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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 $\dfrac{1}{1}$, $\dfrac{3}{2}$,$\dfrac{7}{5}$,$\dfrac{17}{12}$,$\dfrac{41}{29}$, $\dots$ 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.

Alma Arjuna
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Gaurav Tiwari
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  • It seems to me as iteration method for finding fixed point for the function $f(x)=\frac{x+2}{x+1}$. (Since for $x=\frac mn$ you have $\frac{m+2n}{m+n}=\frac{x+2}{x+1}$.) – Martin Sleziak Aug 28 '11 at 14:00
  • I googled and found that this question is some-how connected to Continued Fraction. @Martin It Means that we're to prove $\dfrac{\sqrt{2}+2}{\sqrt{2}+1}$ is a better approximation to $\dfrac{\sqrt{2}}{1}$ than $\dfrac{m}{n}=\dfrac{\sqrt{2}}{1}$. Am I understanding wrong? – Gaurav Tiwari Aug 28 '11 at 14:05
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    @gaurav: How do you know that the sequence you obtain starting with $1/1$ converges to $\sqrt2$? A priori, the upper and lower approximations could converge to numbers strictly larger and strictly lower than $\sqrt2$, respectively. My point is, the first part of the problem should give you a "rate of convergence" that you may want to use in order to prove the second part. – Andrés E. Caicedo Aug 28 '11 at 14:20
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    @gaurav: $\frac{\sqrt2+2}{\sqrt2+1}=\sqrt2=\frac{\sqrt2}1$.\ You're correct about the connection with continued fractions, but I think that you need to be familiar with theory of continuous fractions first if you want to use them for this problem. – Martin Sleziak Aug 28 '11 at 14:26
  • @martin Is there any better online resource than this Wikipedia article? Or I [would] have to learn it from my high school book, Higher Algebra by Hall & Knight. – Gaurav Tiwari Aug 28 '11 at 14:34
  • @gaurav: I know only very little about continued fractions. I have used Khinchin's book to learn some basics about them.\ This question http://math.stackexchange.com/questions/14617/proving-the-continued-fraction-representation-of-sqrt2 seems to be more-or-less different formulation of your question, so you might have a look at it. – Martin Sleziak Aug 28 '11 at 15:02
  • @martin You [all] are wonderful. I did never think that Math.SE be such a cool place. Forgive me If I did anything wrong in my very first question. If Ragib 's answer was extremely easy to understand, André 's one was straight to the topic. Thanks! – Gaurav Tiwari Aug 28 '11 at 15:36

6 Answers6

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Hint $ $ Notice that, $ $since $\,\color{blue}{\dfrac{m}n}\:\color{#c00}{\dfrac{2\:\!n}m}=2,\,$ one fraction is less than $\,\sqrt{2}\,$ and the other greater. Further note that their $\rm\color{#0a0}{mediant}$ $\,\dfrac{\color{blue}m+\color{#c00}{2n}^{\phantom{|^{|^|}}}\!\!\!}{\color{blue}n+\color{#c00}m}\,$ lies (strictly) between them - see below.

If $\large \,0<\color{blue}{\frac{a}b}<\color{#c00}{\frac{c}d}$ then their mediant $\large \,\color{#0a0}{\frac{a+c}{b+d}}\,$ lies between them, since it is the slope of the $\rm\color{#0a0}{diagonal}$ of the parallelogram with sides being the vectors $\color{blue}{(b,a)},\ \color{#c00}{(d,c)}.\:$

$\quad$ enter image description here

Bill Dubuque
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Suppose $\frac{m}{n} $ is slightly bigger than $\sqrt{2}$, so that we can write $\frac{m}{n}= \sqrt{2}(1+\epsilon)$ where $\epsilon >0$ is small.

Then $$\frac{m+2n}{m+n} = \frac{ \frac{m}{n} +2}{\frac{m}{n} +1} = \frac{ \sqrt{2}(1+\epsilon)+2}{\sqrt{2}(1+\epsilon) + 1} = \sqrt{2} \left(1- \left(\frac{\sqrt{2}-1}{\sqrt{2}+1+\sqrt{2}\epsilon}\right)\epsilon \right)$$

Note that $\sqrt{2}+1+\sqrt{2}\epsilon> \sqrt{2}+1 $. Also, since $1<\sqrt{2}< \frac{3}{2}$, we have $\frac{\sqrt{2}-1}{\sqrt{2}+1} < \frac{1}{4}$ so ,$$\frac{\sqrt{2}-1}{\sqrt{2}+1+\sqrt{2}\epsilon}<\frac{1}{4}.$$

Thus, $\frac{m+2n}{m+n}$ is slightly smaller than $\sqrt{2}$ and it's difference from $\sqrt{2}$ is smaller in magnitude than the previous estimate, and decreases by at least a factor of $4$ with each iteration.

In a similar manner you can show the other case.

EDIT: I strengthened the estimates to address the rate of convergence issues Andres Caicedo brought up in a comment above.

Ragib Zaman
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We do the "opposite directions" and "better approximation" parts, including an estimate of how much better.

We are intended to assume that $m$ and $n$ are positive, and indeed that they are positive integers. In the argument below, we do not need $m$ and $n$ to be integers, but we do assume they are positive. Some assumption needs to be made, since $m=-1$, $n=1$ quickly leads to disaster!

