Question

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

Answer 1

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.