Question -
Answer -
(i) DirectMethod:
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) Methodof 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) Methodof 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 = 0and x≠ 0
x =0 and x≠0
This is a contradiction.
Hence, p is true.