**Question:**

Show that the statement

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

(i) direct method

(ii) method of contradiction

(iii) method of contrapositive

**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.

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