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Can someone help me out here? I'm trying to prove this the same way as whether if $\sqrt 2$ is irrational, but i'm not sure what am i doing. Can someone show me how to prove this?

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    $\begingroup$ if this is not a duplicate then I am irrational... $\endgroup$ – Riemann-bitcoin. Feb 2 '17 at 23:37
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    $\begingroup$ @Riemann-bitcoin. And if I cannot find said duplicate, I hope to be transcendental. $\endgroup$ – Simply Beautiful Art Feb 2 '17 at 23:38
  • $\begingroup$ @C.Chan What are you trying, specifically? $\endgroup$ – Carl Schildkraut Feb 2 '17 at 23:41
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    $\begingroup$ Already answered at $\sqrt a$ is either an integer or an irrational number. Guess that's good news for Riemann-bitcoin ;-) $\endgroup$ – dxiv Feb 2 '17 at 23:41
  • $\begingroup$ a ... i am a idiot... but i didnot read the question name... $\endgroup$ – Riemann-bitcoin. Feb 2 '17 at 23:44
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In fact, every integer which is not a square has an irrational square root. Can you see how to generalize the result for $\sqrt{2}$ to prove this?

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Once again, here is a proof that if $n$ is a positive integer that is not a square of an integer, then $\sqrt{n}$ is irrational.

Let $k$ be such that $k^2 < n < (k+1)^2$. Suppose $\sqrt{n}$ is rational. Then there is a smallest positive integer $q$ such that $\sqrt{n} = p/q$.

Then $\sqrt{n} = \sqrt{n}\frac{\sqrt{n}-k}{\sqrt{n}-k} = \frac{n-k\sqrt{n}}{\sqrt{n}-k} = \frac{n-kp/q}{p/q-k} = \frac{nq-kp}{p-kq} $.

Since $k < \sqrt{n} < k+1$, $k < p/q < k+1$, or $kq < p < (k+1)q$, so $0 < p-kq < q$. We have thus found a representation of $\sqrt{n}$ with a smaller denominator, which contradicts the specification of $q$.

Note: This is certainly not original - but I had fun working it out based on the proof I know that $\sqrt{2}$ is irrational.

Note 2: It is interesting that this does not use any divisibility properties.

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For $\forall n \in \mathbb{N}$, either $ \sqrt{n} \in \mathbb{N}$ or $\sqrt{n} \in \mathbb{R} \setminus \mathbb{Q}$.

There are two possible cases:

  1. If $n = p^2$, $p \in \mathbb{N} \Rightarrow \sqrt{n} = p ∈ N$

  2. If $n \ne p^2$ or there is no such $p \in \mathbb{N}$ that would satisfy $n = p^2$, then $\sqrt{n}$ is irrational. Let's suppose the contrary, i.e. there $\exists p,q \in \mathbb{N}$ and $\gcd(p,q) = 1$ such that $$\sqrt{n}=\frac{p}{q} \Leftrightarrow n\cdot q^2 = p^2$$ From $\gcd(p,q) = 1 \Rightarrow \gcd(p^2,q^2) = 1$. Using Bézout’s theorem $\exists z,t \in \mathbb{Z}$ such that $$z·p^2 + t·q^2 = 1 \Leftrightarrow z·n·q^2 + t·q^2 = 1 \Leftrightarrow q^2·(z·n + t) = 1$$ which means $q^2$ divides $1$, but this is possible only if $q = 1$. As a result $n = p^2$ - contradiction with the initial assumption.

This proves the statement (also available here).

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Take $\sqrt a=\frac{p}{q}$ with gcd(p,q)=1 and a is prime square both sides... $$ a=\frac{p^2}{q^2} $$ then $$ 2q^2=p^2 $$ then $$ a|p^2 $$ a contradiction

then think about what would happen if a was not prime

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    $\begingroup$ did you read the question? $\endgroup$ – Arnaldo Feb 2 '17 at 23:39
  • $\begingroup$ I know how to prove sqrt(2) is irrational, its just im doing the same way for this question, but i'm confused. $\endgroup$ – Pikaninja Feb 2 '17 at 23:41
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    $\begingroup$ i guess that makes me irrational... $\endgroup$ – Riemann-bitcoin. Feb 2 '17 at 23:46

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