Types of Relations
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Q1
Let X be a family of sets and R be a relation on X defined by 'A is disjoint from B'. Then R is
Marks:1Answer:
symmetric.
Explanation:
Clearly, the relation is symmetric but it is neither reflexive nor transitive. 
Q2
Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}. The relation R is
Marks:1Answer:
not symmetric.
Explanation:
(2, 3) R but (3, 2) R
R is not symmetric

Q3
Let R be a relation on a set A such that R = R^{1}, then R is
Marks:1Answer:
symmetric.
Explanation:
Let (a, b) R. Then,
(a, b) R (b, a) R^{1} [by defn of R^{1}]
(b, a) R [ R = R^{1}]
So, R is symmetric. 
Q4
Let R = {(2, 3), (3, 4)} be a relation defined on the set A = {1, 2, 3, 4}. The minimum number of ordered pairs required to be added in R so that the new relation becomes an equivalence relation is
Marks:1Answer:
8
Explanation:
Given R = {(2,3), (3,4)} ...(i)
To make it reflexive, we add the following ordered pairs in R:
R = {1, 1), (2, 2), (3, 3), (4, 4), (2, 3), (3,4) ...(ii)
Hence 4 ordered pairs are added.
To make it symmetric, we again add the following ordered pairs in R from eq^{n} (ii)
{(1,1), (2,2), (3,3), (4,4), (2,3) (3,2), (3,4) (4,3)} ...(iii)
Hence two more ordered pairs are added.
Finally to make it transitive, we add the following ordered pairs in R from eq^{n} (iii)
{(1,1), (2,2), (3,3), (4,4), (2,3) (3,2), (3,4) (4,3), (2,4), (4,2)}
Hence, two more ordered pairs are added.
Total 8 ordered pairs must be added to make the relation R an equivalence relation.

Q5
For real number x any y, we write xRy $\iff xy+\sqrt{2}$ is an irrational number. Then the relation R is
Marks:1Answer:
reflexive only
Explanation:
$Wehave\forall x\in R,xx+\sqrt{2}=\sqrt{2}$, an irrational number.
Hence, xRx $\forall $ x, i.e. R is reflexive.
R cannot be symmetric because if x = $\sqrt{2}$, y = 1, then
$xy+\sqrt{2}=2\sqrt{2}1,anirrationalnumber,i.e.,xRy.$
$Butyx+\sqrt{2}=1\sqrt{2}+\sqrt{2}=1,arationalnumber.$
$Thus,xRy\overline{)\Rightarrow}yRx$
Similarly, we may show that R is not transitive.