- #1
Keen94
- 41
- 1
1. Find P'={x I ~px} for the given open sentences px.
#25. x2≥4.
(Problem from 1.10, Principles of Mathematics by Allendoerfer and Oakley.
Solution offered at the back of the book: {x I -4<x<4}.
If P={x∈ℝ I px} then P'={x∈ℝ I ~px}[/B]
x2≥4 ⇒ x≤-2 or x≥2.
P={x∈ℝ I x≤-2 or x≥2}. P'={x∈ℝ I -2<x<2}[/B]
The original proposition is true when a number is equal to or less than -2. It is equally true when it is equal to or greater than 2. If we negate the proposition then the elements of this set will be the ones not found in the original set. This leaves the interval (-2,2). I don't understand why the interval would be (-4,4) as the solution found at the back of the book suggests. BTW First Post!
#25. x2≥4.
(Problem from 1.10, Principles of Mathematics by Allendoerfer and Oakley.
Solution offered at the back of the book: {x I -4<x<4}.
Homework Equations
If P={x∈ℝ I px} then P'={x∈ℝ I ~px}[/B]
The Attempt at a Solution
x2≥4 ⇒ x≤-2 or x≥2.
P={x∈ℝ I x≤-2 or x≥2}. P'={x∈ℝ I -2<x<2}[/B]
The original proposition is true when a number is equal to or less than -2. It is equally true when it is equal to or greater than 2. If we negate the proposition then the elements of this set will be the ones not found in the original set. This leaves the interval (-2,2). I don't understand why the interval would be (-4,4) as the solution found at the back of the book suggests. BTW First Post!
