0
$\begingroup$

Let $$A=\{ \frac{\sqrt{m} -\sqrt{n}}{\sqrt{m}+\sqrt{n}} | m,n\in \Bbb{N} \}$$ I think that we must find sequence of $A$ and find limit of sequence,let $a_m =\frac{\sqrt{ k ^2 m^2} -\sqrt{ m^2}}{\sqrt{k ^ 2 m^2}+\sqrt{ m^2}}=\frac{(k-1)m}{(k+1)m}$ that $k\in\Bbb{N}$ the limit of $a_m$ is $\frac{k-1}{k+1}$ and let $B=\{\frac{k-1}{k+1} |k\in\Bbb{N}\}$ then $B\subseteq A^\prime$,($A^\prime$ is set of limit points of$ A$),the answer is interval $[-1,1]$,

$\endgroup$
  • $\begingroup$ I find sequence. $\endgroup$ – amir bahadory Jul 1 '15 at 10:39
0
$\begingroup$

Some ideas:

First: for all $\;m,n\in\Bbb N\;$ :

$$-1=\frac{-\sqrt n}{\sqrt n}\le\frac{-\sqrt n}{\sqrt m+\sqrt n}\le\frac{\sqrt m-\sqrt n}{\sqrt m+\sqrt n}\le\frac{\sqrt m}{\sqrt m}=1$$

so any limit point of $\;A\;$ indeed has to be in $\;[-1,1]\;$ .

Now, if $\;\alpha\in[-1,1]\;$ take a peek at

$$\frac{\sqrt m-\sqrt n}{\sqrt m+\sqrt n}-\alpha=\frac{\sqrt m(1-\alpha)-\sqrt n(1+\alpha)}{\sqrt m+\sqrt n}\le\frac{\sqrt m}{\sqrt n}(1-\alpha)$$

In order to make the last part above less than some predetermined $\;\epsilon >0\;$ it is then enough to take

$$\sqrt\frac nm>\frac{1-\alpha}\epsilon$$

$\endgroup$
0
$\begingroup$

Hint:
Note that $$\frac{\sqrt{m} -\sqrt{n}}{\sqrt{m}+\sqrt{n}}=1-\frac{2}{1+\sqrt{\frac{m}{n}}},$$ and that $\mathbb{Q}$ is dense in $\mathbb{R}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.