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Suppose $k>1$ is an integer, and k is not a square number, then $\sqrt{k}$ is not a rational number.

Proof:

Let $\sqrt{k}=\frac{p}{q}$, and $(p,q)=1$,So $q^2|p^2$, $p\neq 1$, $k$ is not an integer.When $q=p=1$, and $k>1$.

And $q=1$, then $\sqrt{k}=p$, and $k$ is a square number.

I do think it's so easy...Where is wrong?

How to do it.

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marked as duplicate by vadim123, Jonas Meyer, Amzoti, anorton, Brandon Carter Jul 23 '13 at 3:13

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.

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Sorry, but this is nonsense. The English doesn't even make sense. Can you rewrite your proof carefully? –  Billy Jul 23 '13 at 2:44
    
What do you mean "$k$ is not an integer"? –  marty cohen Jul 23 '13 at 2:51
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1 Answer 1

up vote 2 down vote accepted

In short: I think you started correctly, but made some odd jumps. I'm going to offer some critique of the proof--please don't take anything here personally. :)

A proof is more than just a string of symbols--rather, it must be a clearly (yet concisely) written work that allows the reader a peak inside your mind when you proved the proposition.

So, I would suggest:

  • Whenever you start a proof by contradiction, make sure to denote it.

  • Always "introduce" your variables. Are you assuming $k$ is an integer? A complex number? What about $p$ and $q$?

  • Don't use notation unless it really helps. Sometimes it's easier to say (and nearly always easier to read) "$p$ and $q$ coprime," rather than "$(p, q) = 1$".
  • Show your algebra, or at least mention that you're doing some. Don't make me think about why $\sqrt{k} = \frac{p}{q}$ implies $q^2|p^2$--show me from the definition of divides.

So, a start to the proof could be:

Proof:
Assume proposition is false. That is, there exists a $k \in \Bbb{Z}^+$ such that $k>1$ is not a perfect square and $\sqrt{k}$ is rational.

As $\sqrt{k}$ is rational, there exist coprime integers $p$ and $q$ ($q\ne0$) such that: $$\sqrt{k} = \frac{p}{q}$$

Rearranging the above, we find that: $$p = q\sqrt{k}$$ It follows: $$p^2 = q^2k$$ ...

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