Question

1.Show that the statement p: “If x is a real number such that x3 + 4x = 0, then x is 0” is true by

Answer 1

(i) Direct method: Let x3 + 4x = 0, x ∈ R

⇒ x(x2 + 4) = 0, x ∈ R

⇒ x = 0 ∵ x2 + 4 = 0

⇒ x = ±√(-4) ∉ R

⇒ p is a true statement.

(ii) Method of contradiction: Let x ≠ O, x ∈ R i.e., Let x be a nonzero real number

⇒ x2 > 0 (y square of non-zero real number is always positive)

⇒ x2 + 4 > 0 + 4

⇒ x2+ 4 > 4 x2 + 4 ≠ 0

⇒ x(x2 + 4) ≠ 0

⇒ x3 + 4x = 0, which is a contradiction for x3 + 4x = 0.

Hence x = 0 ⇒ p is true .

(iii) Method of contrapositive ∵ p: “If x is a real number such that x3 + 4x = 0, then x is O”. Let q : “x is a real number and x3 + 4x = 0'' r:”x is 0'' ∴ p is q

⇒ r Its contrapositive is ~ r

⇒ ~ q i.e., “I-x is non-zero real number, then x3 + 4x is also non-zero”. Now, x ∈ R x 0

⇒ x2 >0

⇒ x2 + 4 > 0 + 4

⇒ x2 + 4 > 4

⇒ x2 + 4 ≠ 0

⇒ x(x2 + 4)≠ 0

⇒ x3 + 4x ≠ 0 i.e., ~ r

⇒~ q is always true. Hence q ⇒ r is true, i.e., p is true