Show that the statement p: “If x is a real number such that x3 + 4x = 0, then x is 0” is true by (i) direct method (ii) method of contradiction (iii) method of contrapositive

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p: “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. ∴ x3 + 4x = 0 x (x2 + 4) = 0 ⇒...
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p: “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. ∴ x3 + 4x = 0 x (x2 + 4) = 0 ⇒ x = 0 or x2 + 4 = 0 However, 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 or x2 = – 4 However, 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 ≠ 0 This shows that statement q is not true. Thus, it has been proved that ∼r ⇒ ∼q Therefore, the given statement p is true. read less
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