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 $n^2 > 9$

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Do you know what the question is asking? What have you tried? Do you want help or just the answer? – Arjang Mar 3 '13 at 10:42

Hint:
For (i) Assume $p \in \mathbb{Q}, \ q\in \mathbb{R} \setminus \mathbb{Q}$ and $p+q\in \mathbb{Q}$ as $\mathbb{Q}$ is a field, is $p+q-p$ in $\mathbb{Q}$?

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.

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Be careful with your "axiom" for (ii), check it with $a=1,b=-2,c=1,d=-2$. – Hagen von Eitzen Mar 3 '13 at 10:55

(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.

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Prove by contradiction – chndn Mar 3 '13 at 11:06
Thanks. I've updated my answer to use contradiction in both cases. – azimut Mar 3 '13 at 16:05
@DominicMichaelis: Thanks for spotting this! It's corrected now. – azimut Mar 3 '13 at 22:17

(i) Let us assume that the given statement, p, is false.
Therefore sqrt a + b/c = d/e , where root a is irrational and b,c,d,e are integers
d/e - b/c is rational and root a is irrational
Hence our assumption is wrong and the statement correct.

(ii) Let us assume that n is real with n > 3, but n^2 > 9 is not true
That is n^2 < 9

n is a real no. and n > 3
=> n^2 > 9 , which is a contradiction
Hence the statement is correct

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