(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.
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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.
