TITLE: Approximation to $ \sqrt{2}$
QUESTION [25 upvotes]: I'm a first year Undergraduate student from India. Our professor is going to start a Real Analysis course in September and I was preparing for the initials. I tried and solved many problems, but this one has me confused. Probably the main reason for the confusion is that my book has cited it as Hardy's problem.

If $\dfrac {m}{n}$ is a good approximation to $\sqrt{2}$, prove that $\dfrac{m+2n}{m+n}$ is a better one, and that the errors in the two cases are in opposite direction. Apply this result to show that the limit of the sequence $\dfrac{1}{1}$, $\dfrac{3}{2}$,$\dfrac{7}{5}$,$\dfrac{17}{12}$,$\dfrac{41}{29}$,.... is $ \sqrt{2}$.

I need help regarding the first part of the problem, since the second part is obvious. The simpler the language, the better it is for me.

REPLY [2 votes]: A somewhat surreal and lazy solution using Mathematica to make things easier: 
(* your two guesses *) 
guess1 = m/n 
guess2 = (m+2*n)/(m+n) 

(* if you square your guesses and subtract from 2, you get signed 
closeness to 2; squaring again eliminates the sign *) 

dist1 = (guess1^2-2)^2 
(-2 + m^2/n^2)^2 

dist2 = (guess2^2-2)^2 
(m^2 - 2*n^2)^2/(m + n)^4 

(* We want dist2 < dist1; under what cases can that fail? *) 
Reduce[{dist1 <= dist2, m>0, n>0}, Reals] 
n > 0 && m == Sqrt[2]*Sqrt[n^2] 

So, the only case where dist1 is even EQUAL to dist2 is when m/n is 
Sqrt[2], which is, of course, impossible.