Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $n^2 > 9$

Please help

share|improve this question
7  
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
2  
Please use the (homework) tag if this is a homework question. If it's not, can you give us a little background as to how you encountered the question? –  Eric Stucky Mar 3 '13 at 10:44
5  
-1. That's not the way to ask for help. You must show some effort, otherwise we will think that you are just trying to cheat on your homework. (A lot of people is doing just this on this site in those days, and if this goes on it will lead Math.SE to its demise.) –  Giuseppe Negro Mar 3 '13 at 11:57
add comment

3 Answers

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.

share|improve this answer
2  
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
add comment

(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.

share|improve this answer
    
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
add comment
up vote 0 down vote accepted

(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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.