For problems involving relations, carefully consider each property (reflexive, sym metric, transitive) separately. List out the required ordered pairs and enumerate the possible relations
Let \( A = \{1, 2, 3\} \).
Step 1: Reflexive Property
For the relation to be reflexive:
\( (1,1), (2,2), (3,3) \in R \)
Step 2: Transitive Property
For the relation to be transitive:
\( (1,2) \text{ and } (2,3) \in R \implies (1,3) \in R \)
Step 3: Symmetric Property
The relation is not symmetric because:
\( (2, 1) \in R \text{ but } (3,2) \notin R \)
Relations:
\( R_1 = \{(1,1), (2,2), (3,3), (1,2), (2,3), (1,3)\} \)
\( R_2 = \{(1,1), (2,2), (3,3), (1,2), (2,3), (1,3), (2,1)\} \)
\( R_3 = \{(1,1), (2,2), (3,3), (1,2), (2,3), (1,3), (3,2)\} \)
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: