Question

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

(i) direct method

(ii) method of contradiction

(iii) method of contrapositive

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Solution

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

$r:x$ 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.

$∴x_{3}+4x=0$

$x(x_{2}+4)=0$

$⇒x=0orx_{2}+4=0$

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 $x_{3}+4x=0$ and let $x$ is not $0$.

Therefore $x_{3}+4x=0$

$x(x_{2}+4)=0$

$x=0$ 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,

$∼r:xisnot0$

It can be seen that $(x_{2}+4)$ will always be positive.

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

Let us consider the product of $x$ with $(x_{2}+4)$

$∴x(x_{2}+4)=0$

$⇒x_{3}+4x=0$

This shows that statement $q$ is not true.

Thus it has been proved that

$∼r⇒∼q$

Therefore the given statement $p$ is true.

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