Chapter 5: Relations

We are given $A = \{1, 2\} \implies n(A) = 2$ and $B = \{3, 4, 5\} \implies n(B) = 3$.

The number of elements in the Cartesian product $A × B$ is: $$n(A × B) = n(A) × n(B) = 2 × 3 = 6$$

A relation from $A$ to $B$ is defined as any subset of $A × B$.

Since the number of subsets of a set with $m$ elements is $2^m$, the total number of relations from $A$ to $B$ is: $$2^{n(A × B)} = 2^6 = 64$$

Therefore, the correct option is **(b)**.

By definition of Cartesian product $B × A = \{(x, y) : x \in B, y \in A\}$:

The first elements of the ordered pairs belong to set $B$: $$B = \{a, b\}$$

The second elements of the ordered pairs belong to set $A$: $$A = \{1, 5\}$$

Thus, $A = \{1, 5\}$ and $B = \{a, b\}$.

Therefore, the correct option is **(a)**.

Given $A = \{1, 2, 3, 4\}$. We partition $A$ into odd and even elements:

Odd elements: $O = \{1, 3\}$, Even elements: $E = \{2, 4\}$.

The difference $a - b$ is even if and only if both $a$ and $b$ have the same parity (both odd or both even).

1. Both odd (pairs from $O × O$): $\{(1, 1), (1, 3), (3, 1), (3, 3)\} \implies 2 × 2 = 4$ pairs.

2. Both even (pairs from $E × E$): $\{(2, 2), (2, 4), (4, 2), (4, 4)\} \implies 2 × 2 = 4$ pairs.

Total elements in $R = 4 + 4 = 8$.

Therefore, the correct option is **(b)**.

We use the algebraic property of Cartesian products: $$(A × B) \cap (C × D) = (A \cap C) × (B \cap D)$$

Setting $C = B$ and $D = A$: $$(A × B) \cap (B × A) = (A \cap B) × (B \cap A)$$

Since $A = \{1, 2, 3\}$ and $B = \{2, 3, 4\}$, their intersection is: $$A \cap B = \{2, 3\}$$

Thus, $(A × B) \cap (B × A) = \{2, 3\} × \{2, 3\} = \{(2, 2), (2, 3), (3, 2), (3, 3)\}$.

Therefore, the correct option is **(d)**.

Given $n(A) = 3$ and $n(B) = 4$.

The cardinality of the Cartesian product is the product of their cardinalities: $$n(A × B) = n(A) × n(B) = 3 × 4 = 12$$

Therefore, the correct option is **(c)**.

**Step 1: Analyzing the condition $x^2 = y^2$**

$$x^2 - y^2 = 0 \implies (x - y)(x + y) = 0 \implies y = x or y = -x$$

**Step 2: Roster Form of $R$**

For every integer $x \in \mathbb{Z}$, the related elements are $x$ and $-x$.

$$R = \{(0, 0), (1, 1), (1, -1), (-1, 1), (-1, -1), (2, 2), (2, -2), (-2, 2), (-2, -2), \dots\}$$

In set-builder / roster notation: $$R = \{(x, x), (x, -x) : x \in \mathbb{Z}\}$$

**Step 3: Domain and Range**

• **Domain of $R$**: The set of all first coordinates is all of $\mathbb{Z}$ (the set of all integers).

• **Range of $R$**: The set of all second coordinates is also all of $\mathbb{Z}$.

**Step 1: Finding ordered pairs in $R$**

The defining condition is $4x - y = 0 \implies y = 4x$, where both $x, y \in A = \{1, 2, \dots, 16\}$.

• For $x = 1 \implies y = 4(1) = 4 \in A$

• For $x = 2 \implies y = 4(2) = 8 \in A$

• For $x = 3 \implies y = 4(3) = 12 \in A$

• For $x = 4 \implies y = 4(4) = 16 \in A$

• For $x ≥ 5 \implies y = 4x ≥ 20 \notin A$.

Therefore, $$R = \{(1, 4), (2, 8), (3, 12), (4, 16)\}$$

**Step 2: Identifying Domain, Range, and Codomain**

• **Domain of $R$**: $\{x : (x, y) \in R\} = \{1, 2, 3, 4\}$

• **Range of $R$**: $\{y : (x, y) \in R\} = \{4, 8, 12, 16\}$

• **Codomain of $R$**: The entire target set $A = \{1, 2, 3, \dots, 16\}$.

**Step 1: Analyzing the condition $a - b \in \mathbb{Z}$**

Every element in $A$ is an integer ($1, 2, 3, 4, 5, 6$).

The difference between any two integers is always an integer ($a - b \in \mathbb{Z}$ for all $a, b \in A$).

