Relations

First, find the elements of set $B$ by solving the defining quadratic equation: \[ x^2 - 5x + 6 = 0 \implies (x-2)(x-3) = 0 \implies x = 2 \text{ or } x = 3 \] Since both roots are real numbers, write $B$ in roster form: \[ B = \{2, 3\} \] Now, construct the Cartesian product $A \times B$ by systematically pairing every element of $A = \{1, 2, 3\}$ with every element of $B = \{2, 3\}$: \[ A \times B = \{(1, 2), (1, 3), (2, 2), (2, 3), (3, 2), (3, 3)\} \] Calculate the cardnalities: $|A| = 3$ and $|B| = 2$. Using the product theorem: \[ |A \times B| = |A| \times |B| = 3 \times 2 = 6 \] This perfectly matches the number of ordered pairs listed explicitly above. Final Answer: $A \times B = \{(1, 2), (1, 3), (2, 2), (2, 3), (3, 2), (3, 3)\}$ with cardinality $6$.

• First coordinates: $1, 2, 3 \implies \{1, 2, 3\} \subseteq A$
• Second coordinates: $x, y \implies \{x, y\} \subseteq B$

Let us use logical containment steps to show that an arbitrary ordered pair $(r, s)$ belonging to the left-hand side implies it belongs to the right-hand side: \[ \text{Let } (r, s) \in A \times (B \cap C) \] By the definition of a Cartesian product, the first element belongs to the first set, and the second element belongs to the second set: \[ r \in A \quad \text{and} \quad s \in (B \cap C) \] Apply the definition of a set intersection to the second component: \[ r \in A \quad \text{and} \quad (s \in B \text{ and } s \in C) \] Using the logical laws of conjunction, we can distribute the statement $r \in A$ across both terms inside the parentheses: \[ (r \in A \text{ and } s \in B) \quad \text{and} \quad (r \in A \text{ and } s \in C) \] Convert these individual paired statements back into Cartesian product membership statements: \[ (r, s) \in A \times B \quad \text{and} \quad (r, s) \in A \times C \] By the definition of an intersection, since the ordered pair belongs to both product sets simultaneously, it belongs to their intersection: \[ (r, s) \in (A \times B) \cap (A \times C) \] Thus, the left-hand side is a subset of the right-hand side. Reversing the steps shows they are equal. Final Answer: Proved.

• For $x = 1$: $y = 1 + 2 = 3 \in A \implies (1, 3) \in R$
• For $x = 2$: $y = 2 + 2 = 4 \in A \implies (2, 4) \in R$
• For $x = 3$: $y = 3 + 2 = 5 \in A \implies (3, 5) \in R$
• For $x = 4$: $y = 4 + 2 = 6 \notin A \implies$ invalid pair.
• For $x = 5$: $y = 5 + 2 = 7 \notin A \implies$ invalid pair.

• Domain: The set of all first coordinates $\implies \text{Domain}(R) = \{1, 2, 3\}$.
• Range: The set of all second coordinates $\implies \text{Range}(R) = \{3, 4, 5\}$.

Step 1: Compute the cardinality of the Cartesian product $A \times B$: \[ |A \times B| = |A| \times |B| = 3 \times 2 = 6 \] Step 2: A relation is defined as any subset of the Cartesian product. Therefore, the total number of relations equals the total number of subsets that can be formed from $A \times B$: \[ \text{Total Relations} = 2^{|A \times B|} = 2^6 = 64 \] Step 3: To find the number of non-empty relations, subtract the single empty relation ($\emptyset$) from the total count: \[ \text{Non-empty Relations} = 2^6 - 1 = 64 - 1 = 63 \] Final Answer: (a) Total Relations $= 64$, (b) Non-empty Relations $= 63$.

For an ordered pair $(x, y)$ to be real and well-defined, the logarithmic function must exist. This requires the argument of the logarithm to be strictly positive: \[ x^2 - 4x + 3 > 0 \] Factor the quadratic expression: \[ (x-1)(x-3) > 0 \] Using the wavy-curve method, the product is strictly positive outside the roots $x = 1$ and $x = 3$. This gives the domain intervals: \[ x \in (-\infty, 1) \cup (3, \infty) \] Therefore, the domain of the relation is the set of all real numbers except the closed interval $[1, 3]$. Final Answer: Domain $\in (-\infty, 1) \cup (3, \infty)$.

