# 1.(ii)    Show that the statement             p: “If $x$ is a real number such that $x^3 + 4x = 0$, then $x$ is 0” is true by            (ii) method of contradiction

If $x$ is a real number such that $x^3 + 4x = 0$, then $x$ is 0 : (if p then q)

p: x is a real number such that $\dpi{100} x^3 + 4x = 0$.

q: x is 0.

In order to prove the statement “if p then q”

Contradiction:  By assuming that p is true and q is false.

So,

p is true:  There exists a real number x such that $x^3 + 4x = 0$

q is false: $\dpi{100} x \neq 0$

Now, $\dpi{100} x^3 + 4x = 0 \implies x(x^2 + 4) = 0$

$\dpi{100} \implies x = 0\ or\ (x^2 + 4)= 0$

$\dpi{100} \implies x = 0\ or\ x^2 = -4\ (not\ possible)$

Hence, x = 0

But we assumed $\dpi{100} x \neq 0$. This contradicts our assumption.

Therefore q is true.

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