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.

Explanation

Given statement : if x is a real number such that `x^3+4x=0` , then ` x=0`
Comonent p:x is a real number such that `x^3+4x=0`
Component `q:x=0`
(i) Direct method : `because x ` is a real number
and `x^3+4x=0`
`therefore x(x^2+4)=0`
`rArrx=0` or ` x^2+4=0`
But `x^2+4` cannot be zero.
`therefore x=0`
(ii) Method of contradication :
Let x is a real number and `x^3+4x=0`
let `xne0`
Now `x(x^2+4)=0`
`rArr(x^2+4)=0`
`rArrx=-` or `x^2+4=0`
But `x^2+4` cannot be zero.
`therefore x=0`
(iii) Method of contrapositive :
let `xne0`
`rArrx` is a real number and `xne0`
`therefore x^2gt0`
`rArrx^2+4gt0`
`becausexne0` and `x^2+4gt0`
`rArrxne0` and `x^2+4ne0`
`rArrx(x^2+4)ne0`
`rArrx^3+4xne0`
therefore, `xne 0 rArrx^3+4xne0`
`rArrx^3+4x=0rArr x=0`