Show that the statement $p:$ 'If $x$ is a real number such that $x^{3}+4x=0$,then $x$ is $0$' is true by the method of contrapositive.

(A) The statement $p$ is of the form 'If $q$,then $r$',where:
$q: x$ is a real number such that $x^{3}+4x=0$
$r: x=0$
To prove $p$ is true by the contrapositive method,we must show that $\sim r \Rightarrow \sim q$.
Assume $\sim r$ is true,which means $x \neq 0$.
We need to show that $\sim q$ is true,i.e.,$x^{3}+4x \neq 0$.
Since $x \neq 0$,we have $x^{2} > 0$.
Therefore,$x^{2}+4 > 4$,which implies $x^{2}+4 > 0$.
Since $x \neq 0$ and $(x^{2}+4) > 0$,their product $x(x^{2}+4)$ must be non-zero.
Thus,$x^{3}+4x \neq 0$,which is $\sim q$.
Since $\sim r \Rightarrow \sim q$ is true,the original statement $p$ is true.