The given compound statement is of the form “if p then q”
p: x ∈ R such that x3 + 4x = 0
q: x = 0
(i) Direct method:
We assume that p is true, then
x ∈ R such that x3 + 4x = 0
⇒ x ∈ R such that x(x2 + 4) = 0
⇒ x ∈ R such that x = 0 or x2 + 4 = 0
⇒ x = 0 ⇒ q is true.
So, when p is true, q is true.
Thus, the given compound statement is true.
(ii) Method of contradiction :
We assume that p is true and q is false, then
x ∈ R such that x3 + 4x = 0
⇒ x ∈ R such that x(x2 + 4) = 0
⇒ x ∈ R such that x = 0 or x2 + 4 = 0
⇒ x = 0.
which is a contradiction. So, our assumption that x ≠ 0 is false. Thus, the given compound statement is true.
(iii) Method of contrapositive: We assume that q is false, then x ≠ 0
x ∈ R such that x3 + 4x = 0
⇒ x ∈ R such that x = 0 or x2 + 4 = 0
∴ statement q is false, so x ≠ 0. So, we have,
x ∈ R such that x2 = –2
Which is not true for any x ∈ R.
⇒ p is false
So, when q is false, p is false.
Thus, the given compound statement is true.