Look at $$\frac{m+2n}{m+n}-\sqrt{2}.$$ This is equal to $$\frac{m+2n-m\sqrt{2}-n\sqrt{2}}{m+n},$$ which in turn is equal to $$-\frac{(\sqrt{2}-1)(m-n\sqrt{2})}{m+n}.$$ Divide top and bottom by $n$. We get that the above expression is equal to $$-\frac{(\sqrt{2}-1)(\frac{m}{n}-\sqrt{2})}{1+\frac{m}{n}}.$$ So we conclude that $$\frac{m+2n}{m+n}-\sqrt{2}=\left(-\frac{\sqrt{2}-1}{1+\frac{m}{n}}\right)\left(\frac{m}{n}-\sqrt{2}\right).$$

Note that the "multiplication factor" $-\frac{\sqrt{2}-1}{1+\frac{m}{n}}$ is negative. That means that if $\frac{m}{n}-\sqrt{2}$ is negative, then $\frac{m+2n}{m+n}-\sqrt{2}$ is positive, and if $\frac{m}{n}-\sqrt{2}$ is positive, then $\frac{m+2n}{m+n}-\sqrt{2}$ is negative. Thus the approximations alternate between too big and too small.

Note also that the multiplication factor $-\frac{\sqrt{2}-1}{1+\frac{m}{n}}$ has absolute value less than $\sqrt{2}-1$, which is less than $0.5$. So the absolute value of the error when we approximate $\sqrt{2}$ by $\frac{m+2n}{m+n}$ is less than half the absolute value of the error when we approximate $\sqrt{2}$ by $\frac{m}{n}$.

Note that we can make a better estimate of the rate of approach to $\sqrt{2}$, if we assume that we start with $m=n=1$. For then, forever, our approximation will be bigger than $1$, so the multiplication factor has absolute value $(\sqrt{2}-1)/(1+m/n)$, which is less than $(\sqrt{2}-1)/2$.

André Nicolas
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    You can do the same with simpler formulas, by squaring and subtracting 2: if the error of the $k$-th element of the sequence is $e_k = \left(\dfrac{m}{n}\right)^2 - 2 = \dfrac{m^2-2n^2}{n^2}$, the error for the next element is $e_{k+1} = \left(\dfrac{m+2n}{m+n}\right)^2 - 2 = \dfrac{2n^2-m^2}{(m+n)^2} = -e_k \left(\dfrac{n}{m+n}\right)^2$. Assuming $m+n > 0$ the sequence $|e_k|$ is monotonic decreasing and converging to zero, so $e_k$ is also converging to zero. For the particular sequence starting at $1/1$, the numerators of $e_k$ alternate between 1 and -1. – Paolo Bonzini Aug 30 '11 at 07:11
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    @Paolo Bonzini: That is a nice way to look at it. It could serve as a start to showing the connections between solutions of the Pell Equations $x^2-2y^2=\pm 1$ and certain "good" approximations of $\sqrt{2}$. – André Nicolas Aug 30 '11 at 09:13
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First, let $x_k = \frac{m}{n}$, then $x_{k+1} = \frac{x_k +2}{x_k+1}$.

Notice that $x_{k+1} - x_k = \frac{2-x_k^2}{1+x_k}$ and that $x_{k+1}^2 - 2 = \frac{2-x_k^2}{(1+x_k)^2}$.

Thus if $0< x_k <\sqrt{2}$, then $x_{k+1} > \sqrt{2}$. Also from here

$$ \vert x_{k+1}^2 - 2 \vert < \vert x_k^2 - 2 \vert $$ for $x_k > 0$. Thus $x_k$ converges to $\sqrt{2}$ and $x_k - \sqrt{2}$ is an alternating sequence.

Sasha
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    In general, it does not follow from $|a_{k+1}-2|<|a_k-2|;(k\in\Bbb N)$ that the $a_k$ converge to $2$. As a counterexample, consider $$a_k:=2+(-1)^k\left(1+\frac1{2^{k+1}}\right).$$ – John Bentin Mar 18 '25 at 20:59
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Notice $\dfrac{m+2n}{m+n} = 1+\dfrac1{1+\dfrac mn}$, so consider the function $f(x) \equiv 1+\dfrac1{1+x}$. Have a look

graph for 1+1/(1+x)

Two things are easy to notice here.

  • The value $\sqrt2$ is a fixed point of $f$, that is, $f(\sqrt2) = \sqrt2$.
  • Over the positive integers, $f$ is positive and strictly decreasing.

This alone immediately justifies the switching error direction. In fact, we have $$0 < a < \sqrt 2 < b \implies 0 < f(b) < \sqrt 2 < f(a)$$

To conclude, notice

$$\begin{aligned} f(x)-\sqrt2 &= 1+\frac1{1+x}-\sqrt2\\ &= \frac{2+x-\sqrt2-\sqrt2 x}{1+x}\\ &= \frac{(\sqrt2-1)(\sqrt2-x)}{1+x} \end{aligned}$$

thus, for $x>0$, we have

$$\begin{aligned} 0 < \dfrac{f(x)-\sqrt2}{\sqrt2-x} &= \dfrac{(\sqrt2-1)(\sqrt2-x)}{(1+x)(\sqrt2-x)}\\ &= \dfrac{\sqrt2-1}{1+x} < 1 \end{aligned}$$ so the error is, in fact, decreasing.

Alma Arjuna
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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.