Check the validity of the statement given below by the method specified against it.
$q:$ If $n$ is a real number with $n > 3$,then $n^{2} > 9$ (by contradiction method).

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(N/A) The given statement $q$ is: If $n$ is a real number with $n > 3$,then $n^{2} > 9$.
To prove this by contradiction,we assume the negation of the statement.
Let us assume that $n$ is a real number with $n > 3$,but $n^{2} \leq 9$.
Since $n > 3$ and $n$ is a real number,we can square both sides of the inequality $n > 3$ without changing the inequality sign.
$n^{2} > 3^{2}$
$n^{2} > 9$
This contradicts our assumption that $n^{2} \leq 9$.
Since the assumption leads to a contradiction,the original statement must be true. Thus,if $n$ is a real number with $n > 3$,then $n^{2} > 9$.

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