Question

Let Q(x, y) be the predicate "If x < y then x2 < y2" with domain for both x and y being the set R of real numbers. a. Explain why Q(x, y) is false if x = -2 and y = 1

Answer 1

Predicate Q(x, y) is false for x = -2 and y = 1 because, although -2 is less than 1, the square of -2 is not less than the square of 1.To show this, we substitute the values x = -2 and y = 1 into the predicate: "If -2 < 1 then (-2)^2 < 1^2".

Since -2 is not less than 1, the statement "-2 < 1" is false, and therefore, the entire predicate "If -2 < 1 then (-2)^2 < 1^2" is false.

Step-by-step explanation:

The predicate Q(x, y) is "If x < y then x2 < y2". To determine if this is true when x = -2 and y = 1, we need to first check if the first condition (x < y) is true, and then if the second condition (x2 < y2) is also true. For x = -2 and y = 1, it is evident that -2 < 1, so the first condition is true. However, when we calculate the squares of both x and y we get (-2)2 = 4 and 12 = 1. This means that 4 is not less than 1, so the second condition is false. Therefore, predicate Q(-2, 1) is false.