Prove that $ \left\lfloor{\frac xn}\right\rfloor= \left\lfloor{\lfloor{x}\rfloor\over n}\right\rfloor$ where $n \ge 1, n \in \mathbb{N}$ and $\lfloor{.}\rfloor$ represents Greatest Integer $\mathbf{\le x}$ or floor function    I tried to prove it by writing $x = \lfloor{x}\rfloor + \{x\} $ where $ \{.\}$ represents Fractional Part function and $ 0 \le \{x\} &lt; 1$ So we get,  $ \lfloor{\frac xn}\rfloor= \lfloor{{\lfloor x\rfloor\over n}+ {\{x\}\over n}}\rfloor \tag{1}$  Then I tried to use the property,  $\lfloor{x+y}\rfloor =\begin{cases} \lfloor x\rfloor + \lfloor y\rfloor&amp; \text{if $0\le \{x\} + \{y\}$} &lt; 1 \tag{2}\\ 1+ \lfloor x\rfloor + \lfloor y\rfloor &amp; \text{if $1\le \{x\} + \{y\}$} &lt; 2 \\ \end{cases} $  So if I can prove $(1)$ = first case of $(2) $ I’ll have ,  $ \lfloor{\frac xn}\rfloor= \lfloor{{\lfloor x\rfloor\over n}}\rfloor+ \lfloor{\{x\}\over n}\rfloor = \lfloor{\lfloor{x}\rfloor\over n}\rfloor$ as the second term will come out to be zero. However, I am unable to prove this.     Can someone help me out with this proof by showing me how $\mathbf(1)$= first case of $\mathbf (2)$ and proving the question using this method and also giving a clear proof using a simpler method
43532
How to prove or disprove $\forall x\in\Bbb{R}, \forall n\in\Bbb{N},n\gt 0\implies \left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor=\left\lfloor\frac{x}{n}\right\rfloor$.  So we want to prove $\left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor\ge\left\lfloor\frac{x}{n}\right\rfloor$ and $\left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor\le\left\lfloor\frac{x}{n}\right\rfloor$  Since $\lfloor x\rfloor\le x$, we can just start from here and prove  $\left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor\le\left\lfloor\frac{x}{n}\right\rfloor$  But for $\left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor\ge\left\lfloor\frac{x}{n}\right\rfloor$, I have no idea how to start.
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
49360
I've been asked to solve this and I've tried a few things but I have trouble eliminating $x$. I first tried taking the natural log: $$x\ln \left( 2\right) =2\ln \left( x\right)$$ $$\dfrac {\ln \left( 2\right) }{2}=\dfrac {\ln \left( x\right) }{x}$$ I don't know what to do from here so I decided to try another method: $$2^{x}=2^{\log _{2}\left( x^{2}\right) }$$ $$x=\log _{2}\left( x^{2}\right)$$ And then I get stuck here, I'm all out of ideas. My guess is I've overlooked something simple&hellip;
35976
$\left \lfloor n\log_2 n^2 \right \rfloor + \left \lfloor \log_2(\left \lfloor n\log_2n^2 \right \rfloor) \right \rfloor \leq \left \lfloor (n+1)\log_2 (n+1)^2 \right \rfloor + 1$  How to show this? I tried using $\left \lfloor k \right \rfloor \leq k$, but I'm not sure that that does anything...  LHS $\leq \left \lfloor (n+1)\log_2 (n+1)^2 \right \rfloor  +  \left \lfloor \log_2(\left \lfloor n\log_2n^2 \right \rfloor) \right \rfloor $ &lt; $\left \lfloor (n+1)\log_2 (n+1)^2 \right \rfloor + 1$ because...ugh...  $\left \lfloor \log_2(\left \lfloor n\log_2n^2 \right \rfloor) \right \rfloor \leq \log_2(\left \lfloor n\log_2n^2 \right \rfloor) &lt; 1$?  I don't think the last inequality is true.  Help please?  I think it's supposed to hold for all natural numbers or at least all natural numbers after a certain index.
67944
Prove that $[2x]+[2y] \geq [x]+[y]+[x+y]$ whenever $x$ and $y$ are real numbers.  The $[]$ symbol is the greatest integer or floor function.  I have proved this fact by cases, but I stumbled upon what I believe to be another way to prove the above inequality, and I was wondering if my sequence of statements are legitimate.  I make use of two lemmas that I have proved  Lemma 1.   If $x$ is a real number and m is an integer, then $[x+m] = [x]+m$.  Lemma 2.   $\displaystyle [x]+\left[x+\frac{1}{2} \right] = [2x]$.  My proof begins with an "obvious" statement   $$x+y \leq x+y+1$$  I then take the floor of the inequality to get  $$[x+y] \leq [x+y]+1$$             (1)  which is true in virtue of lemma 1.   Furthermore, if I add the following statements   $$[x+1/2] \leq x + 1/2$$ $$[y+1/2] \leq y + 1/2$$  I procure  $$\left[x+\frac{1}{2} \right]+\left[y+\frac{1}{2}\right] \leq x+y+1$$  which by definition of the floor function renders the equation  $$[x+y]+1 = \left[x+\frac{1}{2} \right]+\left[y+\frac{1}{2} \right]$$        (2)  Substituting (2) for (1), I have   $$[x+y] \leq \left[x+\frac{1}{2} \right]+\left[y+\frac{1}{2} \right]$$  I then add $[x]$ and $[y]$ to the above inequality to produce  $$[x]+[y]+[x+y] \leq [x]+\left[x+\frac{1}{2} \right]+[y]+\left[y+\frac{1}{2} \right]$$  And in virtue of Lemma 2, the right hand side of the inequality becomes  $$[x]+[y]+[x+y] \leq [2x]+[2y].$$  I personally don't see anything wrong, except maybe for the implication made to establish (2).   Solving these kinds of problems is solely for personal gratification, so I will greatly appreciate feedback.  Thanx.
