
⇒ R is symmetric.
∀ x, y, z ∈ N
If (x, y) ∈ R and (y, z) ∈ R
⇒ x + y = 10 and y + z = 10
⇒ x + z = 10, may or may not be true.
Let us take x = 6, y = 4 and z = 6 x + y = 10
⇒ 6 + 4 = 10
⇒ 10 = 10, which is true. y + z = 10
⇒ 4 + 6 = 10
⇒ 10 = 10, which is true. x + z = 10
⇒ 6 + 6 = 10
⇒ 12 = 10, which is not true That is, 12 ≠ 10
⇒ x + z ≠ 10
⇒ (x, z) ∉ R
So, if (x, y) ∈ R and (y, z) ∈ R, and then (x, z) ∉ R
∀ x, y, z ∈ N
⇒ R is not transitive.
Each of the following defines a relation on N :
xy is square of an integer, x, y ∈ N
Determine which of the above relations are reflexive, symmetric and transitive.
Recall that for any binary relation R on set A. We have, R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.
We have
xy is the square of an integer. x, y ∈ N.
This relation is defined on N (set of Natural Numbers) The relation can also be defined as
R = {(x, y): xy = a^{2}, a = √(xy), a ∈ N} on N
∀ x ∈ N
We should have, (x, x) ∈ R
⇒ xx = a^{2}, where a = √(xx)
⇒ x^{2 }= a^{2}, where a = √(x^{2}) which is true every time.
Take x = 1 and y = 4 xy = a^{2}
⇒ 1 × 4 = (√(1 × 4))^{2 }[∵ a = √(xy)]
⇒ 4 = (√4)^{2}
⇒ 4 = (2)^{2}
⇒ 4 = 4
So, ∀ x ∈ N, then (x, x) ∈ R
⇒ R is reflexive.
∀ x, y ∈ N If (x, y) ∈ R
⇒ xy = a^{2}, where a = √(xy)
Now, replace x by y and y by x. We get yx = a^{2}, which is as same as xy = a^{2 }where a = √(yx)
⇒ yx = a^{2}
⇒ (y, x) ∈ R
So, if (x, y) ∈ R, and then (y, x) ∈ R ∀ x, y ∈ N
⇒ R is symmetric.
∀ x, y, z ∈ N
If (x, y) ∈ R and (y, z) ∈ R
⇒ xy = a^{2 }and yz = a^{2}
⇒ xz = a^{2}, may or may not be true. Let us take x = 8, y = 2 and z = 50 xy = a^{2}, where a = √(xy)
⇒ (8)(2) = (√(8 × 2))^{2}
⇒ 16 = (4)^{2}
⇒ 16 = 16, which is true. yz = a^{2}
⇒ (2)(50) = (√(2 × 50))^{2}
⇒ 100 = (10)^{2}
⇒ 100 = 100, which is true xz = a^{2}
⇒ (8)(50) = (√(8 × 50))^{2}
⇒ 400 = (20)^{2}
⇒ 400 = 400
We won’t be able to find a case to show a contradiction.
⇒ xz = a^{2}
⇒ (x, z) ∈ R
So, if (x, y) ∈ R and (y, z) ∈ R, and then (x, z) ∈ R
∀ x, y, z ∈ N
⇒ R is transitive.
Each of the following defines a relation on N :
x + 4y = 10, x, y ∈ N
Determine which of the above relations are reflexive, symmetric and transitive.
Recall that for any binary relation R on set A. We have, R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz. We have
x + 4y = 10, x, y ∈ N
This relation is defined on N (set of Natural Numbers) The relation can also be defined as
R = {(x, y): 4x + y = 10} on N
∀ x ∈ N
We should have, (x, x) ∈ R
⇒ 4x + x = 10, which is obviously not true everytime. Take x = 4.
4x + x = 10
⇒ 16 + 4 = 10
⇒ 20 = 10, which is not true. That is 20 ≠ 10.
So, ∀ x ∈ N, then (x, x) ∉ R
⇒ R is not reflexive.
∀ x, y ∈ N If (x, y) ∈ R
⇒ 4x + y = 10
Now, replace x by y and y by x. We get
4y + x = 10, which may or may not be true. Take x = 1 and y = 6
4x + y = 10
⇒ 4(1) + 6 = 10
⇒ 4 + 6 = 10
⇒ 10 = 10 4y + x = 10
⇒ 4(6) + 1 = 10
⇒ 24 + 1 = 10
⇒ 25 = 10, which is not true.
⇒ 4y + x ≠ 10
⇒ (y, x) ∉ R
So, if (x, y) ∈ R, and then (y, x) ∉ R ∀ x, y ∈ N
⇒ R is not symmetric.
∀ x, y, z ∈ N
If (x, y) ∈ R and (y, z) ∈ R
Then, (x, z) ∈ R We have
4x + y = 10
⇒ y = 10 – 4x Where x, y ∈ N So, put x = 1
⇒ y = 10 – 4(1)
⇒ y = 10 – 4
⇒ y = 6 Put x = 2
⇒ y = 10 – 4(2)
⇒ y = 10 – 8
⇒ y = 2
We can’t take y >2, because if we put y = 3
⇒ y = 10 – 4(3)
⇒ y = 10 – 12
⇒ y = –2
But, y ≠ –2 as y ∈ N
So, only ordered pairs possible are R = {(1, 6), (2, 2)}
This relation R can never be transitive. Because if (a, b) ∈ R, then (b, c) ∉ R
⇒ R is not reflexive.
Exercise 1.2
Show that the relation R defined by R = {(a, b): a – b is divisible by 3; a, b ∈ Z} is an equivalence relation.
We have,
R = {(a,b) : a–b is divisible by 3; a, b ∈ Z} To prove : R is an equivalence relation Proof :
To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive. Reflexivity : For Reflexivity, we need to prove that-
(a, a) ∈ R Let a ∈ Z
⇒ a – a = 0
⇒ a – a is divisible by 3 (∵ 0 is divisible by 3).
⇒ (a, a) ∈ R
⇒ R is reflexive
Symmetric : For Symmetric, we need to prove that- If (a, b) ∈ R, then (b, a) ∈ R
Let a, b ∈ Z and (a, b) ∈ R
⇒ a – b is divisible by 3
⇒ a – b = 3p(say) For some p ∈ Z
⇒ –( a – b) = –3p
⇒ b – a = 3 × (–p)
⇒ b – a ∈ R
⇒ R is symmetric
Transitive : : For Transitivity, we need to prove that- If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R
Let a, b, c ∈ Z and such that (a, b) ∈ R and (b, c) ∈ R
⇒ a – b = 3p(say) and b – c = 3q(say) For some p, q ∈ Z
⇒ a – c = 3 (p + q)
⇒ a – c = 3 (p + q)
⇒ (a, c) ∈ R
⇒ R is transitive
Since, R is reflexive, symmetric and transitive
⇒ R is an equivalence relation.
Show that the relation R on the set Z of integers, given by R = {(a, b) : 2 divides a – b}, is an equivalence relation.
We have,
R = {(a, b) : a – b is divisible by 2; a, b ∈ Z} To prove : R is an equivalence relation Proof :
To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive. Reflexivity : For Reflexivity, we need to prove that-
(a, a) ∈ R Let a ∈ Z
⇒ a – a = 0
⇒ a – a is divisible by 2
⇒ (a, a) ∈ R
⇒ R is reflexive
Symmetric : For Symmetric, we need to prove that- If (a, b) ∈ R, then (b, a) ∈ R
Let a, b ∈ Z and (a, b) ∈ R
⇒ a – b is divisible by 2
⇒ a – b = 2p For some p ∈ Z
⇒ b – a = 2 × (–p)
⇒ b – a ∈ R
⇒ R is symmetric
Transitive : : For Transitivity, we need to prove that- If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R
Let a, b, c ∈ Z and such that (a, b) ∈ R and (b, c) ∈ R
⇒ a – b = 2p(say) and b – c = 2q(say) , For some p, q ∈ Z
⇒ a – c = 2 (p + q)
⇒ a – c is divisible by 2
⇒ (a, c) ∈ R
⇒ R is transitive
Now, since R is symmetric, reflexive as well as transitive-
⇒ R is an equivalence relation.
Prove that the relation R on Z defined by (a, b) ∈ R ⇔ a – b is divisible by 5
is an equivalence relation on Z.
We have,
R = {(a, b) : (a – b) is divisible by 5} on Z.
We want to prove that R is an equivalence relation on Z. Proof :
To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive. Reflexivity : For Reflexivity, we need to prove that-
(a, a) ∈ R Let a ∈ Z
⇒ a – a = 0
⇒ a – a is divisible by 5.
∴ (a, a) ∈ R so R is reflexive
Symmetric : For Symmetric, we need to prove that-
If (a, b) ∈ R, then (b, a) ∈ R Let (a, b) ∈ R
⇒ a – b = 5p(say) For some p ∈ Z
⇒ b – a = 5 × (–p)
⇒ b – a is divisible by 5
⇒ (b, a) ∈ R, so R is symmetric
Transitive : : For Transitivity, we need to prove that- If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R
Let (a, b) ∈ R and (b, c) ∈ R
⇒ a – b = 5p(say) and b – c = 5q(say), For some p, q ∈ Z
⇒ a – c = 5 (p + q)
⇒ a – c is divisible by 5.
⇒ R is transitive
∴ R being reflexive, symmetric and transitive on Z.
⇒ R is equivalence relation on Z.
Let n be a fixed positive integer. Define a relation R on Z as follows :
(a, b) ∈ R ⇔ a – b is divisible by n.
Show that R is an equivalence relation on Z.
R = {(a, b) : a – b is divisible by n} on Z.
To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive. Reflexivity : For Reflexivity, we need to prove that-
(a, a) ∈ R Let a ∈ Z
⇒ a – a = 0 × n
⇒ a – a is divisible by n
⇒ (a, a) ∈ R
⇒ R is reflexive
Symmetric : For Symmetric, we need to prove that- If (a, b) ∈ R, then (b, a) ∈ R
Let (a, b) ∈ R
⇒ a – b = np For some p ∈ Z
⇒ b – a = n(–p)
⇒ b – a is divisible by n
⇒ (b, a) ∈ R
⇒ R is symmetric
Transitive : : For Transitivity, we need to prove that- If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R
Let (a, b) ∈ R and (b, c) ∈ R
⇒ a – b = np and b – c = nq For some p, q ∈ Z
⇒ a – c = n (p + q)
⇒ a – c = is divisible by n
⇒ (a, c) ∈ R
⇒ R is transitive
∴ R being reflexive, symmetric and transitive on Z.
⇒ R is an equivalence relation on Z
Let Z be the set of integers. Show that the relation R = {(a, b) : a, b∈ Z and a + b is even} is an equivalence relation on Z.
We have,
Z = set of integers and
R = {(a, b) : a, b ∈ Z and a + b is even} be a relation on Z. To prove: R is an equivalence relation on Z.
