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Find the limit points of the set $\{m-\sqrt2 n:m,n\in \mathbb{N}\}$ in real numbers. Clearly every postive real number is a limit point what about negative reals?.

neelkanth
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    The proof you used for positive integers works perfectly for negative integers also! – Elaqqad Apr 01 '15 at 08:59
  • Or equivalently, notice that if $x$ is a limit point not in the initial set $A$, then $x$ is the limit of some sequence $(a_k){k\geq 1}$ where $\lbrace a_k \rbrace$ is not finite. It follows that we may further assume that the $a_k$ are all distinct. It follows that we may further assume that there are sequences $(m_k)$ and $(n_k)$ tending to $+\infty$ such that $a_k=m_k-\sqrt{2}n_k$. Then, $x-n=\lim{k}(a_k-n)$ is also a limite point for any $n\in{\mathbb Z}$. – Ewan Delanoy Apr 01 '15 at 12:29
  • Yes Thanking You...... – neelkanth Apr 02 '15 at 08:07
  • for positive reals proof is given in problems in mathematical analysis-1 by W.J.Kaczor and W.T.Nowak...Thanking You... – neelkanth Apr 02 '15 at 12:36
  • Sir can you give simplest proof this problem? – neelkanth Apr 02 '15 at 16:20

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Continued Fraction approximations alternate between too big and too small. Furthermore, for Continued Fraction approximations of $\sqrt2$, we have $$ \left|\,\frac pq-\sqrt2\,\right|\lt\frac1{q^2} $$ for arbitrarily large $q$. This means that $$ -\frac1q\lt\,p-q\sqrt2\lt\frac1q $$ with arbitrary sign. To get within $\varepsilon$ of $-r$, we just need to find $p$ and $q$ so that $q\gt\frac1\varepsilon$ and $p-q\sqrt2\lt0$. Then $$ -\varepsilon\lt p-q\sqrt2\lt0 $$ Let $k=\left\lfloor\frac{-r}{p-q\sqrt2}\right\rfloor$, then $$ -r\lt kp-kq\sqrt2\lt-r+\varepsilon $$ This shows that $m-n\sqrt2$ is dense in the negative reals.

robjohn
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