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

Question

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.

Answer 1

Hint: use the fact that $Q$ and $Q^c$ both are dense in $R$ and also use $Archimidean property$.

Answer 2

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.

Answer 3

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.

Answer 4

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!

• https://en.wikipedia.org/wiki/Mediant_%28mathematics%29

Answer 5

$\{x > \sqrt{2}\}$ is connected once is the subinteveral $(\sqrt{2},+\infty)$. By the connectness, there is $y$ as searched.