74423
If $$A=\frac{1}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}\;,$$ Then $\lfloor A \rfloor\;\;,$ Where $\lfloor x \rfloor $ Represent floor fiunction of $x$   My Try:: Using $\bf{A.M\geq H.M\;,}$ We get  $$\frac{1980+1891+1982+....+2012}{33}&gt;\frac{33}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}$$  So $$\frac{1}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}&lt;\frac{1980+1981+....+2012}{(33)^2}=\frac{1996}{33}\approx 60.5&lt;61$$  Now how can i prove that the above expression $A$ is $&gt;60$  Help me, Thanks
97590
I'm trying to understand properties of the greatest integer function and I am struggling to find the value of $\lfloor x+y \rfloor$  where $x \in \mathbb{R}$, $y \in \mathbb{Z}$, and prove that it is correct value.  I don't really know how to prove this, but I have been dividing it into different cases. I think that it equals $\lfloor x \rfloor + y$ when $x,y$ are both positive but not sure how to prove it. Depending on if one or both $x$ and $y$ are negative, and their ultimate sum, I get different values. I am having trouble determining when exactly this happens though and then proving the results. Any help would be great, thanks!
54613
$$\left\lfloor \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{1024}}\right\rfloor =?$$  I can't find $$? \leq  \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{n}} \leq ?$$ I remeber there was a sharp bound , Am i right  ? I am thankful if someone help me to find the bound  , or give another idea Last question : Is it possible to find $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{1024}}$  by integral  ?
37889
I'm studying calculus right now but I'm stuck at solving a limit involving floor function.  The problem is to find $$\lim_{x\to 0}x \left\lfloor\frac{1}{x} \right\rfloor$$ where $\lfloor\cdot\rfloor$ denotes the floor function.  My first thought was to let $x=1/t$ so when ${x\to 0+}$ then ${t\to \infty}$ so it seems $\lim_{t\to \infty}[t]/t$ doesn't exist. But I can't go any further and don't know whether my thought is correct. It seems $t=N+\delta$ doesn't help because t goes to infinity. Can it be proved by the epsilon-delta method or something else? Thank you for your help.
20720
Prove : $$ \text{If } \; x-\lfloor x \rfloor + \frac{1}{x} - \left\lfloor \frac{1}{x} \right\rfloor = 1 \text{, then } x \text{ is irrational.}$$  I think the way to go here is to falsely assume that $x$ is rational, thus having $x=\frac mn$ with $m,n$ integers and go with that, we also know that $x$ with the celling belongs to the integers and that $\frac 1x$ with the celling gotta be a $0$. Still can't prove it. Thank in advance !
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 $[0,1]$.
42725
I can intuitively see that this is true, but I'm having a very hard time proving it. I'm actually not even quite sure where to begin. I tried using the inequalities that define the floor function, and I'm pretty sure this is the way to go, but I'm failing to make any kind of progress.  I appreciate any help.
8724
Let $S(n)$ be $S(n)=\left\{k\;\left|\;\left\{\frac{n}{k}\right\}\right.\geq \frac{1}{2}\right\}$,where $\{x\}$ is the fractional part of $x$  Prove that :  \begin{align} \sum_{k\in S(n)} \varphi(k)=n^2 \end{align}  maybe the fact \begin{align} \sum_{k \leq 2n}\left\lfloor{\frac{n}{k}+\frac{1}{2}}\right\rfloor \varphi(k)=\frac{3n^2+n}{2}\end{align}  and \begin{align} \sum_{k \leq 2n}\left\lfloor{\frac{n}{k}}\right\rfloor \varphi(k)=\frac{n^2+n}{2}\end{align}  can help, but I can't prove these two equations    Edited:Sorry that my question is to prove the two equation before, because I've already known how to relate the problem with these two equations, and I've also known how to prove the second equation from  but I can't solve the first one with the similar method.
101338
How can I prove that $$\displaystyle x-\frac {x^2} 2 &lt; \ln(1+x)$$ for any $x&gt;0$  I think it's somehow related to Taylor expansion of natural logarithm, when:  $$\displaystyle \ln(1+x)=\color{red}{x-\frac {x^2}2}+\frac {x^3}3 -\cdots$$  Can you please show me how? Thanks.
78586
Calculate the following limit. $$ \lim_{x\to 0} x\left\lfloor\frac{1}{x}\right\rfloor $$ Where $\left\lfloor x \right\rfloor$ represents greatest integer function or floor function, i.e greatest integer less than or equal to $x$.  Thanks.
88464
Does $\lim_{x \to 0+} \left(x\lfloor \frac{a}{x} \rfloor\right)=a?$   I'm going to say this statement is false, and try to use the properties of limits  $$\lim_{x \to 0+} \left(x\lfloor \frac{a}{x} \rfloor\right)=\lim_{x \to 0+} x.\lim_{x \to 0+}\lfloor \frac{a}{x}\rfloor$$ =$0.\infty$ which is undefined.  Or is it $\infty $ because x ends up being a little bigger than 0?
51546
Evaluate $\lim_{x \to 0} (x\lfloor\frac{1}{x}\rfloor)$  I'm trying to solve it by using the squeeze theorem but I'm stuck.  I'm looking for a function $g(x)$ such that $g(x) \leq x\lfloor\frac{1}{x}\rfloor \leq x(\frac{1}{x}) = 1$  Can anybody give me a hint?