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
(i) The given statement is of the form if p then q
Here
p: x is aa real number such that x3 + 4x = 0.
q: x is 0
In Direct Method,
we assume p is true, and prove q is true
So,
Let x be a real number such that x3 + 4x = 0, prove that x = 0
p is true
Consider
x3 + 4x = 0 where x is real
⇒ x(x2 + 4) = 0
⇒ x = 0 or x2 + 4 = 0
⇒ x = 0 or x2 = – 4
⇒ x = 0 or x =
x = is not possible because it is given that x is real.
Hence x = 0 only
⇒ q is true
Hence proved
(ii) p: If x is a real number such that x3 + 4x = 0, then x is 0
Let us assume
x3 + 4x = 0
but x ≠ 0
Solving x3 + 4x = 0 where x is real
⇒ x(x2 + 4) = 0
⇒ x = 0 or x2 + 4 = 0
⇒ x = 0 or x2 = – 4
⇒ x = 0 or x =
x = is not possible because it is given that x is real number
Hence, only solution is x = 0 but we take x ≠ 0
Hence we get a contradiction
Hence our assumption is wrong
Hence x is real number such that x3 + 4x = 0 then x is 0.
(iii) Let p: if x is real number such that x3 + 4x = 0
q: x is 0
The above statement is of the form if p then q
By method of Contrapositive
By assuming that q is false, prove that p must be false
Let q is false
i.e. x is not equal to 0
i.e. x ≠ 0
i.e. x × (positive number) ≠ 0 × (positive number)
i.e. x × (x2 + 4) ≠ 0 × (x2 + 4)
i.e. x3 + 4x ≠ 0
i.e. p is false
Hence x is real number such that x3 + 4x = 0 then x is 0.