How should I prove $\forall x \in \mathbb{Q}$ where $x > \sqrt 2$ , $\exists y \in \mathbb{Q}$ where $\sqrt{2} < y < x$ How should I prove the below statement?


$\forall x \in \mathbb{Q}$ where $x > \sqrt 2$ , $\exists y \in
> \mathbb{Q}$ where $\sqrt{2} < y < x$


I tried to prove it by contradiction but I could not find a clear path.
 A: We can remove the need to think about or refer to $\mathbb R$:
Let $x^2 > 2$, then we show there exists $y$, $y^2 > 2$ and $x > y$.
New write $x = a/b$ and $2 = 2/1$, we can put $y = (a+2)/(b+1)$ and we're done!


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*https://en.wikipedia.org/wiki/Mediant_%28mathematics%29
A: Take $\{a_n\} \subset \mathbb{Q}$ such that $a_n \xrightarrow{n \to \infty} \sqrt{2}$. If you have trouble finding such a sequence (depending on how you defined $\mathbb{R}$, you might have), just take $a_n = \sqrt{2}_n + 10^{-n}$ where by $\sqrt{2}_n$ we denote the decimal expression of $\sqrt{2}$ truncated at the $n$-th decimal. (this is to say we round up the last digit)
Now, taking $\varepsilon = x - \sqrt{2}$ using the limit property you can find a $N$ such that $a_N - \sqrt{2} < \varepsilon$, so $a_N<x$ but obviously $a_N > \sqrt{2}$ and you are done.
A: We can prove the more general statement that between two real number always exists a rational number : 

given  two real numbers $a<b$, there exists a rational number $y$ such that $a<y<b$.

This a consequence of the Archimedean Principle and the Well- Ordering of reals numbers. You can see a proof here.
A: Hint: use the fact that $Q$ and $Q^c$ both are dense in $R$ and also use $Archimidean property$.
A: $\{x > \sqrt{2}\}$ is connected once is the subinteveral $(\sqrt{2},+\infty)$. By the connectness, there is $y$ as searched.
