If x is a real number such that x3 + 4x = 0 then x is 0”.Let q: x is a real number such that x3 + 4x = 0 r: x is 0.i To show that statement p is true we assume that q is true and then show that r is true.Therefore let statement q be true.∴ x2 + 4x = 0 x x2 + 4 = 0⇒ x = 0 or x2+ 4 = 0However since x is real it is 0.Thus statement r is true.Therefore the given statement is true.ii To show statement p to be true by contradiction we assume that p is not true.Let x be a real number such that x3 + 4x = 0 and let x is not 0.Therefore x3 + 4x = 0 x x2+ 4 = 0 x = 0 or x2 + 4 = 0 x = 0 orx2 = – 4However x is real. Therefore x = 0 which is a contradiction since we have assumed that x is not 0.Thus the given statement p is true.iii To prove statement p to be true by contrapositive method we assume that r is false and prove that q must be false.Here r is false implies that it is required to consider the negation of statement r.This obtains the following statement.∼r: x is not 0.It can be seen that x2 + 4 will always be positive.x ≠ 0 implies that the product of any positive real number with x is not zero.Let us consider the product of x with x2 + 4.∴ x x2 + 4 ≠ 0⇒ x3 + 4x ≠ 0This shows that statement q is not true.Thus it has been proved that∼r ⇒∼qTherefore the given statement p is true.

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