Solution:
Here, A = {1, 3}, then \(\ I: A→ A \) = {(1, 1), (3, 3)}
We know, universal relation on \(\ R = A×A \) = {(1, 1), (1, 3), (3, 1), (3, 3)}
2. If A = {1, 2}, then write down all the relations on A.
Solution:
Here, A = {1, 2}, so, A×A = {(1, 1), (1, 2), (2, 1), (2, 2)}
The total number of relations on \(\ A = 2^{(2)(2)} = 2^4 = 16 \)
Now, \(\ P(A×A)\) = {\(\ \phi \), {(1, 1)}, {(1, 2)}, {(2, 1)}, {(2, 2)}, {(1, 1), (1, 2)}, {{(1, 1), (2, 1)}, {(1, 1), (2, 2)}, {(1, 2), (2, 1)}, {(1, 2), (2, 2)}, {(2, 1), (2, 2)}, {(1, 1), (1, 2), (2, 1)}, {(1, 1), (1, 2), (2, 2)}, {(1, 1), (2, 1), (2, 2)}, {(1, 2), (2, 1), (2, 2)}, {(1, 1), (1, 2), (2, 1), (2, 2)}}
3. If A = {1, 2, 3}, then find the elements of the relation R = { \(\ (x, y): x = y \) and \(\ x, y \in A \)} on A.
Solution:
Here, A = {1, 2, 3}
R = { \(\ (x, y): x = y \) and \(\ x, y \in A \)}
= {(1, 1), (2, 2), (3, 3}
4. If \(\ Z^+ \) is the set of positive integers and \(\ R = Z^+ → Z^+ \) is a relation defined as R = {\(\ (a, b): a, b \in Z^+ \) and \(\ a – b > 2\)}. Is it a finite relation? Represent R in roster form.
Solution:
R = {\(\ (a, b): a, b \in Z^+ \) and \(\ a – b > 2\)
R = {(4,1), (5, 1), (6, 1), …., (5,1), (6, 2), (6, 3), ... (15, 2), …..}
The relation R is an infinite relation.
5. A= {2, 3, 4, 5} and B = {3, 6, 7, 10} be two sets and a relation R is defined as R = {\(\ (x, y): x \) completely divides \(\ y \) where \(\ x \in A \) and \(\ y \in B\)} . Write the relation R in the tabular form. Also represent R by arrow diagram and matrix table.
Solution:
Here, A = {2, 3, 4, 5} and B = {3, 6, 7, 10} \)
R = {(2, 6), (2, 10), (3, 3), (3, 6), (5, 10)}
Arrow Diagram:
| \( \ \downarrow{x}|{y}\rightarrow \) | 3 | 6 | 7 | 10 |
|---|---|---|---|---|
| 2 | 0 | 1 | 0 | 1 |
| 3 | 1 | 1 | 0 | 0 |
| 4 | 0 | 0 | 0 | 0 |
| 5 | 0 | 0 | 0 | 1 |
6. Determine \(\ R^{-1} \) of the relation R given in Question 5 above. Also find the d(\(\ R^{-1} \)) and r(\(\ R^{-1} \)).
