Use Archimedean Property to prove that sup$\lbrace \cos(\frac{1}{n}): n \in \mathbb{N} \rbrace = 1$. (Use the fact that $\cos x &gt; x - x^2/2$ for $x \neq 0$  Observe that $\cos 1 \in A$ and $|A| &lt; 2$, so by Axiom of Completeness, this set has a supremum. Let $\alpha = \sup A$. Assume $\alpha &gt; 1$. I'm stuck after this point, any guidance would be appreciated.
121262
I am trying to prove that the sup { $\cos (1/x): x\in\mathbb{N}$ } = $1$ using the Archimedean property.  Here is my proof so far:   Suppose  $y\in$ { $\cos (1/x): x\in\mathbb{N}$ }. Then, since $\cos(1/x)$ is bounded by $1$, we have that $y&lt; 1$ for all $y$ in the set so $1$ is an upper bound.  Now, suppose $z$ is an upper bound such that $z&lt;1$. Then by the Archimedean Property, we have that $\frac {1}{n}$ $&lt; 1-z$ where $n\in\mathbb{N}$. Here is where I am not sure how to continue. I know that $cos(x)&gt;1-\frac{x^2}{2}$ for all $x\neq 0$, so I think I may need to modify the way I have chosen by upper bound.
1570
$1.20 (a)$: If $x,y \in \mathbb{R}$ and $x \gt 0$, then there is a positive integer n such that $nx \gt y$.   The proof in the book:     Let $A$ be the set of all $nx$, where $n$ runs through the positive integers. If (a) were false, then $y$ would be an upper bound of $A$. But then $A$ has a least upper bound in $R$. Put $\alpha = \sup A$. Since $x \gt 0$, $\alpha -x \lt \alpha$, and $\alpha -x$ is not an upper bound of $A$. Hence $\alpha -x \lt mx$ for some positive integer $m$. But then $\alpha \lt (m+1)x \in A$, which is impossible, since $\alpha$ is an upper bound of $A$. $\blacksquare$   My question: for $A$ to have an upper bound in $R$ it also has to be non-empty. The author doesn't say that anywhere in the proof. Why? The proof must be incomplete, right?
4722
How to prove that upper limit of $\cos (x^2)-\cos (x+1)^2$ is $2$.   Can anyone show proof of this?
112564
$|a_{n+1}-a_n| \leq \frac{1}{2} |a_{n}-a_{n-1}|$. I'm trying to show that $(a_n)$ is convergent by showing that $(a_n)$ is Cauchy.  let $|a_2 - a_1 |  =x$.   $|a_3-a_2| \leq \frac{1}{2}x$  $|a_4-a_3| \leq \frac{1}{2}|a_3-a_2|\leq \frac{1}{4}x$  In general, for every $n$:  $|a_{n+1}-a_n| \le \frac {1}{2^{n-1}}x$  Let $\epsilon &gt;0$  for every $m,n &gt; 0 $, say $m&gt;n$ the following holds:  $$|a_m-a_n| = |a_m-a_{m-1} + a_{m-1} - a_{m-2} \dots a_{n+1}- a_n| \le$$  $$|a_m-a_{m-1}| + |a_{m-1} - a_{m-2}| + \dots |a_{n+1}-a_n| \leq \frac{x}{2^{m-2}} + \frac{x}{2^{m-3}} + \dots + \frac{x }{2^{n-1}}$$.  I can say that for big enough $n,m$, $|a_m-a_n|&lt;\epsilon$ and therefore, this sequence is Cauchy? It sounds not good enough to me yet I don't know what else to do here.
124084
I'm having difficulties making progress in proving: $$\forall \varepsilon &gt; 0, \ \exists q \in Q  \text{ where } 0 &lt; |r - q| &lt; \varepsilon $$  To clarify, $r$ is a real number and $q$ is a rational number.   Is there some theorem I should be using? This exercise is presented in the same section/chapter as the Completeness Axiom (each nonempty set has a least upper bound or supremum), the Archimedean Property of Real Numbers ($ \exists n \in Z^{+}$ such that $na&gt;b $ for positive real numbers $a$ and $b$), and a theorem stating there is a rational and irrational number between any two distinct real numbers.  I'm just not seeing the connection (if any at all). Any help in the right direction would be much appreciated. Thank you!
