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Claim: If $m/n$ is a good approximation of $\sqrt{2}$ then $(m+2n)/(m+n)$ is better.

My attempt at the proof:

Let d be the distance between $\sqrt{2}$ and some estimate, s.

So we have $d=s-\sqrt{2}$

Define $d'=m/n-\sqrt{2}$ and $d''=(m+2n)/(m+n)-\sqrt{2}$

To prove the claim, show $d''<d'$

Substituting in for d' and d'' yields:


This result doesn't make sense to me, and I was wondering whether there is an other way I could approach the proof or if I am missing something.

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Did you consider that one or both of the $d',d''$ may be negative ? – Epsilon Nov 19 '12 at 6:48
The problem with your approach is that you are assuming that both $m/n$ and $(m+2n)/(m+n)$ are larger than $\sqrt2$, while this needs not be the case. In fact, the approximations will alternate. For example: If we start with $m/n=1<\sqrt2$, then $(m+2n)/(m+n)=3/2>\sqrt2$, while if $m/n=3/2$, then $(m+2n)/(m+n)=7/5<\sqrt2$. – Andrés E. Caicedo Nov 19 '12 at 6:49
up vote 3 down vote accepted

Hint: Compare $\left|\dfrac{m^2}{n^2}-2\right|$ with $\left|\dfrac{(m+2n)^2}{(m+n)^2}-2\right|$. We need to take absolute values, because if one approximation is too big, the other turns out to be too small, and vice-versa.

Bring the expressions to the denominators $n^2$ and $(m+n)^2$ respectively. So the first becomes $\left|\dfrac{m^2-2n^2}{n^2}\right|$.

Make sure to expand the squares in the second one. The second one will simplify an awful lot: I will leave the pleasure to you. The result will jump out.

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I'm a little concerned that we're comparing the differences of the squares. This is fine if $\frac{m}{n}$ is an overestimate of $\sqrt{2}$ in which case $$\left|\frac{m}{n} + \sqrt{2}\right| > \left|\frac{m+2n}{m+n} + \sqrt{2}\right|$$ and the inequality holds, but is there anything we can do when $\frac{m}{n}$ is an underestimate? – EuYu Nov 19 '12 at 7:11
@EuYu I would suggest trying what is suggested. If the square is closer to 2, the square root will be closer to the square root of 2. – Mark Bennet Nov 19 '12 at 7:22

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