Proof about approximations near $\sqrt{2} + 1$? If $x = \sqrt2 + 1$ then $x = \frac{1}{x} + 2$. 


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*Given this, how can I prove that if $y > 1$ is some approximation to $x = \sqrt2 + 1$ then $ ( \frac{1}{y} + 2 )$ is a closer approximation to $x$ than $y$ is?

*Why one would expect, at the intuitive level, that $ ( \frac{1}{y} + 2 )$ is a closer approximation to $x$ than $y$ is?


I guess you are expected to notice that, since $x>1$ and $y>1$,
\begin{align}
\left|\left(\frac1y+2\right)-x\right|&=\left|\frac1y+2-\left(\frac1x+2\right)\right|\\ \ \\
&=\left|\frac1y-\frac1x\right|\\ \ \\
&=\frac{|y-x|}{xy}<|y-x|.
\end{align}
