Let \(Z\) be the set of all integers, and consider the set: \[ X = \{(x, y) \mid x^2 + 2y^2 = 3, x, y \in Z\}, \quad Y = \{(x, y) \mid x > y, x, y \in Z\} \] Then the number of elements in \(X \cap Y\) is:
Step-by-step Solution:
Given the set \( X \) and \( Y \), we have the following: Set \( X \): \[ X = \left\{ (x, y) : x^2 + 2y^2 = 3 \right\} \] This represents the set of points on the ellipse \( \frac{x^2}{3} + \frac{y^2}{3/2} = 1 \). \[\] The solutions to this equation, given integer values for \( x \) and \( y \), are: \[ X = \{ (1, 1), (-1, 1), (1, -1), (-1, -1) \} \] Set \( Y \): \[ Y = \{ (x, y) : x > y, x, y \in \mathbb{Z} \} \] This set contains all pairs of integers \( (x, y) \) where \( x > y \). \[\] Thus, the intersection \( X \cap Y \) is: \[ X \cap Y = \{ (1, -1) \} \]