Proof :
To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive. Reflexivity : For Reflexivity, we need to prove that-
(a, a) ∈ R Let a ∈ Z
⇒ a + a is even
⇒ (a, a) ∈ R
⇒ R is reflexive
Symmetric: For Symmetric, we need to prove that- If (a, b) ∈ R, then (b, a) ∈ R
Let a, b ∈ Z and (a, b) ∈ R
⇒ a + b is even
⇒ b + a is even
⇒ (b, a) ∈ R
⇒ R is symmetric
Transitive : : For Transitivity, we need to prove that- If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R
Let (a, b) ∈ R and (b, c) ∈ R For some a, b, c ∈ Z
⇒ a + b is even and b + c is even
[if b is odd, then a and c must be odd ⇒ a + c is even, If b is even, then a and c must be even ⇒ a + c is even]
⇒ a + c is even
⇒ (a, c) ∈ R
⇒ R is transitive
Hence, R is an equivalence relation on Z
m is said to be related to n if m and n are integers and m – n is divisible by 13. Does this define an equivalence relation?
To check that relation is equivalence, we need to check that it is reflexive, symmetric and transitive. Reflexivity : For Reflexivity, we need to prove that-
(a, a) ∈ R Let m ∈ Z
⇒ m – m = 0
⇒ m – m is divisible by 13
⇒ (m, m) ∈ R
⇒ R is reflexive
Symmetric : For Symmetric, we need to prove that- If (a, b) ∈ R, then (b, a) ∈ R
Let m, n ∈ Z and (m, n) ∈ R
⇒ m – n = 13p For some p ∈ Z
⇒ n – m = 13 × (–p)
⇒ n – m is divisible by 13
⇒ (n – m) ∈ R,
⇒ R is symmetric
Transitive:: For Transitivity, we need to prove that- If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R
Let (m, n) ∈ R and (n, q) ∈ R For some m, n, q ∈ Z
⇒ m – n = 13p and n – q = 13s For some p, s ∈ Z
⇒ m – q = 13 (p + s)
⇒ m – q is divisible by 13
⇒ (m, q) ∈ R
⇒ R is transitive
Hence, R is an equivalence relation on Z.
Let R be a relation on the set A of ordered pairs of non-zero integers defined by (x, y) R (u, v) iff xv = yu.
Show that R is an equivalence relation.
(x, y) R (u, v) ⇔ xv = yu Proof :
To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive. Reflexivity : For Reflexivity, we need to prove that-
(a, a) ∈ R
∵ xy = yu
∴ (x, y) R (x, y)
Symmetric : For Symmetric, we need to prove that- If (a, b) ∈ R, then (b, a) ∈ R
Let (x, y) R (u, v)
TPT (u, v) R (x, y)
Given xv = yu
⇒ yu = xv
⇒ uy = vx
∴ (u, v) R (x, y)
Transitive : : For Transitivity, we need to prove that- If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R
Let (x, y) R (u, v) and (u, v) R (p, q) …(i) TPT (x, y) R (p, q)
TPT (xq = yp
From (1) xv = yu & uq = vp xvuq = yuvp
xq = yp
∴ R is transitive
Since R is reflexive, symmetric & transitive
⇒ R is an equivalence relation.
Show that the relation R on the set A = {x ∈ Z ; 0 ≤ x ≤ 12}, given by R = {(a, b) : a = b}, is an equivalence relation. Find the set of all elements related to 1.
We have,
A = {x ∈ Z : 0 ≤ x ≤ 12} be a set and R = {(a, b) : a = b} be a relation on A Now,
Proof :
To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive.
Reflexivity : For Reflexivity, we need to prove that- (a, a) ∈ R
Let a ∈ A
⇒ a = a
⇒ (a, a) ∈ R
⇒ R is reflexive
Symmetric : For Symmetric, we need to prove that- If (a, b) ∈ R, then (b, a) ∈ R
Let a, b ∈ A and (a, b) ∈ R
⇒ a = b
⇒ b = a
⇒ (b, a) ∈ R
⇒ R is symmetric
Transitive : : For Transitivity, we need to prove that- If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R
Let a, b & c ∈ A
and Let (a, b) ∈ R and (b, c) ∈ R
⇒ a = b and b = c
⇒ a = c
⇒ (a, c) ∈ R
⇒ R is transitive
Since, R is being reflexive, symmetric and transitive, so R is an equivalence relation. Also, we need to find the set of all elements related to 1.