Solution:
Here, R = {(2, 6), (2, 10), (3, 3), (3, 6), (5, 10)}
So, \(\ R^{-1} \) = {(6, 2), (10, 2), (3, 3), (6, 3), (10, 5)}
d(\(\ R^{-1} \)) = {3, 6, 10} and r(\(\ R^{-1} \)) = {2, 3, 5}
7. A = {3, 6, 8, 9} is a set and \(\ aRb\) iff \(\ a – b \) is divisible by 3 for \(\ a, b \in A \). then,
(i) Write R as a set and draw the arrow diagram
(ii) Find d(R) and r(R)
(iii) Can we find the inverse of R
Solution:
Here, A = {3, 6, 8, 9}
(i) R = {\(\ aRb\) iff \(\ a – b \) is divisible by 3 for \(\ a, b \in A \)}
= {(3, 3), (3, 6), (3, 9), (6, 3), (6, 6), (6, 9),(8, 8), (9, 3), (9, 6), (9, 9)}
Arrow Diagram:
(ii) d(R) = {3, 6, 8, 9} and r(R) = {3, 6, 8, 9}
(iii) \(\ R^{-1} \) = {{(3, 3), (6, 3), (9, 3), (3, 6), (9, 6), (8,, 9), (3, 9), (6, 9), (9, 9)} }
8. A = {1, 2, 3, 4}
and B = {a, b, c, d} be two sets. Choose which of the followings are relations from A to B- (i) {(1, a), (1, b), (2, c), (4, d)}
(ii) {(1, 1), (1, a), (3, c)}
(iii) A×B
(iv) {(a, 1), (b, 2), (c, 3)}
(v) {\(\ \phi \)}
(vi) {(1, c), (2, c), (3, c), (4, c)}
Solution:
Here, A = {1, 2, 3, 4} and B = {a, b, c, d}
So, A × B = {(1, a), (1, b), (1, c), (1, d), (2, a), (2, b), (2, c), (2, d), (3, a), (3, b), (3, c), (3, d), (4, a), (4, b), (4, c), (4, d)}
(i) Since, {(1, a), (1, b), (2, c), (4, d)} \(\ \in A × B \)
So, it is a relation from A to B.
(ii) Since, {(1, 1), (1, a), (3, c)} \(\ ∉ A × B \)
So, this is not a relation from A to B.
(iii) Since, \(\ A × B ⊆ A × B \)
So, it is the universal relation from A to B. (iv) Since, {(a, 1), (b, 2), (c, 3)} \(\ ∉ A × B \)
So, this is not a relation from A to B.
(v) Since, {\(\ \phi\)} \(\ ∉ A × B \)
So, it is not a relation from A to B.
(vi) Since, {(1, c), (2, c), (3, c), (4, c)} \(\ \in A × B \)
So, it is a relation from A to B
9. A relation R is defined on the set of natural numbers N as \(\ aRb \) where \(\ a = b^2 \) for \(\ a, b \in N \). Write the relation R. also write \(\ R^{-1} \) in the set builder method.
Solution:
Here, R = {\(\ (a, b): a = b^2 ~\forall~ a, b \in N \) }
= {(1, 1), (4, 2), (9, 3), (16, 4), (25, 5), ….}
\(\ R^{-1} \) = {\(\ (a, b): a^2 = b ~\forall~ a, b \in N\)}
= {(1, 1), (2, 4), (3, 9), (4, 16), (5, 25), ….}
10. If A = {1, 2, 3, 4, 6} and R is a relation on A defined as R = {\(\ (x, y): y \) is exactly divisible by \(\ x\) where \(\ x, y\in A\)}, then
(i) How many elements are there in R and list them?
(ii) Find \(\ R^{-1} \)
(iii) Find d(R), r(R), d(\(\ R^{-1} \)) and r(\(\ R^{-1} \))
Solution:
Here, A = {1, 2, 3, 4, 5}
(i) R = {\(\ (x, y): y \) is exactly divisible by \(\ x ~\forall~ x, y \in A \)}
= (1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}
There are 12 elements.
(ii) \(\ R^{-1} \) = {(1, 1), (2,1), (3,1), (4,1), (6,1), (2, 2), (4,2), (6,2), (3, 3), (6,3), (4, 4), (6, 6)}
(iii) Now, d(R) = {1,2,3,4,6}
r(R) = {1,2,3,4,6}
r(\(\ R^{-1} \)) = {1,2,3,4,6}
11. Let A = {10, 11, 12, 13} and B = {1, 2, 3, 4} be two sets. Write the following relations from A to B as defined in each case-
(i) \(\ R_1 \) = {\(\ (a, b): a – b \) is odd for \(\ a \in A \) and \(\ b \in B \) }
(ii) \(\ R_2 \) = {\(\ (a, b): a + b \) is a multiple of 4 for \(\ a\in A \) and \(\ b \in B \)}
(iii) \(\ R_3 \) = {\(\ (a, b): a – b \in N \) for \(\ a< 10 \) where \(\ a\in A\) and \(\ b\in B \) }
Solution:
Here, A = {10, 11, 12, 13} and B = {1, 2, 3, 4} then
(i) \(\ R_1 \) = {(10, 1), (10, 3), (11, 2), (11, 4), (12, 1), (12, 3), (13, 2), (13, 4)}
(ii) \(\ R_2 \) = {(10, 2), (11, 1), (12, 4), (13, 3)}
(iii) \(\ R_3 \) = \(\ {} \) or \(\ \phi \)
12. \(\ R_1 \) and \(\ R_2 \) be two relations on the set of integers Z defined as below- \(\ R_1 \) = {\(\ (a, b): a^2 = b^2 ~\forall~ a, b \in Z \) }
\(\ R_2 \) = {\(\ (a, b): a – b \) is positive where \(\ a, b \) are primes and both are smaller than 10}.
