Question -
Answer -
p:“If x is a real number such that x3 +4x = 0, then x is 0”.
Let q: x isa real number such that x3 + 4x = 0
r: x is0.
(i) To show that statement p istrue, we assume that q is true and then show that r istrue.
Therefore, let statement q betrue.
∴ x3 + 4x = 0
x (x2 +4) = 0
⇒ x = 0 or x2 + 4 = 0
However, since x isreal, it is 0.
Thus, statement r istrue.
Therefore, the given statement istrue.
(ii) To show statement p tobe true by contradiction, we assume that p is not true.
Let x be a realnumber such that x3 + 4x = 0 andlet x is not 0.
Therefore, x3 +4x = 0
x (x2 +4) = 0
x = 0or x2 + 4 = 0
x = 0or x2 = – 4
However, x isreal. Therefore, x = 0, which is a contradiction since we haveassumed that x is not 0.
Thus, the given statement p istrue.
(iii) To prove statement p tobe true by contrapositive method, we assume that r is falseand prove that q must be false.
Here, r is falseimplies 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 ≠ 0implies that the product of any positive real number with x isnot zero.
Let us consider the productof x with (x2 + 4).
∴ x (x2 + 4) ≠ 0
⇒ x3 + 4x ≠ 0
This shows that statement q isnot true.
Thus, it has been proved that
∼r ⇒ ∼q
Therefore, the givenstatement p is true.