1. **Problem Statement:** Given the set $A = \{3,6,9,12\}$ and a binary relation $R$ on $A$ defined by $R = \{(a,b) \mid a,b \in A, (a+b) \text{ divisible by } 6\}$. 2. **Find all elements of $R$.** We check all pairs $(a,b)$ in $A \times A$ where $a+b$ is divisible by 6. - $(3,3): 3+3=6$ divisible by 6 - $(3,9): 3+9=12$ divisible by 6 - $(6,6): 6+6=12$ divisible by 6 - $(6,12): 6+12=18$ divisible by 6 - $(9,3): 9+3=12$ divisible by 6 - $(9,9): 9+9=18$ divisible by 6 - $(12,6): 12+6=18$ divisible by 6 - $(12,12): 12+12=24$ divisible by 6 So, $R = \{(3,3),(3,9),(6,6),(6,12),(9,3),(9,9),(12,6),(12,12)\}$. 3. **Determine if $R$ is an equivalence relation.** An equivalence relation must be reflexive, symmetric, and transitive. - Reflexive: For all $a \in A$, $(a,a) \in R$? - $(3,3)$ in $R$? Yes. - $(6,6)$ in $R$? Yes. - $(9,9)$ in $R$? Yes. - $(12,12)$ in $R$? Yes. So, reflexive. - Symmetric: If $(a,b) \in R$, then $(b,a) \in R$? - $(3,9)$ in $R$, is $(9,3)$ in $R$? Yes. - $(6,12)$ in $R$, is $(12,6)$ in $R$? Yes. So, symmetric. - Transitive: If $(a,b) \in R$ and $(b,c) \in R$, then $(a,c) \in R$? Check example: - $(3,9)$ and $(9,3)$ in $R$, is $(3,3)$ in $R$? Yes. - $(6,12)$ and $(12,6)$ in $R$, is $(6,6)$ in $R$? Yes. So, transitive. Therefore, $R$ is an equivalence relation. 4. **List all equivalence classes of $A$.** Equivalence classes group elements related to each other. - Class of 3: $[3] = \{x \in A \mid (3,x) \in R\} = \{3,9\}$ - Class of 6: $[6] = \{6,12\}$ - Class of 9: $[9] = \{3,9\}$ (same as $[3]$) - Class of 12: $[12] = \{6,12\}$ (same as $[6]$) So, equivalence classes are $\{3,9\}$ and $\{6,12\}$. 5. **Determine the partition of $A$.** The partition is the set of distinct equivalence classes: $$\{\{3,9\}, \{6,12\}\}$$ --- 6. **Evaluate** $$\frac{\sqrt{-38}}{\text{trunc}[1.3] - 1.53 \Big\lfloor \sqrt{1} \Big\rfloor}$$ - $\sqrt{-38}$ is imaginary: $\sqrt{-38} = i\sqrt{38}$ - $\text{trunc}[1.3] = 1$ - $\sqrt{1} = 1$, so $\lfloor \sqrt{1} \rfloor = 1$ - Denominator: $1 - 1.53 \times 1 = 1 - 1.53 = -0.53$ So, $$\frac{i\sqrt{38}}{-0.53} = -\frac{i\sqrt{38}}{0.53}$$ 7. **Evaluate** $$\text{trunc} \frac{1.8}{4} + \bigl( |-4.1| \bigr) [6 - \sqrt{9.6}]$$ - $\frac{1.8}{4} = 0.45$ - $\text{trunc}(0.45) = 0$ - $|-4.1| = 4.1$ - $\sqrt{9.6} \approx 3.098$ (approximate) - $6 - 3.098 = 2.902$ Calculate: $$0 + 4.1 \times 2.902 = 11.8882$$ **Final answers:** - i) $R = \{(3,3),(3,9),(6,6),(6,12),(9,3),(9,9),(12,6),(12,12)\}$ - ii) $R$ is an equivalence relation because it is reflexive, symmetric, and transitive. - iii) Equivalence classes: $\{3,9\}$ and $\{6,12\}$ - iv) Partition of $A$: $\{\{3,9\}, \{6,12\}\}$ - b) $\displaystyle -\frac{i\sqrt{38}}{0.53}$ - c) $\approx 11.8882$