In the set of real numbers, the relation \( R \) defined by \( R = \{(a, b) : a \leq b^2 \} \) is:
Step 1: Check if the relation is reflexive: For reflexivity, \( a \leq a^2 \) must hold for all real numbers. This does not hold for all values of \( a \) (e.g., for \( a = -1 \), \( -1 \leq (-1)^2 \) is false), so the relation is not reflexive.
Step 2: Check if the relation is symmetric: The relation is not symmetric because if \( a \leq b^2 \), it does not imply that \( b \leq a^2 \) in general. For example, if \( a = 2 \) and \( b = 1 \), then \( 2 \leq 1^2 \) holds, but \( 1 \leq 2^2 \) does not hold.
Step 3: Check if the relation is transitive: The relation is transitive. If \( a \leq b^2 \) and \( b \leq c^2 \), then we can prove \( a \leq c^2 \) holds. Therefore, the correct answer is (A) not reflexive and symmetric, but transitive.
Let $ A $ be the set of all functions $ f: \mathbb{Z} \to \mathbb{Z} $ and $ R $ be a relation on $ A $ such that $$ R = \{ (f, g) : f(0) = g(1) \text{ and } f(1) = g(0) \} $$ Then $ R $ is:
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: