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Well, I solved it, and I would like to know if there is anything that can be corrected or improved here. I think that the proof ended up too long, and with too many letters. Surely there is a better way to write it. Alternate solutions are welcome too. Thank you.

Proof: Suppose by contradiction that exists $s \in \Bbb Q$ such that $s^2 = 6$. Then, we have $s = \dfrac{p}{q}$, with $p,q \in \Bbb Z, q \neq 0, \gcd(p,q) = 1$. The standard strategy is to contradict the part about the $\gcd$. We have: $$\begin{align} \left(\frac{p}{q}\right)^2 &= 6 \\ p^2 &= 6q^2 \\ p^2 &= 2(3q^2)\end{align}$$ so $p^2$ is even, and it follows that $p$ is even, and so exists $m \in \Bbb Z$, with $p = 2m$. Proceeding, we have: $$\begin{align} (2m)^2 &= 2(3q^2) \\ 4m^2 &= 2(3q^2) \\ 2m^2 &= 3q^2\end{align}$$ From here, we have that $3q^2$ is even. If $q$ is also even, $\gcd(p,q) \neq 1$ and we're finished. Let's see the case that $q$ is odd. Then exists $\ell \in \Bbb Z$, with $q = 2 \ell + 1$. Proceeding in this case: $$\begin{align} 2m^2 &= 3(2 \ell + 1)^2 \\ 2m^2 &= 3(4 \ell^2 + 4\ell + 1) \\ 2m^2 &= 12 \ell^2 + 12 \ell + 3 \\ 2m^2 &= 2(6 \ell^2 + 6 \ell + 1) + 1 \end{align}$$ which is a contradiction, because the left-hand side is even, and the right-hand side is odd. Therefore $q$ must be even, and $\gcd(p,q) \neq 1$. Hence, there is no rational number whose square is $6$.

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  • $\begingroup$ Possible duplicate of math.stackexchange.com/questions/538324/…. $\endgroup$
    – lhf
    Jun 21, 2014 at 4:20
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    $\begingroup$ Here is a way to simplify your argument: As $ 2 $ is prime and $ 2 \mid 3 q^{2} $ but $ 2 \nmid 3 $, it follows immediately that $ 2 \mid q^{2} $, which means that $ 2 \mid q $. Hence, $ \gcd(p,q) \geq 2 $, which contradicts your initial assumption that $ \gcd(p,q) = 1 $. $\endgroup$ Jun 21, 2014 at 4:21
  • $\begingroup$ Nice. I hadn't thought of using this. You could even have posted it as an answer instead of a comment (: $\endgroup$
    – Ivo Terek
    Jun 21, 2014 at 4:23
  • $\begingroup$ Here, the Fundamental theorem of arithmetic comes in handy to see both that $2|n$ follows from $2|n^2$ and that $2 |m$ follow from $2 | 3m$ ;-) $\endgroup$ Jun 21, 2014 at 4:27
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    $\begingroup$ Instead of the rather lengthy proof that $q$ odd implies a contradiction, all you need say is that if $q$ is odd then $3q^2$ is odd, but $2m^2$ is even, so $2m^2=3q^2$ is impossible. Moreover, then very minor changes will give you a generalisation: if $k$ is odd, then there is no rational $s$ such that $s^2=2k$. $\endgroup$
    – David
    Jun 21, 2014 at 4:30

4 Answers 4

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Since you invited alternative solutions, here is one: suppose that $$\sqrt6=\frac{p}{q}\ ,$$ and take the smallest possible positive value for $q$. Then we have also $$\sqrt6=\frac{6q-2p}{p-2q}\ ;$$ but the denominator is positive and less than $q$, which is a contradiction. I leave you to fill in the details.

The interesting thing about this proof is that it really only involves order properties of the integers, and is quite independent of prime factorisation. It can easily be generalised to show that if $k$ and $n$ are positive integers with $k^2<n<(k+1)^2$, then $\sqrt n$ is irrational.

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  • $\begingroup$ That's clever. +1 I'll try to write the general form of it. Thanks again for the idea (: $\endgroup$
    – Ivo Terek
    Jun 21, 2014 at 4:51
  • $\begingroup$ I like your solution. As you have said, it only involves the well-ordering of $ \mathbb{N} $. $\endgroup$ Jun 21, 2014 at 5:08
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Here is a way to simplify your argument: As $ 2 $ is prime and $ 2 \mid 3 q^{2} $ but $ 2 \nmid 3 $, it follows immediately that $ 2 \mid q^{2} $, which means that $ 2 \mid q $. Hence, $ \gcd(p,q) \geq 2 $, which contradicts your initial assumption that $ \gcd(p,q) = 1 $.

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  • $\begingroup$ I figured that I might as well make my comment an answer. =P $\endgroup$ Jun 21, 2014 at 4:26
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This is getting really old.

Copy and paste from another answer of mine.

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 $q$.

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If $x = \sqrt 6$ were a rational number, then the polynomial $x^2 - 6$ would have a rational root. The rational root theorem says that the only possible rational roots of $x^2 - 6 = 0$ are $\pm 1, \pm 2, \pm 3,$ and $\pm 6$. Since none of these possible roots solve the equation, $\sqrt 6$ must be irrational.

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