Question

Using conradiction method, check the validity of the following statement if n is a real number with $n>3,then,n^2>9.$

Answer 1

Let n be a real number such that n>3 and if possible, let $n^2$ be not greater than 9.

Then ,$n^2\le 9.$

Let $n^2=(9-a),$ where a is a real number such that $a\ge 0.$

Then $n=\sqrt{n^2}=\sqrt{9-a}\le 3[\because (9-a)\le 9).]$

But, this is a contradiction since n>3.

Since the contradiction arises by assuming that $n^2$ is not greater than 9. therefore $n^2>9.$

Hence, the given statement is valid.