Q. 14.5( 15 Votes )

# Show that the statement p: “If x is a real number such that x^{3} + 4x = 0, then x is 0” is true by

(i) direct method

(ii) method of contradiction

(iii) method of contrapositive

Answer :

(i) The given statement is of the form if p then q

Here

p: x is aa real number such that x^{3} + 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 x^{3} + 4x = 0, prove that x = 0

p is true

Consider

x^{3} + 4x = 0 where x is real

⇒ x(x^{2} + 4) = 0

⇒ x = 0 or x^{2} + 4 = 0

⇒ x = 0 or x^{2} = – 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 x^{3} + 4x = 0, then x is 0

Let us assume

x^{3} + 4x = 0

but x ≠ 0

Solving x^{3} + 4x = 0 where x is real

⇒ x(x^{2} + 4) = 0

⇒ x = 0 or x^{2} + 4 = 0

⇒ x = 0 or x^{2} = – 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 x^{3} + 4x = 0 then x is 0.

(iii) Let p: if x is real number such that x^{3} + 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 × (x^{2} + 4) ≠ 0 × (x^{2} + 4)

i.e. x^{3} + 4x ≠ 0

i.e. p is false

Hence x is real number such that x^{3} + 4x = 0 then x is 0.

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