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I need to prove there is no rational number p/q whose square is 2.I think there may be an easy way to prove this by contradiction,But i cannot solve it.Please help

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marked as duplicate by Dominic Michaelis, T. Bongers, Stefan4024, mrf, Norbert Nov 4 '13 at 8:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
A much more general answer you can find here –  Dominic Michaelis Nov 4 '13 at 6:17
    
@DominicMichaelis Im looking for a specific answer –  techno Nov 4 '13 at 6:26
    
as I see you have been to lazy to click on the link, the highest voted answer gives a link to a pdf whichs very first proof is the irrationality of $\sqrt{2}$ and then proceed to generalize. –  Dominic Michaelis Nov 4 '13 at 6:28
    
@DominicMichaelis I have seen some pdf's but they say like according to section 2.1 we can say.... so i need extra proof –  techno Nov 4 '13 at 6:32

2 Answers 2

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Suppose there does exist a rational number $m/n$ whose square is $2$.

This means there are integers $m,n$ such that $$2=\frac{m^2}{n^2}$$

Take $\frac{m}{n}$ to be in lowest terms (this means that $m,n$ have no common factors greater than $1$).

Hence $$m^2 = 2n^2$$

$m^2 =2n^2$ is even, therefore, $m$ is even. Hence, we can write $m=2k$, where $k$ is an integer. Then $$m^2=4k^2=2n^2$$.

Consequently $n^2=2k^2$. So $n^2$ is even, and again, this means $n$ is also even.

We have now shown that both $m$ and $n$ are even. However this means that the fraction $\frac{m}{n}$ is not in lowest terms. This is a contradiction. Therefore, $\sqrt{2}$ is not rational.

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Thanks :).But my question comes here how can you say (m/n)^2=2 –  techno Nov 4 '13 at 6:21
    
Because assume there are m,n such that (m/n)^2 = 2 - you'll never be able to get m/n in lowest terms. Perhaps think of it as a proof by infinite descent instead of a proof by contradiction. No matter how much you cancel down m/n, it'll always have a multiple of 2 in the numerator and the denominator. This is clearly absurd. –  LTS Nov 4 '13 at 6:23
    
Thinking about it as a proof by infinite descent also helps, as it leads you to think that m and n must be tending to $\infty$, since we can cancel it down an infinite number of times, and are never done. This makes sense since $\sqrt{2}$ is an infinite non-repeating decimal, so any such fraction $m/n$ would have to be infinitely long to contain all the digits. –  LTS Nov 4 '13 at 6:26
    
Im a math noob. All these wont get into my brain.Your answer seems to give better explanation and the final contradiction is stated clearly,so i think i will choose your answer :) –  techno Nov 4 '13 at 6:30
    
en.wikipedia.org/wiki/Square_root_of_2#Proofs_of_irrationality Read through this very slowly, don't worry if it seems difficult. I remember finding it difficult to make sense of at first. –  LTS Nov 4 '13 at 6:34

Suppose there is a rational number $p/q$ , written in lowest terms such that

$$ \frac{p^{2}}{q^{2}} = 2$$.

Then $ p^{2} = 2q^{2} \implies p^{2}$ is even. Hence $p$ is even.

Thus, $\,p = 2k \implies p^{2} = 4k^{2} \implies q^{2} = 2k^{2} \implies q$ is even.

This is a contradiction (to what? Do you see?)

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You are saying p and q are relatively prime right? –  techno Nov 4 '13 at 6:17
    
@techno Yeah. That is the meaning of being in lowest terms. –  Vishal Nov 4 '13 at 6:18
    
This might be stupid.But how can you say p2/q2=2 –  techno Nov 4 '13 at 6:19
    
I did not get. What are you asking? If you are asking how I got $p^{2}q^{2} = 2$, then I did not claim that anywhere. If you meant $p^{2}/q^{2} = 2$, then that was the question. A rational number whose square is $2$. –  Vishal Nov 4 '13 at 6:21
    
I meant the second one,i dont know to type in the math :).Plus my bad this was already given in the question.A complete math noob here :) –  techno Nov 4 '13 at 6:24

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