• Part (a): The identity relation requires every element of set $A$ to be mapped only to itself. Since the elements are $1, 2,$ and $3$, we construct the diagonal ordered pairs: \[ \mathcal{I}_A = \{(1, 1), (2, 2), (3, 3)\} \] No additional non-diagonal elements can be included in an identity relation.
• Part (b): The universal relation is the complete Cartesian product $A \times A$. It includes every possible pair that can be formed from the elements: \[ R_U = \{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)\} \]

For an ordered pair $(x, y)$ to belong to the relation $R$, the real coordinates must satisfy the algebraic condition: \[ x^2 + y^2 + 1 = 0 \implies x^2 + y^2 = -1 \] Let us analyze the values on the left-hand side. For any real numbers $x, y \in \mathbb{R}$, squares of real numbers are always non-negative ($x^2 \ge 0$ and $y^2 \ge 0$). Thus, their sum must be non-negative: \[ x^2 + y^2 \ge 0 \] However, the equation requires the sum to equal $-1$, which is strictly negative. This is a mathematical contradiction over the real field. Therefore, no ordered pairs of real numbers satisfy the condition, so the relation contains zero elements. \[ R = \emptyset \] This proves that $R$ is an empty relation. Final Answer: Proved ($R = \emptyset$).

Let us check the relation against the mathematical definition of an identity relation: An identity relation $\mathcal{I}_A$ on a set $A$ must contain exactly the ordered pairs $(a, a)$ for every $a \in A$, and no other pairs. For $A = \{1, 2, 3\}$, the required identity relation is: \[ \mathcal{I}_A = \{(1, 1), (2, 2), (3, 3)\} \] Let us look at the given relation $R$: it contains all the required diagonal elements, but it also contains the non-diagonal ordered pair $(1, 2)$. Because of the presence of $(1, 2)$, $R \neq \mathcal{I}_A$. Therefore, it is not an identity relation. (Note: It is, however, classified as a reflexive relation). Final Answer: No, $R$ is not an identity relation due to the presence of the pair $(1, 2)$.

1. Reflexive Test: Check if $(a, a) \in R$ for any integer $a \in \mathbb{Z}$. Calculate the defining difference: $a - a = 0$. Since $0$ is perfectly divisible by $3$ ($0 = 3 \times 0$), the condition is met. Thus, $R$ is Reflexive.
2. Symmetric Test: Assume $(a, b) \in R$. This means $a - b$ is a multiple of $3$: \[ a - b = 3k \quad \text{where } k \in \mathbb{Z} \] Now analyze the reversed pair $(b, a)$ by factoring out $-1$: \[ b - a = -(a - b) = -3k = 3(-k) \] Since $-k$ is also an integer, $b - a$ is a multiple of 3, so $(b, a) \in R$. Thus, $R$ is Symmetric.
3. Transitive Test: Assume $(a, b) \in R$ and $(b, c) \in R$ simultaneously. This gives two equations: \[ a - b = 3k_1 \quad \text{and} \quad b - c = 3k_2, \quad \text{where } k_1, k_2 \in \mathbb{Z} \] To find the relationship between $a$ and $c$, add the two equations together: \[ (a - b) + (b - c) = 3k_1 + 3k_2 \implies a - c = 3(k_1 + k_2) \] Since $k_1 + k_2$ is an integer, the difference $a - c$ is a multiple of $3$, which means $(a, c) \in R$. Thus, $R$ is Transitive.

• We have the ordered pair $(1, 2) \in R$. Are there any ordered pairs in $R$ that start with the number 2? No.
• We have the ordered pair $(3, 4) \in R$. Are there any ordered pairs in $R$ that start with the number 4? No.

• If $a = 2$: $2 \le 2^2 \implies 2 \le 4$ (True)
• If $a = 0.5$: Let us check a fractional value between 0 and 1: \[ 0.5 \le (0.5)^2 \implies 0.5 \le 0.25 \]

1. Reflexive: For any real number $a \in \mathbb{R}$, $a \le a$ is always true. Therefore, $(a, a) \in R$ for all $a$, so the relation is reflexive.
2. Anti-Symmetric: Assume that $(a, b) \in R$ and $(b, a) \in R$ simultaneously. This gives two mathematical inequalities: \[ a \le b \quad \text{and} \quad b \le a \] The only way for both inequalities to be true at the same time is if the two numbers are exactly equal: \[ a = b \] This satisfies the definition of anti-symmetry.
3. Transitive: Assume that $(a, b) \in R$ and $(b, a) \in R$. This gives: \[ a \le b \quad \text{and} \quad b \le c \implies a \le c \] This means $(a, c) \in R$, so the relation is transitive.

By definition, the equivalence class $[a]$ is the set of all elements in the domain that are related to $a$ under the relation $R$: \[ [0] = \{x \in \mathbb{Z} : (x, 0) \in R\} \] Substitute the given definition of the relation into this set format: \[ x - 0 = 4k \implies x = 4k, \quad \text{where } k \in \mathbb{Z} \] This means the equivalence class $[0]$ is the set of all integers that are multiples of $4$. Write this in roster form: \[ [0] = \{\dots, -8, -4, 0, 4, 8, \dots\} \] Notice that this relation partitions the integers into exactly four disjoint equivalence classes: $[0], [1], [2],$ and $[3]$. Final Answer: $[0] = \{4k : k \in \mathbb{Z}\} = \{\dots, -4, 0, 4, \dots\}$.

