Mock Test 2 — Relations

Let $R$ be a relation defined on the set of all natural numbers $\mathbb{N}$ by the rule $aRb \iff a \mid b$ ($a$ is a factor of $b$). This relation is classified as a:

The total number of symmetric relations that can be defined on a finite set containing exactly 4 elements is:

Let $R = \{(1, 2), (2, 3)\}$ be a relation on the set $A = \{1, 2, 3\}$. To make $R$ transitive, what is the minimum number of ordered pairs that must be added to it?

If a relation $R$ is symmetric and transitive simultaneously, then its inverse relation $R^{-1}$ must be:

The equivalence relation $a \equiv b \pmod 5$ defined on the set of all integers partitions $\mathbb{Z}$ into how many disjoint equivalence classes?

The total number of binary relations on a set with $n$ elements that are both reflexive and symmetric simultaneously is given by the formula:

Let $R$ and $S$ be two equivalence relations on a non-empty set $A$. Then their intersection $R \cap S$ is always:

If the domain of a relation $R$ is identical to its range ($\text{Domain}(R) = \text{Range}(R)$), then the relation:

The composition relation $R \circ R^{-1}$ for an identity relation $\mathcal{I}_A$ evaluates to:

If a relation $R$ on a set $A$ is reflexive, then the identity relation $\mathcal{I}_A$ and $R$ maintain which relationship?

Find the total number of distinct equivalence relations that can be formed on a set containing exactly 4 elements (the $4^{\text{th}}$ Bell number).

Let $A = \{1, 2, 3\}$. Find the number of distinct reflexive relations that can be defined on $A$.

Let $R = \{(x, y) \in \mathbb{Z} \times \mathbb{Z} : x^2 + y^2 \le 2\}$. Find the total number of ordered pairs inside this relation set.

Assertion (A): The empty relation $\emptyset$ on a non-empty set $A$ is symmetric and transitive, but it is not reflexive.
Reason (R): For a relation to be reflexive, every element must be related to itself, which requires the presence of all diagonal pairs $(a, a)$, but the empty set contains no elements at all.

Assertion (A): If $R$ and $S$ are two equivalence relations on a set $A$, then their union $R \cup S$ is always an equivalence relation.
Reason (R): The union of two transitive relations can fail to be transitive because it can introduce chaining paths without including the required shortcut endpoints.