Since the relation is given by, R = {(a, b) : a = b}, and 1 is an element of A, R = {(1, 1) : 1 = 1}
Thus, the set of all element related to 1 is 1.
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.
We have, L is the set of lines.
R = {(L1, L2) : L1 is parallel to L2} be a relation on L Now,
Proof :
To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive. Reflexivity : For Reflexivity, we need to prove that-
(a, a) ∈ R
Since a line is always parallel to itself.
∴ (L1, L2) ∈ R
⇒ R is reflexive
Symmetric : For Symmetric, we need to prove that- If (a, b) ∈ R, then (b, a) ∈ R
Let L1, L2∈ L and (L1, L2) ∈ R
⇒ L1 is parallel to L2
⇒ L2 is parallel to L1
⇒ (L1, L2) ∈ R
⇒ R is symmetric
Transitive: For Transitivity, we need to prove that- If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R
Let L1, L2 and L3∈ L such that (L1, L2) ∈ R and (L2, L3) ∈ R
⇒ L1 is parallel to L2 and L2 is parallel to L3
⇒ L1 is parallel to L3
⇒ (L1, L3) ∈ R
⇒ R is transitive
Since, R is reflexive, symmetric and transitive, so R is an equivalence relation. And, the set of lines parallel to the line y = 2x + 4 is y = 2x + c For all c∈ R where R is the set of real numbers.
Show that the relation R, defined on 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?
R = {(P1, P2): P1 and P2 have same the number of sides} Proof :
To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive. Reflexivity: For Reflexivity, we need to prove that-
(a, a) ∈ R
R is reflexive since (P1, P1) ∈ R as the same polygon has the same number of sides with itself. Symmetric: For Symmetric, we need to prove that-
If (a, b) ∈ R, then (b, a) ∈ R Let (P1, P2) ∈ R.
⇒ P1 and P2 have the same number of sides.
⇒ P2 and P1 have the same number of sides.
⇒ (P2, P1) ∈ R
∴ R is symmetric.
Transitive: For Transitivity, we need to prove that- If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R
Now, (P1, P2), (P2, P3) ∈ R
⇒ P1 and P2 have the same number of sides. Also, P2 and P3 have the same number of sides.
⇒P1 and P3 have the same number of sides.
⇒ (P1, P3) ∈ R
∴ R is transitive.
Hence, R is an equivalence relation.
And, now the elements in A related to the right-angled triangle (T) with sides 3, 4 and 5 are those polygons which have three sides (since T is a polygon with three sides).
Hence, the set of all elements in A related to triangle T is the set of all triangles.
Let O be the origin. We define a relation between two points P and Q in a plane if OP = OQ. Show that the relation, so defined is an equivalence relation.
Let A be set of points on the plane.
Let R = {(P, Q) : OP = OQ} be a relation on A where O is the origin.
To prove R is an equivalence relation, we need to show that R is reflexive, symmetric and transitive on A. Proof :
To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive. Reflexivity : For Reflexivity, we need to prove that-
(a, a) ∈ R Let p ∈ A
Since OP = OP ⇒ (P, P) ∈ R
⇒ R is reflexive
Symmetric : For Symmetric, we need to prove that- If (a, b) ∈ R, then (b, a) ∈ R
Let (P, Q) ∈ R for P, Q ∈ R Then OP = OQ
⇒ Op = OP
⇒ (Q, P) ∈ R
⇒ R is symmetric
Transitive: For Transitivity, we need to prove that- If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R
Let (P, Q) ∈ R and (Q, S) ∈ R
⇒ OP = OQ and OQ = OS
⇒ OP = OS
⇒ (P, S) ∈ R
⇒ R is transitive
Thus, R is an equivalence relation on A
Let R be the relation defined on the set A = {1, 2, 3, 4, 5, 6, 7} by R = {(a, b) : both a and b are either odd or even}. Show that R is an equivalence relation. Further, show that all the elements of the subset {1, 3, 5, 7} are related to each other, and all the elements of the subset {2, 4, 6} are related to each other, but no element of the subset {1, 3, 5, 7} is related to any element of the subset {2, 4, 6}.
