I got the following problem in a chapter of approximations:

If $\frac{m}{n}$ is an approximation to $\sqrt{2}$ then prove that $\frac{m}{2n}+\frac{n}{m}$ is a better approximation to $\sqrt{2}.$(where $\frac{m}{n}$ is a rational number)

I am stuck in this problem. Any help will be appreciated.

  • 2
    $\begingroup$ You might want to look up Newton's method for the square root. $\endgroup$ Commented May 16, 2016 at 15:34

4 Answers 4


The claim is false without additional assumptions on $\frac mn$. For example if $\frac mn=\frac1{10}$ then your next guess is $10.05$ which is further away from $\sqrt 2$. The claim is true, however, if we impose a condition like $\frac mn>1$.

Write $x=\frac mn$. You want to show that $\frac x2 +\frac1x$ is closer to $\sqrt 2$ than $x$ is. In other words, show: $$ \left|{\left(\frac x2+\frac1x\right)-\sqrt2\over x-\sqrt2}\right|<1.\tag1 $$ By algebra we have: $$ \left|{\left(\frac x2+\frac1x\right)-\sqrt2\over x-\sqrt2}\right|=\left|{x-\sqrt 2\over 2x}\right|\tag2 $$ (the identity is true even without the absolute values.)

If we assume $x>1$, we can argue by cases that the RHS of (2) is less than one:

  • If $1<x\le\sqrt 2$ then $|x-\sqrt2|=\sqrt 2-x<2<2x=|2x|$.

  • If $x>\sqrt 2$ then by the triangle inequality $|x-\sqrt2|\le x+\sqrt2< x+x=|2x|$.

  • $\begingroup$ Thanks, it was a great help. $\endgroup$
    – user333900
    Commented May 17, 2016 at 2:28

$\frac{m}{n}$ approximately equals $\sqrt 2$

Suppose $\epsilon$ is our the error in our estimate. i.e. $\frac{m}{n} + \epsilon = \sqrt 2$ or

$(\frac{m}{n} + \epsilon)^2 = 2$

$\frac{m^2}{n^2} + 2\frac{m}{n}\epsilon + \epsilon^2 = 2$

Now we need to solve for $\epsilon.$

If $\epsilon$ is small then then $\epsilon^2$ is very small. Which is a little fuzzy. Certainly we can say that $\epsilon^2 < 2\frac{m}{n}\epsilon$ and if we solved for $\epsilon^*$ in this equation:

$2\frac{m}{n}\epsilon^* = 2 - \frac{m^2}{n^2}$


$|(\frac{m}{n} + \epsilon^*) - \sqrt 2| < |\frac{m}{n} - \sqrt2|$

$\epsilon^* = \frac{n}{m} - \frac{m}{2n}\\ \frac{m}{n} + \epsilon^* = \frac{n}{m} + \frac{m}{2n}$


Hint 1 : Putting $x=\frac{m}{n}$, try to show that if $x$ is an approximation to $\sqrt{2}$ then $f(x)=\frac{x}{2}+\frac{1}{x}$ is a better approximation to $\sqrt{2}$.

Hint 2: $\frac{f(x)-f(\sqrt{2})}{x-\sqrt{2}}=f'(c)$ for some $c$ between $x$ and $\sqrt{2}$ (this is the mean value theorem)

If you don't know about derivatives, you can also compute that $\frac{f(x)-f(\sqrt{2})}{x-\sqrt{2}}=\frac{1}{2}-\frac{1}{\sqrt{2}x}$

  • $\begingroup$ "you need to show that if $x$ is an approximation to $\sqrt{2}$ then prove that $f(x)=\frac{x}{2}+\frac{1}{x}$ is a better approximation to $\sqrt{2}$" But this will prove difficult to show since this is not always true. $\endgroup$
    – Did
    Commented May 16, 2016 at 19:15
  • $\begingroup$ Pray tell when it is not true? The starting real value is supposed to be reasonably near to 1.4142....Or you mean one has to start with an approximate with its real value more than $\sqrt2$ ? $\endgroup$
    – Narasimham
    Commented May 16, 2016 at 19:39
  • $\begingroup$ @Narasimham If you are commenting on questions by others to ask something to the author of a comment, you need to @ them. Re your comment, you seem to begin to realize there is a problem, which is good, but are still denying there is one, which is not so good. $\endgroup$
    – Did
    Commented May 16, 2016 at 19:40
  • $\begingroup$ A comment without @ is redirected to the commenter of immediately above placed lines irrespective of all others. My question is to you, about what you find missing in the answers of those who all say the same thing in different ways around the Newton-Raphson method or algorithm. I wish to see your reply if possible. $\endgroup$
    – Narasimham
    Commented May 16, 2016 at 19:55
  • $\begingroup$ @Narasimham Redirection: no, apparently not. Missing step: no, not all the answers on the page proclaim wrongly that each point produced by the Newton's algorithm would be always closer to the target than the preceding one. $\endgroup$
    – Did
    Commented May 16, 2016 at 20:04

You can use the iterative method: $a_{n+1}=\frac{a_n+\frac{2}{a_n}}{2}$. Thereby $a_{n+1}$ is more accurate than $a_n$. If $a_n=\frac{m}{n}$:


This is equivalent to:

$a_{n+1}={\frac{m}{2n} + \frac{n}{m}}$

  • $\begingroup$ "Thereby $a_{n+1}$ is more accurate than $a_n$." Why? $\endgroup$
    – Did
    Commented May 16, 2016 at 18:49

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