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 contra positive.
(i) Direct method.
(ii) Method of contradiction.
(iii) Method of contra positive.
(i) Direct method: x3 + 4x = 0 or x(x2 + 4) = 0
Now x2 + 4 ≠ 0 and x ∈ R and hence x = 0.
(ii) Method of Contradiction: Let x ≠ 0 and let x = p, p ∈ R is a root of x3 + 4x = 0. Therefore p3 + 4p = 0
(or) p(p2 + 4) = 0 as p = 0 Thus p2 + 4 = 0 which is not possible. Therefore, our supposition is wrong. Hence p = 0 or x = 0
(iii) Contra positive method: p is not true
Let x = 0 is not true Let x = p ≠ 0
Therefore p3 + 4p = 0
P being the root of x2 + 4 = 0
(or) p(p2 + 4) = 0
Now p = 0, also (p2 + 4) = 0
Which implies p(p2 + 4) = 0
Now p = 0, also p2 + 4 = 0
p(p2 + 4) ≠ 0 if p is not true.
Hence x = 0 is the root of x3 + 4x = 0
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