Show that the statementp : “If x is a real number such that x3 + x

1.	Show that the statementp : “If x is a real number such that x3 + x = 0, then x is 0” is true by(i) Direct method(ii) method of Contrapositive(iii) method of contradiction
Answer» (i) Direct Method: Let us assume that ‘q’ and ‘r’ be the statements given by q: x is a real number such that x³+ x = 0. r: x is 0. The given statement can be written as: if q, then r. Let q be true. Then, x is a real number such that x³+ x = 0 x is a real number such that x(x²+ 1) = 0 x = 0 r is true Thus, q is true Therefore, q is true and r is true. Hence, p is true. (ii) Method of Contrapositive: Let r be false. Then, R is not true x ≠ 0, x∈R x(x²+ 1) ≠ 0, x ∈ R q is not true Thus, -r = -q Hence, p : q and r is true (iii) Method of Contradiction: If possible, let p be false. Then, P is not true -p is true -p (p => r) is true q and –r is true x is a real number such that x³+ x = 0 and x ≠ 0 x = 0 and x ≠ 0 This is a contradiction. Hence, p is true.

Show that the statementp : “If x is a real number such that x3 + x = 0, then x is 0” is true by(i) Direct method(ii) method of Contrapositive(iii) method of contradiction

Answer»

(i) Direct Method:

Let us assume that ‘q’ and ‘r’ be the statements given by

q: x is a real number such that x³+ x = 0.

r: x is 0.

The given statement can be written as:

if q, then r.

Let q be true. Then, x is a real number such that x³+ x = 0

x is a real number such that x(x²+ 1) = 0

x = 0

r is true

Thus, q is true

Therefore, q is true and r is true.

Hence, p is true.

(ii) Method of Contrapositive:

Let r be false. Then,

R is not true

x ≠ 0, x∈R

x(x²+ 1) ≠ 0, x ∈ R

q is not true

Thus, -r = -q

Hence, p : q and r is true

(iii) Method of Contradiction:

If possible, let p be false. Then,

P is not true

-p is true

-p (p => r) is true

q and –r is true

x is a real number such that x³+ x = 0 and x ≠ 0

x = 0 and x ≠ 0

This is a contradiction.

Hence, p is true.