• It is Reflexive because $(a, a) \in \mathcal{I}_A$ for all $a \in A$.
• It is Symmetric because if $(a, a) \in \mathcal{I}_A$, its reversed pair is also $(a, a) \in \mathcal{I}_A$.
• It is Anti-Symmetric because if $(a, b) \in \mathcal{I}_A$ and $(b, a) \in \mathcal{I}_A$, the only elements present are diagonal pairs where $a=b$, which satisfies the condition.
• It is Transitive because if $(a, a) \in \mathcal{I}_A$ and $(a, a) \in \mathcal{I}_A$, chaining them together returns $(a, a) \in \mathcal{I}_A$.

Step 1: Find $R^{-1}$ by reversing the coordinates of each ordered pair in $R$: \[ R^{-1} = \{(2, 1), (4, 3), (6, 5)\} \] Step 2: Find the domain and range of the original relation $R$: \[ \text{Domain}(R) = \{1, 3, 5\} \quad \text{and} \quad \text{Range}(R) = \{2, 4, 6\} \] Step 3: Find the domain and range of the inverted relation $R^{-1}$: \[ \text{Domain}(R^{-1}) = \{2, 4, 6\} \quad \text{and} \quad \text{Range}(R^{-1}) = \{1, 3, 5\} \] Comparing the sets, we can see that $\text{Domain}(R^{-1}) = \text{Range}(R)$ and $\text{Range}(R^{-1}) = \text{Domain}(R)$. The swap property is verified. Final Answer: $R^{-1} = \{(2, 1), (4, 3), (6, 5)\}$.

• Start with $1 \in \text{Domain}(R)$: we have $(1, a) \in R$. Now look for pairs in $S$ that start with $a$: we find $(a, \alpha) \in S$. Chaining them together gives: $(1, \alpha) \in S \circ R$.
• Start with $2 \in \text{Domain}(R)$: we have $(2, b) \in R$. Look for pairs in $S$ that start with $b$: we find $(b, \beta) \in S$ and $(b, \gamma) \in S$. Chaining them together gives two pairs: $(2, \beta) \in S \circ R$ and $(2, \gamma) \in S \circ R$.
• Start with $3 \in \text{Domain}(R)$: we have $(3, a) \in R$. Look for pairs in $S$ that start with $a$: we find $(a, \alpha) \in S$. Chaining them together gives: $(3, \alpha) \in S \circ R$.

Let us use double containment analysis based on the definition of coordinate swapping: \[ \text{Let } (a, b) \in (R^{-1})^{-1} \] By the definition of an inverse relation, if an ordered pair belongs to the inverse of a relation, its reversed pair must belong to the relation itself: \[ (b, a) \in R^{-1} \] Apply the inverse relation definition a second time to this statement: \[ (a, b) \in R \] This shows that any ordered pair belonging to $(R^{-1})^{-1}$ must also belong to $R$. Reversing the logical steps shows that any pair in $R$ belongs to $(R^{-1})^{-1}$. Therefore, the sets are identical. Final Answer: Proved.

• Part (a): Apply the counting formula for reflexive relations: \[ \text{Reflexive Count} = 2^{n^2 - n} \] Substitute $n = 3$ into the exponent: \[ \text{Reflexive Count} = 2^{3^2 - 3} = 2^{9 - 3} = 2^6 = 64 \]
• Part (b): Apply the counting formula for symmetric relations: \[ \text{Symmetric Count} = 2^{\frac{n(n+1)}{2}} \] Substitute $n = 3$ into the formula: \[ \text{Symmetric Count} = 2^{\frac{3(3+1)}{2}} = 2^{\frac{3 \times 4}{2}} = 2^6 = 64 \]

• For $n = 0$: $B_0 = 1$
• For $n = 1$: $B_1 = \binom{0}{0} B_0 = 1 \times 1 = 1$
• For $n = 2$: Compute $B_2$ using the values of $B_0$ and $B_1$: \[ B_2 = \binom{1}{0}B_0 + \binom{1}{1}B_1 = 1(1) + 1(1) = 2 \]
• For $n = 3$: Compute $B_3$ using the values of $B_0, B_1,$ and $B_2$: \[ B_3 = \binom{2}{0}B_0 + \binom{2}{1}B_1 + \binom{2}{2}B_2 \] Substitute the values of the binomial coefficients and the lower Bell numbers: \[ B_3 = 1(1) + 2(1) + 1(2) = 1 + 2 + 2 = 5 \]

1. The relation must be reflexive: This means all $n$ diagonal elements $(a, a)$ must be included in the relation. There is exactly $1$ choice for each diagonal position.
2. The relation must be symmetric: This means for any non-diagonal position $(a, b)$, its symmetrical counterpart $(b, a)$ must mirror its inclusion status. They must either be both included or both excluded. The total number of such non-diagonal pairs is given by: \[ \text{Number of Pairs} = \frac{n^2 - n}{2} \] Each pair has exactly $2$ choices (either both are in the relation, or both are out).