Let $A=\{{m+n\sqrt{2}}:m,n\in Z\}$, where $Z$ stands for the set of all integers .Then which of the following is correct   (a) $A$ is dense in $R$  (b) $A$ has only countably many limits points in $R$   (c) $A$ has no limit point in $R$  (d) only irrational numbers can be limit points of $A$  If m=0 and n=1 .Then $\sqrt{2}\in A$ and $(\sqrt{2}-0.5,\sqrt{2}+0.5)\cap A\sim{\sqrt{2}}=\Phi$ .  Then , only option c is correct. Please correct if I am wrong . I am not able to find any counter examples.
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I am having trouble proving the statement:     Let $$S = \{m + n\sqrt 2 : m, n \in\mathbb Z\}$$ Prove that for every $\epsilon &gt; 0$, the intersection of $S$ and $(0, \epsilon)$ is nonempty.
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The question :     Let $D$ be a nonempty subset of the reals that is bounded above. Is the supremum of $D$ a limit point of $D$?   My Reasoning: I think this is false for these two cases. Case 1:If I look at $D = \{n \in \mathbb{Z} | n \le 0\}$ the supremum is $0$. And since I need a convergent sequence $\{x_n\} \subset D/\{0\}$ the converges to $0$ for it to be a limit point I can say in this case if I look at $\epsilon = \frac{1}{2}$ for the converges of the sequence it will fail to converge and so $0$ isn't a limit point.   Case 2: Also if I look at $D = {0}$ then the supremum is $0$. And $D$ is a subset of the reals. So if I look for a sequence $\{x_n\} \subset D/\{0\}$ I can't make one because $D/\{0\}$ is the empty set.   My question is this. Since the problem asked about an arbitrary subset of the reals $D$, can I define $D$ to give a counterexample like above or have I misunderstood the question?  --Thanks in advance.
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Is $\mathcal{O}_{\mathbb{Q}(\sqrt{5})} = \mathbb{Z}[\phi]$, where $\phi={1+\sqrt{5}\over 2}$ is the golden ratio?  I know that $5 \equiv 1 \mod 4$, so that then $\mathbb{Z}[\sqrt{5}]$ is not closed as far as integers go. But I'm a little confused cause  $\mathcal{O}_{\mathbb{Q}(\sqrt{-3})} = \mathbb{Z}\left[-\frac{1}{2} + \frac{\sqrt{-3}}{2}\right]$? So how do I know when to use $\frac{1}{2}$ rather than $-\frac{1}{2}$? Which one applies for $\mathcal{O}_{\mathbb{Q}(\sqrt{5})}$?
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Possible Duplicate:          I have to show that show that $A=\{ \frac{m}{2^n}:m\in \mathbb {Z},n\in \mathbb {N}\} $ is dense in $\mathbb {R}$.  A set A is dense in $\mathbb {R}$ if $\overline A=\mathbb {R}$.  But also $Y$ is a subset of $X$, we say that $Y$ is dense in $X$, if for every $x\in X$ , there is $y \in Y$  that is arbitary close to $x$.  So ,I have to prove that for every $x \in \mathbb {R}$ ,there is a number $\frac{m}{2^n}$ arbitrarily close to $x$.So $\forall \epsilon,x ,\exists y$ such that $|y-x|&lt;\epsilon$.   I got a little stuck at this point...Could anyone give me a hint?Thanks a lot!
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Let $x_n \in[0,1)$ s.t $x_n - \sqrt{n}\in \mathbb{Z}$. Prove that $X=\{x_n | n\in \mathbb{N}\}$ is dense in $[0,1]$.  It it suffices to prove that $\overline{X}=[0,1].$ I was trying to use the fact that $p+q\sqrt{2}$ $(p,q \in \mathbb{N})$ is dense in $\mathbb{R}$. So I can choose $n=2q^2$ and $x_n=p+q\sqrt{2}$ close enought of $1$. Hence, $x_n=p+q\sqrt{2} - q\sqrt{2} = p \in \mathbb{Z}$. I am pretty sure that this argument is invalid but I feel that the idea can be useful somehow.
