Exercise 1.4 | Class 10 | Advanced Mathematics | Set

1. Give an example of a relation R so that-
(a) R is reflexive, but neither symmetric nor transitive.
Solution:
Let A = {1, 2, 3}, then example of relation R on A is
R = {(1, 1), (2, 2), (3, 3), (3, 2), (2, 1)}
Validation
Reflexivity:
Here, $\ (1,1)\in R, (2,2)\in R, (3,3)\in R$
⸫ R is reflexive.
Symmetry:
Here, $\ (3,2)\in R$ but $\ (2,3)\notin R$
⸫ R is not symmetric.
Transitivity:
Here, $\ (3,2)\in R$ and $\ (2,1)\in R$ but $\ (3,1)\notin R$
⸫ R is not transitive.


(b) R is symmetric, but neither reflexive nor transitive.
Solution:
Let B = {1, 2, 3}, then example of relation R on B is
R = {(1, 2),(2, 1)}
Validation
Reflexivity:
Here, $\ (1,1)\notin R, (2,2)\notin R, (3,3)\notin R$
⸫ R is not reflexive.
Symmetry:
Here, $\ (1,2)\in R$ and $\ (2,1)\in R$
⸫ R is symmetric.
Transitivity:
Here, $\ (1,2)\in R$ and $\ (2,1)\in R$ but $\ (1,1)\notin R$
⸫ R is not transitive.


(c) R is transitive, but neither reflexive nor symmetric.
Solution:
Let C = {1, 2, 3}, then example of relation R on C is
R = {(1, 2), (2, 1), (1, 1), (3, 1)}
Validation
Transitivity:
Here, for $\ (1,2)\in R$ and $\ (2,1)\in R   \exists   (1,1)\in R$
⸫ R is transitive.
Reflexivity:
Here, $\ (2,2)\notin R, (3,3)\notin R$
⸫ R is not reflexive.
Symmetry:
Here, $\ (3,1)\in R$ but $\ (1,3)\notin R$
⸫ R is not symmetric.




2. Choose the correct one-
(a) The universal relation on a non-empty set is-
(i) Reflexive   (ii) Symmetric   (iii) Transitive  (iv) All of them
Answer: all of them
Explanation:
Let A = {1, 2}, then universal relation R on A is
R = A*A = {(1, 1), (1, 2), (2, 1), (2, 2)}
Reflexivity:
Here, $\ (1,1)\in R, (2,2)\in R$
⸫ R is reflexive.
Symmetry:
Here, $\ (1,2)\in R$ and $\ (2,1)\in R$
⸫ R is symmetric.
Transitivity:
Here, for $\ (1,1)\in R$ and $\ (1,2)\in R   \exists   (1,2)\in R$
for $\ (1,2)\in R$ and $\ (2,1)\in R   \exists   (1,1)\in R$
for $\ (2,1)\in R$ and $\ (2,2)\in R   \exists   (2,2)\in R$
⸫ R is transitive.


(b) The identity relation on a non-empty set is-
(i) Reflexive   (ii) Symmetric   (iii) Transitive   (iv) All of them
Answer: all of them
Explanation:
Let A = {1, 2, 3}, then identity relation R on A is
R = {(1, 1),(2, 2), (3,3)}
Reflexivity:
Here, $\ (1,1)\in R, (2,2)\in R, (3,3)\in R$
⸫ R is reflexive.
Symmetry:
Here, for $\ (1,1)\in R   \exists   (1,1)\in R$
for $\ (2,2)\in R   \exists   (2,2)\in R$
for $\ (3,3)\in R   \exists   (3,3)\in R$
⸫ R is symmetric.
Transitivity:
for $\ (1,1)\in R   \nexists   (1,x)\in R  \forall   x\in A$
for $\ (2,2)\in R   \nexists   (2,x)\in R  \forall   x\in A$
for $\ (3,3)\in R   \nexists   (3,x)\in R  \forall   x\in A$
⸫ R is transitive.
Click Here


3. A few relations on the set A = {a, b, c} are given below. Identify which of them are reflexive, symmetric, transitive or none of them.
(i) $\ R_1$ = {(a, b)}
Solution:
Here, $\ R_1$ = {(a, b)}
$\ (a,a), (b,b), (c,c) \notin R_1$
⸫ $\ R_1$ is not reflexive.
Now, for $\ (a,b)\in R_1   \nexists   (b,a)\in R_1$
⸫ $\ R_1$ is not symmetric.
Again, for $\ (a,b)\in R_1   \nexists   (b,c)\in R_1$
⸫ $\ R_1$ is transitive.


