Questions · Page 2 of 2

M.C.Q (1 Marks)

MCQ 511 Mark
Let $A=\{1,2,3,4\}$ and $R$ be a relation in $A$ given by $R=\{(1,1),(2,2),(3,3),(4,4),(1,2)$, $(2,1),(3,1)\}$. Then, $R$ is
  • Reflexive only
  • B
    Symmetric only
  • C
    An equivalence relation
  • D
    None of these
Answer
Correct option: A.
Reflexive only
(a) : Reflexive: $(1,1),(2,2),(3,3),(4,4) \in R$;
$\therefore \quad R$ is reflexive.
View full question & answer
MCQ 521 Mark
Let $L$ denote the set of all straight lines in a plane. Let a relation $R$ be defined by $\alpha R \beta \Leftrightarrow \alpha \perp \beta, \alpha, \beta \in L$. Then, $R$ is
  • A
    Reflexive only
  • Symmetric only
  • C
    Transitive only
  • D
    None of these
Answer
Correct option: B.
Symmetric only
(b) : Given, $\alpha R \beta \Leftrightarrow \alpha \perp \beta \Leftrightarrow \beta \perp \alpha \Rightarrow \beta R \alpha$ Hence, $R$ is symmetric.
View full question & answer
MCQ 531 Mark
The maximum number of equivalence relations on the set $A=\{1,2,3\}$ are
  • A
    1
  • B
    2
  • C
    3
  • 5
Answer
Correct option: D.
5
(d): The smallest equivalence relation is the identity relation $R_1=\{(1,1),(2,2),(3,3)\}$
Then, two ordered pairs of two distinct elements can be added to give three more equivalence relations.
$
R_2=\{(1,1),(2,2),(3,3),(1,2),(2,1)\}
$
Similarly $R_3$ and $R_4$.
Finally the largest equivalence relation, that is the universal relation.
$
\begin{array}{l}
R_5=\{(1,1),(2,2),(3,3),(1,2),(2,1),(1,3),(3,1),(2,3), \\
(3,2)\}
\end{array}
$
View full question & answer
MCQ 541 Mark
Let $T$ be the set of all triangles in the Euclidean plane, and let a relation $R$ on $T$ be defined as $a R b$ if $a$ is congruent to $b \forall a, b \in T$. Then $R$ is
  • A
    reflexive but not transitive
  • B
    transitive but not symmetric
  • an equivalence relation
  • D
    None of these
Answer
Correct option: C.
an equivalence relation
(c) : (i) We know that every triangle is congruent to itself.
$\therefore \quad\left(T_1, T_1\right) \in R$ for all $T_1 \in T$. Thus, $R$ is reflexive.
(ii) Let $\left(T_1, T_2\right) \in R$
$\Rightarrow \quad T_1$ is congruent to $T_2$.
$\Rightarrow \quad T_2$ is congruent to $T_1$.
$\therefore \quad\left(T_2, T_1\right) \in R$
Thus, $R$ is symmetric.
(iii) Let $\left(T_1, T_2\right) \in R$ and $\left(T_2, T_3\right) \in R$.
$\Rightarrow \quad T_1$ is congruent to $T_2$ and $T_2$ is congruent to $T_3$.
$\therefore \quad T_1$ is congruent to $T_3$
$\Rightarrow \quad\left(T_1, T_3\right) \in R$.
Thus, $R$ is transitive.
$\therefore \quad R$ is an equivalence relation.
View full question & answer
MCQ 551 Mark
For the set $A=\{1,2,3\}$, define a relation $R$ on the set $A$ as follows :
$R=\{(1,1),(2,2),(3,3),(1,3)\}$
How many ordered pairs to be added to $R$ to make it the smallest equivalence relation?
  • 1
  • B
    2
  • C
    3
  • D
    4
Answer
Correct option: A.
1
(a): Here, $A=\{1,2,3\}$ and the relation $R=\{(1,1),(2,2)(3,3)(1,3)\}$.
Clearly, $R$ is reflexive but not symmetric as $(1,3) \in R$ but $(3,1) \notin R$.
We shall include $(3,1)$ to the above relation to make it smallest equivalence relation
$
R^{\prime}=\{(1,1),(2,2),(3,3)(1,3),(3,1)\} \text {. }
$
$R^{\prime}$ is certainly transitive as transitivity is not contradicted.
View full question & answer