Approximation of $\sqrt{2}$

Question

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.

Answer 1

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 $|x-\sqrt2|\le x+\sqrt2< x+x=|2x|$.

Answer 2

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 $\frac{f(x)-f(\sqrt{2})}{x-\sqrt{2}}=\frac{1}{2}-\frac{1}{\sqrt{2}x}$

Answer 3

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 $a_n=\frac{m}{n}$:

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

This is equivalent to:

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

Answer 4

$\frac{m}{n}$ 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:

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

then:

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

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