Let P (x), Q (x), R (x) and S (x) denote the following predicates with domain Z:

P (x) : x ≤ 0,

Q (x) : x2 = 1,

R (x) : x is odd

S (x) : x = x + 1.

For each predicate, determine its truth value

Answer: Hi!, first, Z are the integer numbers, so we only will work with them.

P (x) : x ≤ 0

ok, this predicate is true if x is less or equal tan 0, and false if x is greater than 0.

so P (x) is true if { x∈Z, x ≤ 0}

Q (x) : x2 = 1

Q (x) is true only if 2*x = 1. now, this means that if x=1/2 is true, but 1/2 isnt an integer, then Q (x) is false ∀ x ∈ Z.

R (x) : x is odd

R (x) is true if x is odd, we can write odd numbers as x = 2k + 1, where k is a random integer; then:

R (x) is true if x=2k + 1, with k∈Z.

S (x) : x = x + 1

S (x) is true if x = x+1, if we subtract x from both sides of the equality, we get that S (x) is true if 0=1, and this is absurd, then:

S (x) is false ∀ x ∈ Z.