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

Statement p: "If x is a real number such that $x^3 + 4x = 0$, then x is 0."

1. Direct method: $x^3 + 4x = 0 \Rightarrow x(x^2 + 4) = 0$. So either $x = 0$ or $x^2 + 4 = 0 \Rightarrow x^2 = -4$. But $x^2 \ge 0$ for every real x, so $x^2 = -4$ is impossible for real x. Hence the only real solution is $x = 0$. p is true.
2. Method of contradiction: Assume $x^3 + 4x = 0$ but $x \ne 0$. Since $x \ne 0$, divide both sides of $x(x^2+4)=0$ by x: $x^2 + 4 = 0 \Rightarrow x^2 = -4$. This is impossible for a real number x (squares are never negative) — a contradiction. Hence our assumption $x \ne 0$ is false, so $x = 0$. p is true.
3. Method of contrapositive: Contrapositive of p: "If $x \ne 0$, then $x^3 + 4x \ne 0$." …

Direct: factor and use that x² ≥ 0. Contradiction: assume x ≠ 0 and x³+4x=0 simultaneously, derive x²=−4, impossible. Contrapositive: assume x≠0, show x …

• Forgetting that x² + 4 is always strictly positive (never zero) for real x, which is the key fact all three methods rely on. …