Therefore, every pair $(a, b) \in A × A$ satisfies the condition.

Hence, $R = A × A$ (which is the **Universal Relation** on set $A$).

**Step 2: Domain and Range**

• **Domain of $R$**: $A = \{1, 2, 3, 4, 5, 6\}$

• **Range of $R$**: $A = \{1, 2, 3, 4, 5, 6\}$.

**Part (i): Checking Symmetry ($(a, b) \in R \implies (b, a) \in R$)**

• Let $(a, b) \in R$. By definition, $a - b \in \mathbb{Z}$.

• Let $a - b = k$, where $k$ is an integer ($k \in \mathbb{Z}$).

• Then, $b - a = -(a - b) = -k$.

• Since $k \in \mathbb{Z}$, its negative $-k$ is also an integer ($-k \in \mathbb{Z}$).

• Hence, $b - a \in \mathbb{Z} \implies (b, a) \in R$.

• **Conclusion**: Yes, $(a, b) \in R \implies (b, a) \in R$. Thus, the relation $R$ is **Symmetric**.

**Part (ii): Checking Transitivity ($(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$)**

• Let $(a, b) \in R$ and $(b, c) \in R$.

• By definition, $a - b = k_1$ and $b - c = k_2$, where $k_1, k_2 \in \mathbb{Z}$.

• Adding the two equations: $$(a - b) + (b - c) = k_1 + k_2 \implies a - c = k_1 + k_2$$

• Since the sum of two integers $k_1 + k_2$ is always an integer, $a - c \in \mathbb{Z}$.

• Hence, $(a, c) \in R$.

• **Conclusion**: Yes, $(a, b) \in R and (b, c) \in R \implies (a, c) \in R$. Thus, the relation $R$ is **Transitive**.

**Part (i): Domain and Range of "Route"**

• **Domain** (set of origin zones / first elements): $\{R, A, S, H, U\} = Z$

• **Range** (set of destination zones / second elements): $\{A, S, H, U, R\} = Z$

**Part (ii): Number of elements in $B × C$**

• $B = \{R, A, S\} \implies n(B) = 3$

• $C = \{H, U\} \implies n(C) = 2$

• $$n(B × C) = n(B) × n(C) = 3 × 2 = 6$$

**Part (iii): Roster form of $B × C$ & existing routes**

• $$B × C = \{(R, H), (R, U), (A, H), (A, U), (S, H), (S, U)\}$$

• Comparing with $Route = \{(R, A), (R, S), (A, H), (S, H), (S, U), (H, U), (U, R)\}$, the overlapping pairs are: $$(A, H), (S, H), (S, U)$$

• Thus, **3 combinations** are already present in the Route relation.

**Part (iv): Routes not currently available in Route**

• Subtracting existing routes from $B × C$: $$(B × C) \setminus Route = \{(R, H), (R, U), (A, U)\}$$

• The 3 new routes to be added are: **(Railway Station, Hospital)**, **(Railway Station, University)**, and **(Airport, University)**.

• **Checking Assertion (A)**: Here $n(A) = 2$ and $n(B) = 2$. Then $n(A × B) = 2 × 2 = 4$. The number of relations is $2^{n(A × B)} = 2^4 = 16$. Thus, **Assertion (A) is true**.

• **Checking Reason (R)**: A relation is any subset of $A × B$. The total number of subsets of a set with $n$ elements is $2^n$. Thus, **Reason (R) is true**.

• Since Reason (R) directly gives the formula used to calculate Assertion (A), **R is the correct explanation of A**.

Therefore, the correct option is **(a)**.

• **Checking Assertion (A)**: $n(A × B) = n(A) × n(B) = n(B) × n(A) = n(B × A)$. Cardinalities are always equal. Thus, **Assertion (A) is true**.

• **Checking Reason (R)**: In general, $A × B \ne B × A$ unless $A = B$, because an ordered pair $(a, b) \ne (b, a)$ when $a \ne b$. Thus, **Reason (R) is false**.

Therefore, Assertion (A) is true but Reason (R) is false, so the correct option is **(c)**.

• **Checking Reason (R)**: By definition, $Domain(R) = \{a : (a, b) \in R\}$ (first components) and $Range(R) = \{b : (a, b) \in R\}$ (second components). Thus, **Reason (R) is true**.

• **Checking Assertion (A)**: For $R = \{(1, 2), (2, 3)\}$, first components are $\{1, 2\}$ and second components are $\{2, 3\}$. Hence $Domain(R) = \{1, 2\}$ and $Range(R) = \{2, 3\}$. Thus, **Assertion (A) is true**.

• Reason (R) correctly provides the exact rule and definition justifying Assertion (A).

Therefore, the correct option is **(a)**.