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Is my proof that the square root of a positive integer is either an integer or an irrational number correct?  The proof goes like this:  Suppose an arbitrary number n, where n is non-negative. If $\sqrt{n}$ is an integer, then $\sqrt{n}$ must be rational. Since $\sqrt{n}$ is an integer, we can conclude that n is a square number, that is  for some integer a.  Therefore, if n is a square number, then $\sqrt{n}$ is rational.  Suppose now that n is not a square number, we want to show that the square root of any non-square number is irrational.  We prove by contradiction. That is, we suppose that the square root of any non-square number is rational. So $\sqrt{n} = \frac{a}{b}$, where $a,b \in Z^+, b \neq 0$. We also suppose that $a \neq 0$, otherwise $\frac ab = 0$ , and n will be a square number, which is rational.   Hence $n = \frac {a^2}{b^2}$, so $nb^2 = a^2$.    Suppose $b=1$. Then $\sqrt n = a$ , which shows that n is a square number. So $b \neq 1$. Since $\sqrt n &gt; 1$, then $a&gt;b&gt;1$.  By the unique factorization of integers theorem, every positive integer greater than $1$ can be expressed as the product of its primes.  Therefore, we can write $a$ as a product of primes and for every prime number that exists in $a$, there will be an even number of primes in $a^2$. Similarly, we can express $b$ as a product of primes and for every prime number that exists in $b$, there will be an even number of primes in $b^2$.  However, we can also express $n$ as a product of primes. Since $n$ is not a square number, then there exist at least one prime number that has an odd number of primes.  Therefore, there exists at least one prime in the product of $nb^2$  that has an odd number of primes. Since $nb^2=a^2$ , then this contradicts the fact that there is an even number of primes in $a^2$ since a number can neither be even and odd.  Therefore, this contradicts the fact that $\sqrt n$ is rational. Therefore, $\sqrt n$ must be irrational.  Is this sufficient? Or is there any parts I did not explain well?
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Let's consider the  defined by:      Take $ A\subset \mathbb N$; define the sequence $x_n= \dfrac{|A\cap[1,n]|}{n}$, and then if $\lim\limits_{n\to\infty} x_n$ exists, call it $d(A)$, the density of $A$ over $\mathbb N$.   My teacher said that the collection of sets that have a natural density is not an algebra, but I don't know why. I can't even find a set that does not have a natural density.  Please help me )=
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suppose that $\phi(n)$ is Euler function. prove that, $\{\frac{\phi (n)}{n}\}_{n \in \Bbb N}$ is dense in $[0,1]$  (if $A_n=\{1 \leq m \leq n \mid m \in \Bbb N ; \gcd(n,m)=1\}$ then $\phi(n)=|A_n|$)  I think : If $p$ is prime then $\phi(p)=p-1$. for any $ \varepsilon &gt; 0 $ there is a prime number $p$ such that $1-\varepsilon \leq \frac{p-1}{p} \leq 1$. for other elements i don't know what can i do.
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I am not sure if what I am looking for even makes sense (or) exists. Anyway I would be happy if someone can clear my confusion.  The set of real numbers $\mathbb{R}$ is obtained as completion of $\mathbb{Q}$. However, $\mathbb{Q}$ is not the only set which is dense in $\mathbb{R}$. $\mathbb{Q} \backslash \mathbb{Z}$ is also a dense subset of $\mathbb{R}$. I am wondering if it makes sense to talk of the "smallest" dense subset of $\mathbb{R}$. To phrase what I am looking for precisely and what I mean by smallest, I am looking for a dense set $A$ of $\mathbb{R}$ such that if $B$ is a proper subset of $A$ then $B$ is not dense in $\mathbb{R}$. I am able to "see" that the set $A$, I am looking for doesn't exist since any open interval consists of infinite rationals. But I am unable to precisely argue out to myself and convince why $A$ doesn't exist. Could some one throw more light on this?
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This is exercise 1.9 in Lie Groups, Lie Algebras and Representations - Hall.     Suppose $a$ is an irrational real number.  Show that the set $E_a$ of the numbers of the form $e^{2\pi i n a}$, $n \in \mathbb{Z}$, is dense in the unit circle $S_1$.  Hint: Show that if we divided $S^1$ into $N$ equally sized "bins" of length $2\pi/N$, there is at least one bin that contains infinitely many elements of $E_a$.  Then use the fact that $E_a$ is a subgroup of $S^1$.   My proof of this proposition is as follows.  Since $a$ is irrational, you can determine that the set of rotations $E_a$ is infinite.  Since $S^1$ is compact we can find two $r_1, r_2$ that are within $\epsilon$ of each other.  Then $r_1^{-1}r_2$ is a small rotation of size $\epsilon$.  Now, $r_1^{-1}r_2$ generates rotations that are within $\epsilon$ distance of any point of $S^1$.    That said, I don't believe that the hint suggested in the problem uses that technique.  My knowledge of algebra is not all that strong so I was hoping someone could shed some light on what is being suggested there.
