# Show that the statement " If $x$ is real number such that $x^3+4x=0$, then $x$ is 0." is true using contradiction method.
" If $x$ is real number such that $x^3+4x=0$, then $x$ is 0."
Let $p :$ If $x$ is real number such that $x^3+4x=0. and$q:x=0$. To show that the above statement is true using contradiction method, Let us assume that$p$is true. and$q$is not true.$i.e.,x$is a real number such that$x^3+4x=0$and$x\neq 0\Rightarrow\:\:x(x^2+4)=0\Rightarrow\:$either$x=0$or$x^2+4=0$But our assumption is$x\neq 0$and also$x^2+4$cannot be$0$since$x^2 \geq 0\:\: \forall x \in R\thereforex^3+4x$cannot be equal to zero which is contradiction.$\thereforex$should be equal to$0$.$\therefore\$ The given statement is true.