Exercise 1.3 | Class 10 | Advanced Mathematics | Set

1. If A = {1, 3}, then write the identity relation $\ I: A→ A$. Also write the universal relation on A.
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:

Matrix Table :

$\ \downarrow{x}|{y}\rightarrow$	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