Determine the domain and range of each of them.
Solution:
\(\ R_1 \) = {(1,1), (2, 2), (3, 3), …. (1, -1), (2, -2), (3, -3), …}
d(\(\ R_1 \)) = {1, 2, 3, …} = \(\ Z \)
r(\(\ R_1 \)) = {1, 2, 3, …} = \(\ Z \)
Now, \(\ R_2 \) = {(3, 2), (5, 2), (5, 3), (7, 2), (7, 3), (7, 5)}
d(\(\ R_2 \)) = {3, 5, 7}
r(\(\ R_2 \)) = {2, 3, 5}
13. If R and S be two relations on any set A, then show that- (i) If \(\ R \subset S \) then \(\ R^{-1} \subset S^{-1} \)
(ii) \(\ (R\cup T)^{-1} = R^{-1} \cup T^{-1} \)
Solution:
(i) Let, \(\ (x, y)\in R^{-1} \)
\(\ ⇒ (y, x) \in R \)
\(\ ⇒ (y, x) \in S \) [⸪ \(\ R\subset S \)]
\(\ ⇒ (x, y) \in S^{-1}\)
Therefore, \(\ R^{-1}\subset S^{-1} \)
(ii) Let, \(\ (x, y) \in (R\cup T)^{-1} \)
\(\ ⇒ (y, x) \in (R\cup T) \)
\(\ ⇒ (y, x) \in R \) or \(\ (y, x) \in T \)
\(\ ⇒ (x, y) \in R^{-1} \) or \(\ (x, y) \in T^{-1} \)
\(\ ⇒ (x, y) \in R^{-1}\cup T^{-1}) \)
\(\ ⇒ (R\cup T)^{-1} ⊆ R^{-1} \cup T^{-1}) \)
Conversely,
Let, \(\ (x, y) \in R^{-1} \cup T^{-1}) \)
\(\ ⇒ (x, y) \in R^{-1} \) or \(\ (x, y) \in T^{-1} \)
\(\ ⇒ (y, x) \in R \) or \(\ (y, x) \in T \)
\(\ ⇒ (y, x) \in (R\cup T) \)
\(\ ⇒ (x, y) \in (R\cup T)^{-1} \)
\(\ ⇒ R^{-1} \cup T^{-1}) ⊆ R\cup T)^{-1} \)
Therefore, \(\ (R\cup T)^{-1} = R^{-1}\cup T^{-1} \)
14. If R be a relation on the set of natural numbers N and R is defined as R = {\(\ (a, b): 3a + b = 10 \) and \(\ a, b \in N \)}. Write \(\ R^{-1} \) as a set in Roster method.
Solution:
Here, R = {(1, 7), (2, 4), (3, 1)}
Therefore, \(\ R^{-1} \) = {(7, 1), (4, 2), (3, 1)}
15. Let A = {1, 2, 3} be a set and R be a relation on A such that \( aRb \) iff \(\ a ≤ b \) for \(\ a, b \in A \). Find R as a set in Roster form. Also find the smallest and greatest relations on A.
Solution:
Here, A = {1, 2, 3}
and R = {\(\ (a,b): a ≤ b ~\forall~ a, b \in A \)}
= {(1,1), (1,2), (1,3), (2,2), (2,3), (3,3)}
The smallest relation on A is the null relation,
i.e., R= {\(\ \phi \)}
The greatest relation on A is the universal relation,
i.e., R = A×A