Question

Show that the statement
$p$ : "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

Answer 1

$p$ : "If $x$ is a real number such that $x^3+4x=0$ then $x$ is $0$"
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.
$\therefore x^3+4x=0$
$x(x^2+4)=0$
$\Rightarrow x=0or{x}^2+4=0$
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 $x^3+4x=0$ 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,
$\sim r:xisnot0$
It can be seen that $(x^2+4)$ will always be positive.
x$\ne 0$ implies that the product of any positive real number with $x$ is not zero.
Let us consider the product of $x$ with $(x^2+4)$
$\therefore x(x^2+4)\ne 0$
$\Rightarrow x^3+4x\ne 0$
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.