(ii) $\ R_2$ = {(a, a), (c, c), (a, c), (c, a)}
Solution:
Here, $\ R_2$ = {(a, a), (c, c), (a, c), (c, a)}
$\ (b,b)\notin R_2$
⸫ $\ R_2$ is not reflexive.
Now, for $\ (a, c)\in R_2   \exists   (c,a)\in R_2$
⸫ $\ R_2$ is symmetric.
Again, for $\ (a,c)\in R_2$ and $\ (c,a)\in R_2   \exists   (a,a)\in R_2$
⸫ $\ R_1$ is transitive.


(iii) $\ R_3$ = {(a, a), (a, b), (a, c), (b, b), (c, a), (b, a)}
Solution:
Here, $\ R_3$ = {(a, a), (a, b), (a, c), (b, b), (c, a), (b, a)}
$\ (c,c)\notin R_3$
⸫ $\ R_3$ is not reflexive.
Now, for $\ (a, b)\in R_3   \exists   (b,a)\in R_3$
Now, for $\ (a, c)\in R_3   \exists   (c,a)\in R_3$
⸫ $\ R_3$ is symmetric.
Again, for $\ (a,b)\in R 3$ and $\ (b,a)\in R_3   \exists   (a,a)\in R_3$
Again, for $\ (a,c)\in R 3$ and $\ (c,a)\in R_3   \exists   (a,a)\in R_3$
⸫ $\ R_1$ is transitive.


(iv) $\ R_4$ = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)}
Solution:
Here, $\ R_4$ = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)}
$\ (a,a), (b,b), (c,c)\in R_4$
⸫ $\ R_4$ is reflexive.
Now, for $\ (a, b)\in R_4   \exists   (b,a)\in R_4$
for $\ (b,c)\in R_4   \exists   (c,b)\in R_4$
for $\ (a, c)\in R_4   \exists   (c,a)\in R_4$
⸫ $\ R_4$ is symmetric.
Again, for $\ (a,b)\in R_4$ and $\ (b,a)\in R_4   \exists   (a,a)\in R_4$
Again, for $\ (b,c)\in R_4$ and $\ (c,a)\in R_4   \exists   (b,a)\in R_4$
Again, for $\ (a,c)\in R_4$ and $\ (c,a)\in R_4   \exists   (a,a)\in R_4$
⸫ $\ R_1$ is transitive.



4. A few relations on A = {1, 2, 3} are given below. State which of them are reflexive-
(a) R = {(1, 2), (3, 2), (2, 2), (2, 3)}
Solution:
$\ (1,1),(2,2), (3,3)\notin R$
⸫ R is not reflexive.


(b) S = {(3, 1)}
Solution:
$\ (1,1),(2,2), (3,3)\notin S$
⸫ S is not reflexive.


(c) T = {(1, 1), (3, 1), (3, 3), (2, 1), (2, 2)}
Solution:
$\ (1,1),(2,2), (3,3)\in T$
⸫ T is reflexive.



5. State true or false for the following relations defined on A = {1, 2, 3}
(a) $\ R_1$ = {(1, 1), (2, 1), (2, 2,), (3, 2), (2, 3)} is symmetric.
Solution:
Now, for $\ (2,1)\in R_1   \nexists   (1,2)\in R_1$
⸫ $\ R_1$ is not symmetric.
⸫ The statement is false.


(b) $\ R_2$ = {(3, 3)} is symmetric, but not reflexive.
Solution:
Now, for $\ (3, 3)\in R_2$
⸫ $\ R_2$ is symmetric.
but $\ (1, 1), (2,2)\notin R_2$
⸫ $\ R_2$ is not reflexive.
⸫ The statement is true.


(c) $\ R_3$ = {(1, 2)} is anti-symmetric.
Solution:
Now, for $\ (1,2)\in R$ and $\ (2,1)\in R ⇒ 1\neq 2$
⸫ $\ R_3$ is not anti-symmetric.
⸫ The statement is true.


(d) $\ R_4$ = {(1, 1), (3, 2), (2, 3)} is symmetric but not anti-symmetric.
Solution:
$\ (1,1)\in R_4$ ; $\ (3,2)(2,3)\in R_4$
⸫ $\ R_1$ is symmetric.
but for $\ (3,2)\in R$ and $\ (2,3)\in R ⇒ 3\neq 2$
⸫ $\ R_4$ is not anti-symmetric.
⸫ The statement is true.


(e) $\ R_5$ = {(2, 2)} is anti-symmetric, but not reflexive.
Solution:
for $\ (2,2)\in R$ and $\ (2,2)\in R ⇒ 2= 2$
⸫ $\ R_5$ is not anti-symmetric.
but, $\ (1, 1), (3,3)\notin R_5$
⸫ $\ R_5$ is not reflexive.
⸫ The statement is true.


(f) $\ R_6$ = {1, 2), (2, 3), (1, 3), (1, 1), (2, 1)} is transitive.
Solution:
for $\ (1,2), (2,3)\in R_6   \exists   (1,3)\in R_6$
for $\ (1,2),(2,1)\in R_6   \exists   (1,1)\in R_6$
for $\ (1,1)(1,3)\in R_6   \exists   (1,3)\in R_6$
for $\ (2,1),(1,2)\in R_6   \nexists   (2,2)\in R_6$
⸫ $\ R_6$ is not transitive.
⸫ The statement is true.


(g) $\ R_7$ = {(1, 3)} and $\ R_8$ = {(2, 2)} are both transitive.
Solution:
for $\ (1,3)\in R_7   \nexists   (3,x)\in R_7   \forall   x\in A$
⸫ $\ R_7$ is transitive.
similarly,
for $\ (2,2)\in R_8   \nexists   (2,x)\in R_8   \forall   x\in A$
⸫ $\ R_8$ is also transitive.
⸫ The statement is true.


(h) R = A×A is an equivalence relation. But it is not anti-symmetric.
Solution:
R = A×A= {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)}
Reflexivity:
Here, $\ (1,1)\in R, (2,2)\in R, (3,3)\in R$
⸫ R is reflexive.
Symmetry:
Here, for $\ (1,1)\in R   \exists   (1,1)\in R$
Here, for $\ (1,2)\in R   \exists   (2,1)\in R$
Here, for $\ (1,3)\in R   \exists   (3,1)\in R$
for $\ (2,2)\in R   \exists   (2,2)\in R$
for $\ (2,3)\in R   \exists   (3,2)\in R$
for $\ (3,3)\in R   \exists   (3,3)\in R$
⸫ R is symmetric.
Transitivity:
for $\ (1,1)(1,2)\in R   \exists   (1,2)\in R$
for $\ (1,1)(1,3)\in R   \exists   (1,3)\in R$
for $\ (1,2), (2,1)\in R   \exists   (1,1)\in R$
for $\ (1,2), (2,2)\in R   \exists   (1,2)\in R$
for $\ (1,2), (2,3)\in R   \exists   (1,3)\in R$
for $\ (1,3), (3,1)\in R   \exists   (1,1)\in R$
⸫ R is transitive.



6. Demonstrate by an example that the identity relation on a non-empty set is always a reflexive relation. However, a reflexive relation is not necessarily an identity relation.
Solution:
Let A = {1,2,3}
I = {(1,1), (2,2), (3,3)}
Here it is seen that the identity relation I on the set A is reflexive as-
$\ (1,1),(2,2),(3,3)\in R$
2nd part
Let us consider a reflexive relation on set A is such that
R = {(1,1),(2,2),(3,3),(1,3)}
and here the relation R is not an identity relation as $\ (1,3)\in R$



7. Let $\ R$ be a relation defined on the set of natura numbers N as ‘ $\ (x,y)\in R$ iff $\ (x,y):$ x divides y for all $\ x,y \in N$. Show that R is not an equivalence relation.
Solution:
Here, R = {$\ (x,y): x$ divides $\ y ~\forall~ x,y \in N$}
= {(1,1), (1,2), (1,3),.... (2,2), (2,4), (2,6),.... (3,3), (3,6), (3,9),.... (4,4), (4,8), (4,12),....}
Reflexivity:
Let $\ (1,1),(2,2),(3,3),(4,4).... \in R$
⸫ R is reflexive.
Symmetry:
for $\ (1,2)\in R   \nexists   (2,1)\in R$
⸫ R is not symmetric.
⸫ The relation R is not an equivalence relation.



8. Let R be a relation defined as R = {$\ (x,y):~x-y$ is divisible by 5 for $\ x,y \in Z$}. Show that R is an equivalence relation.
Solution:
Here, R = {$\ (x,y):~x-y$ is divisible by 5 for $\ x,y \in Z$}
Reflexivity:
Let $\ x\in R$
⇒$\ x-x=0$; divisible by 5
⸫ $\ (x,x)\in R ~\forall~ a\in R$
⸫ R is reflexive.
Symmetry:
Let $\ (x,y)\in R$
⇒ $\ x-y$; divisible by 5
and $\ y-x$; divisible by 5
⸫ $\ (y,a)\in R$
⸫ R is symmetric.
Transitivity:
Let $\ (x,y)\in R$
⇒ $\ x-y$; divisible by 5
also $\ (y,z)\in R$
⇒$\ y-x$; divisible by 5
then $\ \exists   (x,z)\in R$
⇒$\ x-z$; divisible by 5
⸫ R is transitive.