Given A = {1, 2, 3, 4, 5, 6, 7} and R = {(a, b) : both a and b are either odd or even number} Therefore,
R = {(1, 1), (1, 3), (1, 5), (1, 7), (3, 3), (3, 5), (3, 7), (5, 5), (5, 7), (7, 7), (7, 5), (7, 3), (5, 3), (6, 1), (5, 1), (3,
1), (2, 2), (2, 4), (2, 6), (4, 4), (4, 6), (6, 6), (6, 4), (6, 2), (4, 2)}
To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive. Reflexivity : For Reflexivity, we need to prove that-
(a, a) ∈ R
Here (1,1), (2,2), (3,3), (4,4), (5,5), (6,6), (7,7) all ∈ R
From the relation R it is seen that R is reflexive. Symmetric: For Symmetric, we need to prove that- If (a, b) ∈ R, then (b, a) ∈ R
From the relation R, it is seen that R is symmetric. Transitive: For Transitivity, we need to prove that-
If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R
[I (a, b) are odd and (b, c) are odd then (a, c) are also odd numbers] From the relation R, it is seen that R is transitive too.
Also, from the relation R, it is seen that {1, 3, 5, 7} are related with each other only and {2, 4, 6} are related with each other .
Let S be a relation on the set R of all real numbers defined by S = {(a, b) ∈ R × R : a^{2 }+ b^{2 }= 1}. Prove that S is not an equivalence relation on R.
S = {(a, b) : a^{2 }+ b^{2 }=1}
Proof :
To prove that relation is not equivalence, we need to prove that it is either not reflexive, or not symmetric or not transitive.
Reflexivity : For Reflexivity, we need to prove that- (a, a) ∈ R
Let a = 1/2, a ∈ R Then,
⇒ (a, a) ∉ S
⇒ S is not reflexive
Hence, S is not an equivalence relation on R.
Let Z be the set of all integers and Z0 be the set of all non-zero integers. Let a relation R on Z × Z0 be defined as follows :
(a, b) R (c, d) ⇔ ad = bc for all (a, b), (c, d) ∈ Z × Z0 Prove that R is an equivalence relation on Z × Z0 Answer
We have, Z be set of integers and Z0 be the set of non-zero integers. R = {(a, b) (c, d) : ad = bc} be a relation on Z and Z0.
Proof :
To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive. Reflexivity : For Reflexivity, we need to prove that-
(a, a) ∈ R
(a, b) ∈ Z × Z0
⇒ ab = ba
⇒ ((a, b), (a, b)) ∈ R
⇒ R is reflexive
Symmetric : For Symmetric, we need to prove that- If (a, b) ∈ R, then (b, a) ∈ R
Let ((a, b), (c, d) ∈ R
⇒ ad = bc
⇒ cd = da
⇒ ((c, d), (a, b)) ∈ R
⇒ R is symmetric
Transitive : : For Transitivity, we need to prove that-
If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R
Let (a, b), (c, d) ∈ R and (c, d), (e, f) ∈ R
⇒ ad = bc and cf = de
⇒ _{}
⇒ _{}
⇒ af = be
⇒ (a, c) (e, f) ∈ R
⇒ R is transitive
Hence, R is an equivalence relation on Z × Z0
If R and S are relations on a set A, then prove the following :
R and S are two symmetric relations on set A
Symmetric: For Symmetric, we need to prove that- If (a, b) ∈ R, then (b, a) ∈ R
Let (a, b) ∈ R ⋂ S
⇒ (a, b) ∈ R and (a, b) ∈ S
⇒ (b, a) ∈ R and (b, a) ∈ S [∴ R and S are symmetric]
⇒ (b, a) ∈ R ⋂ S
⇒ R ⋂ S is symmetric
To prove: R ⋃ S is symmetric
Symmetric: For Symmetric, we need to prove that- If (a, b) ∈ R, then (b, a) ∈ R
Let (a, b) ∈ R ⋃ S
⇒ (a, b) ∈ R or (a, b) ∈ S
⇒ (b, a) ∈ R or (b, a) ∈ S [∴ R and S are symmetric]
⇒ (b, a) ∈ R ⋃ S
⇒ R ⋃ S is symmetric
Reflexivity : For Reflexivity, we need to prove that- (a, a) ∈ R
Suppose R ⋃ S is not reflexive.