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Let $\alpha$ be a nonreal algebraic number. I'm interested in conditions that imply that $\mathbb{Z}[\alpha]$ is dense in $\mathbb{C}$. I'm particularly interested algebraic integer $\alpha$.  This is what I know so far:   if there is a $n \in \mathbb{N}$ such that $\alpha^n \in \mathbb{R} \setminus \mathbb{Z}$, then $\mathbb{Z}[\alpha]$ is dense in $\mathbb{C}$; algebraic integers of degree two don't satisfy the condition, although algebraic nonintegers of degree two may.   Many thanks in advance.
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In $\mathbb{R}$ , what is the limit point of the set $G= \{1/n: n \in \mathbb N\}$?  Definition:  A point $p \in X$ is a limit of of E if every neighborhood of $p $ contains a point $q \in E$ $q\neq p$  I do understand $0$ is a limit point but shouldn't it be the case that all the points in [0,1] are limit points by the definition?
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This appears on problem 1 of chapter 1 in Stein &amp; Shakarchi's Real Analysis:     Given an irrational $x$, one can show (using the pigeon-hole principle, for example) that there are infinitely many fractions $p/q$, with relatively prime integers $p$ and $q$ such that $$\left|x-\frac{p}{q}\right|\leq\frac{1}{q^2}.$$   However, prove that the set of all $x\in\mathbb R$ such that there exist infinitely many fractions $p/q$, with relatively prime integers $p$ and $q$ such that $$\left|x-\frac{p}{q}\right|\leq\frac{1}{q^{3}} \text{ (or} \leq 1/q^{\epsilon+2})$$ is a set of measure zero.   The last part of the exercise is easy using the Borel-Cantelli theorem. However I have no idea why the first part is true.
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I have a sequence defined by $$ a_1=1,\quad a_2=1,\quad a_n=\frac{1}{a_{n-1}}+\frac{1}{a_{n-2}}\text{ for } n\ge2. $$ Now, if $\lim_{n\to \infty}a_n=g$ then $\lim_{n\to \infty}a_n=\lim_{n\to \infty}\Bigl(\frac{1}{a_{n-1}}+\frac{1}{a_{n-2}}\Bigr)=\frac{2}{g}$ so $g=\sqrt{2}$ or $g=-\sqrt{2}$ but $a_n&gt;0$ so $g=\sqrt{2}$.  Now, how do I prove that it has an actual  limit? Also, it can be proven that $1\le a_n\le2$, and it's not monotonic because $a_4 \gt a_5 \lt a_6$. Also, it's not monotonic after some $N\in\mathbb N$
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I'm not sure how to solve this one. Thank you!  $2.$ For any $\alpha\in \mathbb R$ we define $$\lfloor \alpha \rfloor = \max_{n\in\mathbb Z}\{\,n\mid n\leq \alpha\,\}$$ and $$\alpha\bmod 1 = \alpha - \lfloor \alpha \rfloor$$ Let $\alpha$ be irrational. (a) Given $n\in\mathbb N$ show that $\{\,k\alpha\bmod1\mid k\in\mathbb N\,\}\cap\left[0,\frac{1}{n}\right]\neq\emptyset$ (b) Prove that $\{\,n\alpha\bmod 1\mid n\in\mathbb N\,\}$ is dense in $[0,1]$.
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Does the condition $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=1$ imply that $a_n$ convergent?  When I say convergent I mean to $L\in \mathbb{R}$ or to $\infty$ or to $-\infty$  I have a gut feeling this statement is false, just can't find a series which contradicts it.  Thanks a lot
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Question: Consider a sequence $\{a_{n}\}$ such that $a_{1},a_{2}&gt;0$, and for all $n \in \mathbb{N}$ we have:  $$a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}$$  Prove that :  $\displaystyle\lim_{n\to\infty}a_{n}$ exists and find this limit.  My work: If this limit exists, let $\displaystyle\lim_{n\to\infty}a_{n}=x&gt;0$; then we have $$x=\sqrt{x}+\sqrt{x}\Longrightarrow x=4$$ But I can't see how to prove the limit of $\{a_{n}\}$ exists. Thank you for you help!