Check the validity of the statements below using contradiction method
(i) p: The sum of an irrational number and a rational number is irrational
(ii) q: If $n$ is a real number with $n > 3$, then $n^2 > 9$
For (ii) Assume $n^2 \leq 9=3^2$. This is equal to: $$n^2 -9 = n^2-3^2 = (n+3) (n-3) \leq 0 $$ Now think about what is necessary to make the left hand side lower equal zero.
(i) Assume that $x$ is irrational and $y$ is rational such that $x + y$ is rational. Then the difference of the two rational numbers $(x + y) - y = x$ is also rational (since rational numbers are closed under subtraction), which is a contradiction to the hypotheses.
(ii) Assume that $n^2 \leq 9$. Then $\left|n\right|\cdot \left|n\right| \leq 9$ and thus $\left|n\right| \leq 3$. Contradiction.
(i) Let us assume that the given statement, p, is false.
(ii) Let us assume that n is real with n > 3, but n^2 > 9 is not true
n is a real no. and n > 3