This means that there is a ∈ R ⋃ S such that (a, a) ∉ R ⋃ S Since a ∈ R ⋃ S,
∴ a ∈ R or a ∈ S
If a ∈ R, then (a, a) ∈ R [∵ R is reflexive]
⇒ (a, a) ∈ R ⋃ S
Hence, R ⋃ S is reflexive
If R and S are transitive relations on a set A, then prove that R ⋃ S may not be a transitive relation on A.
We will prove this using an example. Let A = {a, b, c} be a set and
R = {(a, a) (b, b) (c, c) (a, b) (b, a)} and
S = {(a, a) (b, b) (c, c) (b, c) (c, d)} are two relations on A Clearly R and S are transitive relation on A
Now,
R ⋃ S = {(a, a) (b, b) (c, c) (a, b) (b, a) (b, c) (c, b)}
Here, (a, b) ∈ R ⋃ S and (b, c) ∈ R ⋃ S but (a, c) ∉ R ⋃ S
∴ R ⋃ S is not transitive
Let C be the set of all complex numbers and C0 be the set of all non-zero complex numbers. Let a relation R on C0 be defined as
z1 R z2 _{} is real for all z1, z2∈ C0. Show that R is an equivalence relation.
We have,
We want to prove that R is an equivalence relation on Z. Now,
Proof :
To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive. Reflexivity : For Reflexivity, we need to prove that-
(a, a) ∈ R Let a ∈ C0
And, 0 is real
∴ (a, a) ∈ R, so R is reflexive
Symmetric: For Symmetric, we need to prove that- If (a, b) ∈ R, then (b, a) ∈ R
Let (a, b) ∈ R
⇒ p is real.
And ∵ p is real
⇒ -p is also a real no.
⇒ (b, a) ∈ R, so R is symmetric
Transitive : : For Transitivity, we need to prove that- If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R
Let (a, b) ∈ R and (b, c) ∈ R
⇒ p is real no.
_{} ….(1)
⇒ q is real.
_{} …..(2)
Dividing (1) by (2), we get-
Where, Q is a rational number.
⇒ Q is real number
Now, by componendo dividendo-
⇒ (a, c) ∈ R.
⇒ R is transitive
Thus, R is reflexive, symmetric and, transitive on C0. Hence, R is an equivalence relation on C0.
Write the domain of the relation R defined on the set Z of integers as follows: (a, b) ϵR ⬄ a^{2 }+ b^{2 }= 25
Given a and b are integers, i.e. a,b ∈ Z.
∴ Domain of R = Set of all first elements in the relation.
= Values of ‘a’ which are in the relation.
=Z (Integers)
Range of R=Set of all second elements in the relation.
=Values of ‘b’ which are in the relation.
=Z(Integers)
Since, a^{2}+ b^{2 }= 25 and a, b are integers;
⇒ R= {(5,0), (0,5), (-5,0), (0, -5), (3,4), (4,3), (-3, -4), (-4, -3),
(-3,4), (4, -3), (-4,3), (3, -4)}
⇒ Domain of R= {-5, -4, -3,0,3,4,5}
If R = {(x,y): x^{2 }