<s>[INST] Proof that doesn't exists a rational $s$ such that $s^2 = 6$ 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 $6$.
 [/INST] 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 $q$.
</s>