Show that the statement

Solution:

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

Let $q$ : $x$ is a real number such that $x^{3}+4 x=0$

r: x is 0.

(i) To show that statement p is true, we assume that q is true and then show that r is true.

Therefore, let statement q be true.

$\therefore x^{3}+4 x=0$

$x\left(x^{2}+4\right)=0$

      $\Rightarrow x=0$ or $x^{2}+4=0$

However, since x is real, it is 0.

Thus, statement r is true.

Therefore, the given statement is true.

(ii) To show statement p to be true by contradiction, we assume that p is not true.

Let $x$ be a real number such that $x^{3}+4 x=0$ and let $x$ is not 0 .

Therefore, $x^{3}+4 x=0$

$x\left(x^{2}+4\right)=0$

$x=0$ or $x^{2}+4=0$

$x=0$ or $x^{2}=-4$

However, x is real. Therefore, x = 0, which is a contradiction since we have assumed that x is not 0.

Thus, the given statement p is true.

(iii) To prove statement p to be true by contrapositive method, we assume that r is false and prove that q must be false.

Here, r is false implies that it is required to consider the negation of statement r. This obtains the following statement.

$\sim r: x$ is not $0 .$

It can be seen that $\left(x^{2}+4\right)$ will always be positive.

$x \neq 0$ implies that the product of any positive real number with $x$ is not zero.

$\therefore x\left(x^{2}+4\right) \neq 0$

$\Rightarrow x^{3}+4 x \neq 0$

This shows that statement q is not true.

Thus, it has been proved that

$\sim r \Rightarrow \sim q$

Therefore, the given statement p is true.