<s>[INST] Approximation of $\sqrt{2}$ I got the following problem in a chapter of approximations:
If $\frac{m}{n}$ is an approximation to $\sqrt{2}$ then prove that $\frac{m}{2n}+\frac{n}{m}$ is a better approximation to $\sqrt{2}.$(where $\frac{m}{n}$ is a rational number)
I am stuck in this problem. Any help will be appreciated.
 [/INST] $\frac{m}{n}$ approximately equals $\sqrt 2$
Suppose $\epsilon$ is our the error in our estimate.  i.e. $\frac{m}{n} + \epsilon = \sqrt 2$  or
$(\frac{m}{n} + \epsilon)^2 = 2$
$\frac{m^2}{n^2} + 2\frac{m}{n}\epsilon  + \epsilon^2 = 2$
Now we need to solve for $\epsilon.$
If $\epsilon$ is small then then $\epsilon^2$ is very small.  Which is a little fuzzy.  Certainly we can say that $\epsilon^2 < 2\frac{m}{n}\epsilon$ and if we solved for $\epsilon^*$ in this equation: 
$2\frac{m}{n}\epsilon^* = 2 - \frac{m^2}{n^2}$
then:
$|(\frac{m}{n} + \epsilon^*) - \sqrt 2| < |\frac{m}{n} - \sqrt2|$ 
$\epsilon^* = \frac{n}{m} - \frac{m}{2n}\\
\frac{m}{n} + \epsilon^* = \frac{n}{m} + \frac{m}{2n}$
</s>