55328
Show given any $x\in\mathbb{R}$ show there exists a unique $n\in\mathbb{Z}$ such that $n-1\leq x &lt;n$. I already know that there exists a $n\in\mathbb{N}$ such that $x&lt;n$ by the archimedean property. But to prove that $n-1\leq x$ part I'm not sure. I think I have to create some set such that $n-1$ is a lower bound. But I'm not sure.
20485
Let $x\in \mathbb{R}$ an irrational number. Define $X=\{nx-\lfloor nx\rfloor: n\in \mathbb{N}\}$. Prove that $X$ is dense on $[0,1)$.    Can anyone give some hint to solve this problem? I tried contradiction but could not reach a proof.  I spend part of the day studying this question   and its answers. Only one answer is clear and give clues to solve the problem. This answer is the first one. However, this answer does not answer the question nor directly, nor the proof follows from this answer.   This answer has some mistakes, he use that $[(k_1-k_2)\alpha]=[k_1\alpha]-[k_2\alpha]$ which is not true. Consider $k_1=3, k_2=1, \alpha=\sqrt{2}$ we have $[(k_1-k_2)\alpha]=2\not= 3=[k_1\alpha]-[k_2\alpha] $. We only can assure that $[k_2\alpha]-[k_1\alpha]-1\leq [(k_2-k_1)\alpha]\leq[k_2\alpha]-[k_1\alpha]$.   Who answered said something interesting about additive subgroups of $\mathbb{R}$, but unfortunately the set $X=\{nx-[nx] : n\in \mathbb{N} \}$ is not a subgroup. Considering the additive subgroup $G=\langle X \rangle$, if we prove the part (a) of the link, we get that indeed $G$ is dense on $\mathbb{R}$ but we can not conclude that $X$ is dense on $[0,1)$.  I think this problem has not been solved.  Thanks!
14865
Prove that if $u_n\rightarrow \infty$ then $\frac{1}{u_n} \rightarrow 0$  Could someone please point me in the right direction? I don't know how to show this at all. Any help would be much appreciated :)
105005
Find the supremum of the following set:  $$A:=\left\{{(n-1)\over(2n+3)} : n \in\mathbb{N}\right\}$$  So I have as my answer that $\sup(A) = \frac12$ but need to justify. We were taught to justify in two steps, first show our answer is an upper bound for the set then show it is a least upper bound.   For step $1$: if $a$ is a member of $A$ then $a = (n-1)/(2n+3)$ for some natural number $n$ and $\frac{n-1}{2n+3} \leq \frac{n}{2n} = \frac12$ so $\frac12$ is an upper bound for the set  Step 2 is where I am uncertain. I think I need to show that for any $y&lt;\frac12$ there exists a member of the set between $y$ and $\frac12$ but not sure how to do this. Thanks for all the answers so far!
105641
This is a problem from Rudin, but I wanted to add my own intuition to it. It uses Rudin's definition of Archimedean property. I'd just like to know if my version holds  If $x \in \mathbb R$, $y\in \mathbb R$ and $x&lt;y$, then $\exists p \in \mathbb Q$ such that $x &lt; p &lt; y$  Since $x &lt; y$, then $y-x&gt;0$. Applying Archimedean property, we see $\exists n \in \mathbb Z^+$ such that $n(y-x)&gt;1$. Allow $A=\{i\in \mathbb Z\mid i &gt; nx\}$. Obviously $A$ is bounded by $nx$, so take $m = \inf A$. So we have $m &gt; nx$ and $m-1 \leq nx$ by set and $\inf$ definition. This gives us $m-1 \leq nx &lt; m$. Combining and rearranging with $n(y-x) &gt; 1$ gives us $nx &lt; m \leq nx +1 &lt; ny \implies nx &lt; m &lt; ny$. Thus $x &lt; \frac{m}{n} &lt; y$
74809
Let $m&gt;2$ be an integer, $x_0 = m$ and $x_n = {(x_{n-1})}^2 - 2$ for $n &gt; 0.$ Prove that $x_n=\lceil\tau(n) \rceil$, where $\tau(n) = α^{2^n}$ and $\alpha &gt;1$ satisfies $\alpha + \frac{1}{\alpha} = m$.  This is the problem.  The only thing I can think of is that it could have something to do with $\varphi$, since for $m=3$ we have $x(n)=\lceil\left(\varphi^2\right)^n\cdot \varphi) \rceil$, $\alpha = \varphi^2$. Other than that I have no idea how to move on. Thanks in advance
66011
I've been experimenting with recursive sequences lately and I've come up with this problem:     Let  $a_n= \cos(a_{n-1})$ with $a_0 \in \Bbb{R}$ and $L=[a_1,a_2,...,a_n,...].$   Does there exist an $a_0$ such that $L$ is dense in $[-1,1]?$     I know of $3$ ways of examining whether a set is dense:  $i)$The definition, that is, whether its closure is the set on which it is dense, in our case this means if: $\bar L=[-1,1]$.  ii)$(\forall x \in [-1,1])(\forall \epsilon&gt;0)(\exists b \in L):|x-b|&lt;\epsilon$ $iii)$ $(\forall x \in [-1,1])(\exists b_n \subseteq L):b_n\rightarrow  x$   So far I haven't been able to use these to answer the question. I tried plugging in different values of $a_0$ and see where that leads but I have not found any corresponding promising "pattern" for $a_n$. Any ideas on how to approach this?
