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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    $\begingroup$ Did you consider that one or both of the $d',d''$ may be negative ? $\endgroup$
    – user42761
    Commented Nov 19, 2012 at 6:48
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    $\begingroup$ 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 $(m+2n)/(m+n)=7/5<\sqrt2$. $\endgroup$ Commented Nov 19, 2012 at 6:49

3 Answers 3


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.

  • $\begingroup$ 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? $\endgroup$
    – EuYu
    Commented Nov 19, 2012 at 7:11
  • $\begingroup$ @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. $\endgroup$ Commented Nov 19, 2012 at 7:22

Assume $\dfrac mn\ne\sqrt2;$ otherwise $\dfrac mn$ is $\sqrt2$, not an approximation.

Then $d'\ne0$ so we can compute $\dfrac {d''}{d'}=\dfrac{\dfrac{m+2n}{m+n}-\sqrt2}{\dfrac mn-\sqrt2}= \dfrac n{m+n}\dfrac{m+2n-\sqrt2(m+n)}{m-\sqrt2n}$

$=\dfrac n{m+n}\dfrac{m-\sqrt2n-\sqrt2(m-\sqrt2n)}{m-\sqrt2n}=\dfrac {1}{1+\dfrac mn}\left(1-\sqrt2\right).$

We could assume $\dfrac mn\ge0$ (otherwise $\dfrac mn$ is not "a good approximation of $\sqrt2$"),

and $-1<1-\sqrt2<0$ since $1<\sqrt2<2$.

From here it is easy to see that $|d''|<|d'|,$ and $d''$ and $d'$ have opposite signs.

  • 1
    $\begingroup$ So this technique gets that the sequence is alternating. (+1) $\endgroup$ Commented Jul 15, 2019 at 1:37
  • $\begingroup$ Starting with any rational number $p_0 = m_0/n_0 \gt 0$, the sequence $p_{k+1} = (m_k+2n_k)/(m_k+n_k)$ is well-defined. Can you 'bring it home' in your answer and show that $\lim_{k\to +\infty} p_k = \sqrt 2$? When I see 'good approximation' in the OP's question, I think of the quote $\quad$ Capt. Ross : Please the court, is there a question anywhere in our future? $\quad$/ A Few Good Men (1992) $\endgroup$ Commented Jul 15, 2019 at 11:04
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    $\begingroup$ @CopyPasteIt: sandwich: if $m_0,n_0>0, $ then $m_k,n_k>0,$ and $|p_k-\sqrt2|<(1-\sqrt2)^k$ and $\lim_{k\to\infty}(1-\sqrt2)^k=0$ $\endgroup$ Commented Jul 15, 2019 at 12:48

First Proof of Convergence

Although not explicitly asked for by the OP, a convergent sequence can be defined.

Let $F(x) = \frac{x+2}{x+1}$ and $p_0 \gt 0$ a rational number. Define the sequence

$$\tag 1 p_{k+1} = F(p_k), \quad \text{ for } k \ge 0$$

Now J. W. Tanner's answer tells us that $(p_k)$ is alternating about $\sqrt 2$ with each term getting closer.

To show convergence we can also assume that $p_0 \lt \sqrt 2$, so that we have an increasing sub-sequence

$$\tag 2 {(p_{2k})}_{\, k \ge 0} \text{ with each } p_{2k} \lt \sqrt 2$$

But to skip the odd entries, we can construct the function

$$\tag 3 G(x) = F \circ F(x) = \frac{3x+4}{2x+3}$$

so that

$$\tag 4 p_{2(k+1)} = \frac{3p_{2k}+4}{2p_{2k}+3}$$

The increasing bounded sequence $\text{(2)}$ has a limit $\alpha \gt 0$ and the following is also true,

$$\tag 5 \lim_{k\to +\infty} p_{2k} - p_{2(k+1)} = \lim_{k\to +\infty} p_{2k} - \frac{3p_{2k}+4}{2p_{2k}+3} = 0$$

Using more basic properties of limits, we can also write

$$\tag 5 \alpha - \frac{3\alpha+4}{2\alpha+3} = 0$$

and simple algebra shows that $\alpha = \sqrt 2$.

Using the fact that each next term is better as we alternate around $\sqrt2$, we must conclude that the starting sequence $\text{(1)}$ also converges to it.

Second Proof of Convergence

Following J. W. Tanner's hints (see comments), there is a simpler way to prove convergence via the following identity,

$\tag 6 \text{For every } x \in \Bbb R \setminus \{-1\}, \quad F(x) - \sqrt 2 = (x-\sqrt2) \left(\frac{1}{1+x}\right)\left(1-\sqrt2\right)$

We can use $\text{(6)}$ since the terms of our sequence $(p_k)$ contain only positive numbers. Moreover, to show convergence we can assume that $|p_0 - \sqrt 2| \lt 1$. But then

$\tag 7 \text{For every } k \ge 0, \quad |p_{k+1}-\sqrt2|<(\sqrt2 -1)^k$

and we have convergence.

  • $\begingroup$ I don't understand what you say about good approximation. But it is necessary $-$ if $m/n$ is negative, the next term can be worse (or even infinite). $\endgroup$
    – TonyK
    Commented Jul 15, 2019 at 16:18
  • $\begingroup$ @TonyK Fair enough - I deleted the first two lines... $\endgroup$ Commented Jul 15, 2019 at 16:25
  • $\begingroup$ Did you see my comment to your comment on my post? $\\ \sqrt2$ is also a fixed point of $x\mapsto\dfrac{x+2}{x+1}$ $\endgroup$ Commented Jul 16, 2019 at 0:38
  • $\begingroup$ @J.W.Tanner Yes. Thanks. I tried using your hint, but gave up. So I guess there is some general fixed point theory that points the way to convergence? Not to mention continued-fractions and other formulations... $\endgroup$ Commented Jul 16, 2019 at 1:40
  • $\begingroup$ @J.W.Tanner Added your proof - thanks for the technique! $\endgroup$ Commented Jul 16, 2019 at 12:23

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