a.p(x): x^3 ≥ 1
x^3 ≥ 1 => x ≥ 1 => all the integers equal or greater than 1 = { x ∈ Z | x ≥ 1} = {1, 2, 3, 4, 5, 6, ...}
That is the same that the natural numbers except 0 = N - {0}.
b.q(x): x2 = 2
x^2 = 2 => x = +/- √2, which is not an integer, so the truth set is the empty set = { } =∅
c.r(x): x < x2
x < x^2 => x^2 - x > 0
=> x (x - 1) > 0
=>
1) x > 0 and x - 1 > 0
=> x > 0 and x > 1
=> x > 1
2) x < 0 and x - 1 < 0
=> x < 0 and x < 1 => x < 0
=> the solution is the union of the two sets: x > 1 ∪ x < 0
That is all the integers except 0 and 1.
=> the truth set is { x ∈ Z | x < 0 or x > 1} = Z - { 0,1}