84322
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 $[0,1]$.
37369
Let $s_n$ be a sequence and prove that it is a Cauchy sequence for $|s_{n+1}-s_n|\lt2^{-n}$  I'm not sure how to start doing this, but here's what I tried to do so based on the Cauchy sequence theorem I think i have to find an $N$ such that $n,\ n+1&gt;N\implies |s_{n+1}-s_n|\lt2^{-n}$ but from the hypothesis it seems to be saying that all $n$ and its previous or $n-1$ are greater than $N$ and i don't think that gets me anywhere. I know there are similar posts for this problem, but i don't understand their solutions or how they approached the problem.
1937
How to prove for $x&gt;0$, $\dfrac{x}{1+x^2}&lt;\arctan x&lt;x$?   I started with saying $0&lt;\arctan x&lt;\frac{\pi}{2}$  when $x&gt;0$  I'm just not sure how to proceed with this proof, it works when I take random  values but I cant see how to prove that it works for any $x$.
30580
I'm trying to prove that; If any Cauchy sequence is convergent in an ordered field F, every nonempty subset of F that has an upperbound has a sup in F.  Let A be a nonempty subset of F that is not a singleton and has an upperbound in F. Let $a_0 \notin v(A)$ and $b_0 \in v(A)$. It's written in my book that for every $e \in P_F$, there exists $N \in \omega$ such that $N≧(b_0 - a_0)/e$.  I think this is not accurate since it hasn't showed that such F is Archimedean.. Is such F archimedean? Or in such a condition does there exist such N?  -Definition of a Cauchy sequence; For every $e\in P_F$, there exists $N\in \omega$ such that if $i,j≧N$, then $|x(i) - x(j)| &lt; e$. ($x:\omega \to F$ is a sequence)
27054
I was given the following question: Let $q$ be a real number such that $0 &lt; q &lt; 1$, and let ${a_n}$be a sequence that follows the rule : $|a_{n+1}-a_n| &lt; q|a_n-a_{n-1}|$ for any $n&gt;1$. Prove that ${a_n}$ converges. I already know this can be proven using Cauchy's criterion, but I had a different proof in mind (using Cantor's lemma), and wanted to know if it's valid, or if I might be mistaken. So here it is: First - I prove that the sequence is bounded. Since the sequence is bounded, the set of its members (and the set of any sub sequence of ${a_n}$) has a Supremum and an Infimum. Then, for any natural number $k$, I define the closed interval $A_{k}$, such that $A_{n}=[\inf(a_k), \sup(a_k)]$, where inf and sup of $k$ are the Infimum and Supremum of the set of the members of $a_n$, starting from $a_k$ (meaning, any member with an index greater than or equal to $k$). Naturally, $A_1$ is the &quot;biggest&quot; interval, and because of how the sequence is defined, $A_{n+1}$ is always a proper subset of $A_n$ for any $n$. Now I use Cantor's lemma - because the interval gets infinitely smaller and smaller (after n iterations, it gets at least $q^n$ times smaller), the intersection of all intervals $A$ (meaning, $A$ of $n=\infty$) contains one single point, let's call it $L$. Now, I state that $L$ must be the limit of $a_n$ - for any epsilon, there exists a natural number $k$, such that for any $n&gt;k$, $A_k$ is contained in the epsilon neighborhood of$ L$, and therefore all members of $a_n$ starting from $a_k$ are in the epsilon neighborhood of $L$, which means $L$ must be the limit. So do you think this is a valid proof? Or am I making some false assumptions? I know this proof is a bit longer than using Cauchy's criterion, but I like it better because I feel like it gives a better explanation as to 'why' the series converges. Thanks in advance!