Show that $\exists q\in\mathbb{Q}:d-c>|q-\sqrt{2}|$ where $d>c\in\mathbb{R}$ I'm trying to show that if $d-c>0$, then
$\exists q\in\mathbb{Q}:d-c>|q-\sqrt{2}|$.
In the case where $d-c>\sqrt{2}$, we have:
$$
\exists q\in\mathbb{Q}:\sqrt{2}>q>0
\implies d-c>q>q-\sqrt{2},d-c>\sqrt{2}>\sqrt{2}-q
\implies d-c>|q-\sqrt{2}|
$$
Now, I'm trying to show that the result is valid for any $d,c:d>c$. Any ideas on how to prove that using what I proved above? Or is there any other easier way to prove this?
Thanks! :D 
 A: Define the famous $(q_n)$ by: 
$$q_0 = 1, \text{and} \ \forall \ n \in \mathbb N, q_{n+1} = \frac12 \left(q_n + \frac2{q_n} \right)$$
It can be shown that $\forall \ n$,  $q_n \in \mathbb Q^{+}$ and that $q_n \to \sqrt2$.
Then, for $\epsilon = d - c>0$, $\exists$ $n_0 \in \mathbb N$, $\forall$ $n \ge n_0$, $|q_n - \sqrt2| < d - c$.
Take $q = q_{n_0}$.
A: The standard way of proving this is to explicitly construct such a number. Here is one way one could do this:


*

*If $r = d-c <\sqrt{2}$ then first find (argue for the existence of) a rational number $x$ that is smaller than $r$. 

*Now consider taking steps of length $x$ towards $\sqrt{2}$, i.e. consider the numbers $x,2x,3x,\ldots$. Show that sooner or later you are going to end up inside the interval $(\sqrt{2}-x,\sqrt{2}+x)$. In other words show that there exist a $k\in\mathbb{N}$ such that  $kx\in(\sqrt{2}-x,\sqrt{2}+x)$.

*Finally show that the rational number $q = kx$ has the desired properties.


Another approach is to consider the decimal expansion of $\sqrt{2}$ to $n$ decimal places. This is a rational number $q$. Now show that we can pick $n$ large enough (relative to the value of $r$) such that $|q-\sqrt{2}|<r$.