9. Let R and S be two relations on set A. Examine whether the following statements are true or false:
(a) If R is reflexive then $\ R^{-1}$ is reflexive.
Solution:
Let A = {1,2,3}
then a reflexive relation on A is such that
R = {(1,1),(2,2),(3,3)(1,3)}
and $\ R^{-1}$ = {(1,1),(2,2),(3,3)(3,1)}
since $\ (1,1),(2,2),(3,3) \in R^{-1}$
⸫ $\ R^{-1}$ is reflexive.
⸫ The statement is true.


(b) If R is an equivalence relation then $\ R^{-1}$ is also an equivalence relation.
Solution:
See Example 6 (from textbook)!


(c) If R and S are symmetric then $\ R\cup S$ is also symmetric.
Solution:
Let $\ (a,b)\in R\cup S$
⇒ $\ (a,b)\in R$ or $\ (a,b)\in S$
⇒ $\ (b,a)\in R$ or $\ (b,a)\in S$   [R and S are symmetric]
⇒ $\ (b,a)\in R\cup S$
⸫ The statement is true.


(d) If R and S are reflexive then $\ R\cap S$ is reflexive.
Solution:
Let $\ (a,b)\in R\cap S$
⇒ $\ (a,b)\in R$ and $\ (a,b)\in S$
⇒ $\ (a,a),(b,b)\in R$ and $\ (a,a),(b,b)\in S$   [R and S are reflexive]
⇒ $\ (a,a),(b,b)\in R\cap S$
⸫ The statement is true.


10. Give an example to show that the statement ‘if R and S both are transitive’ is not true. [Hint: R = {(1, 3)}, S = {(3, 2)} are transitive. But, $\ R\cup S$ = {(1, 3), (3, 2)} is not transitive]
Solution:
Let R = {(1,3)} and S = {(3,2)} be two transitive relations.
To Show: $\ R\cup S$ = {(1,3), (3,2)} is not transitive.
for $\ (1,3)\in R, (3,2)\in R   \nexists   (1,2)\in R$
⸫ $\ R\cup S$ is not transitive.



11. Let L be the set of straight lines on the rectangular cartesian plane. If we define a relation R on L as ‘ $\ x$ is perpendicular to $\ y$ for $\ (x,y)\in L$ then state whether or not R is- (i) reflexive (ii) symmetric (iii) transitive (iv) anti-symmetric.
Solution:
Here, R = {$\ (x,y):~x$ is perpendicular to $\ y$ for $\ x,y \in L$}
(i) R is not reflexive as a line cannot be perpendicular to itself, i.e., $\ (x,x)\notin R$
(ii) R is symmetric as the lines are perpendicular to each other, i.e., $\ (x,y), (y,x)\in R$
(iii) R is not transitive as the first line is not perperpendicular to the third line, i.e.,
for $\ (x,y)\in R$ and $\ (y,z)\in R   \nexists   (x,z)\in R$
(iv) R is not anti-symmetric because
for $\ (x,y)\in R$ and $\ (y,x)\in R   ⇒ x\neq y$



12. Change the definition of R in Q.11 as ‘ $\ x$ is parallel to $\ y$’ and verify the conditions.
Solution:
Here, $\ l\in R$
$\ (l,l)\in R$
⸫ R is reflexive.

Let, $\ (l_1,l_2)\in R$ ⇒ $\ (l_2,l_1)\in R$
⸫ R is symmetric.

Let, $\ (l_1,l_2)\in R$ and $\ (l_2,l_3)\in R$
⇒ $\ (l_1,l_3)\in R$
⸫ R is transitive.
⸫ R is an equivalennce relation.
But, for $\ (l_1,l_2)\in R$ and $\ (l_2,l_1)\in R ⇒ l_1\neq l_2$
⸫ R is not anti-symmetric.



13. Draw the graphs of the following relations-
(i) R = {$\ (x, y)\in R×R: y = 2x + 1$}
Solution:
Here, R = {$\ (-1,-1), (0,1), (1,3).....$}
Here, the domain of the relation is the set of real numbers R itself and so the graph will be a continuous curve as shown in the figure below.

(ii) S = {$\ (x, y)\in R×R: y ≥ x – 1$}
Solution:
Here, S = {$\ (-1,-2), (0,-1), (-1,0).....$}
Here, the domain of the relation is the set of real numbers R itself. Now we first draw the straight line $\ y=x-1$ on cartesian plane and the graph of the relation will be the line $\ y=x-1$ and the shaded area as shown in the figure below.