Q.Consider the statement p: If x is a real number such that x³+4x=0, then x=0, prove that p is a true statement, using a) method of contradiction and b) method of contrapositive.

Contradiction assumes $x\ne0$ satisfies $x^3+4x=0$ and derives an impossible real equation; contrapositive directly shows $x\ne0\implies x^3+4x\ne0$.

Statement $p$: If $x\in\mathbb R$ and $x^3+4x=0$, then $x=0$.

a) Method of contradiction. Assume $p$ is false: suppose there exists a real $x$ with $x^3+4x=0$ and $x\ne0$.

Factor: $x^3+4x=x(x^2+4)=0$. Since $x\ne0$ (by assumption), we must have $x^2+4=0$, i.e. $x^2=-4$.

But for any real $x$, $x^2\ge0$, so $x^2=-4$ is impossible. This contradicts our assumption, so the assumption is false. Hence $p$ is true.

b) Method of contrapositive. The contrapositive of 'if $x^3+4x=0$ then $x=0$' is 'if $x\ne0$ then $x^3+4x\ne0$' — logically equivalent to $p$.

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