Page No 342:
Question 1:
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
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.
