Let $x$ is a real number such that $n^x\in\Bbb Z,$ for every positive integer $n.$ Prove that $x$ is an integer.    I got that problem  and it looks difficult, I tried writing $x$ as $\lfloor x\rfloor+\{x\}$ (where $\lfloor x\rfloor$ and $\{x\}$ are the floor and the fractional part of $x,$ respectively), then $$n^x=n^{\lfloor x\rfloor}n^{\{x\}}.$$ Since $n^{\lfloor x\rfloor}&gt;0,$ then $$n^{x-\lfloor x\rfloor}=n^{\{x\}},$$ so that $n^{\{x\}}$ must be a rational number for every positive integer $n,$ but I don't know how to prove that $\{x\}$ must necessarily be $0$ ($0\leq\{x\}&lt;1$).   Any help is really appreciated!
123405
My apologies if this has been asked already, I've searched but couldn't find it...  Let $x$ such that for every $y \in N$, $y^x$ is an integer. Does that necessarily mean that $x$ is an integer?
118599
I am working on an assignment and have to write a proof for the following:  $$\lfloor 4x \rfloor =  \lfloor x \rfloor + \lfloor x+\frac{1}4 \rfloor + \lfloor x+\frac{1}2 \rfloor + \lfloor x+\frac{3}4 \rfloor$$  I have to do by proof by case I know the cases to be as follows  $$0 \le x-\lfloor x \rfloor \lt \frac 1 4$$ $$\frac 1 4 \le x-\lfloor x \rfloor \lt \frac 1 2$$ $$\frac 1 2 \le x-\lfloor x \rfloor \lt \frac 3 4$$ $$\frac 3 4 \le x-\lfloor x \rfloor \lt 1$$  The problem is I don't understand how I can express this.  Case 0 In this case I know that it would be $4n$ here  Case 1 In this case I know that it would be $4n+1$  Case 2 Here it would be $4n+2$  Case 3 Now the last case would end up being  $4n+3$  Can someone help me express this?
93725
Let $x$ be positive, rational, but not an integer. That means $x$ can be written as $\frac{p}{q}$ with $p,q$ coprime, $p,q \neq 0$ and $q \neq 1$. Is $x^x$ always irrational?   I think that this has to be the case, but I can't prove it.  $x^x = (\frac{p}{q})^\frac{p}{q} = \frac{p^\frac{p}{q}}{q^\frac{p}{q}}$  I know that $p^\frac{p}{q}$ and $q^\frac{p}{q}$ can't both be rational, but there are cases where the division of two irrational numbers gives us an rational number. For example $\frac{\sqrt{2}}{\sqrt{2}} = 1$
114441
Let $a \geq 2$ be a positive integer such that $\log_{10}a$ is irrational. Then for any positive integer $N$ there exist infinitely many powers $a^n$ of $a$ whose decimal representation begins with $N$.   I am not sure how to use the fact that $a^n$ begins with $N$. How should I prove this?
40310
As written in title, I want to prove that      If $n$ is an integer, show that if $2+2\sqrt{28n^2+1}$ is an integer than it must be perfect square.    I m struggling in making a start . Please help.
79603
Give a convincing argument that $\lfloor x\rfloor + \lfloor y\rfloor \leq \lfloor x+y\rfloor$ for every pair of numbers $x$ and $y$. Could someone please explain how to prove this? I attempted to say that the largest values that could be added to $x$ and $y$ is $0.99$ and that doing so still made $\lfloor x\rfloor + \lfloor y\rfloor &lt; \lfloor x+y\rfloor$. However, my answer was not accepted.
75170
Let $\{x_n\}_{n=1}^\infty$ be a squence of nonnegative real numbers such that $\displaystyle \lim_{n \to \infty} x_n =x \in \Bbb R$. If $\alpha \in (0, \infty)$, is it true that $\displaystyle \lim_{n \to \infty} x_n^\alpha =x^\alpha$?  I read the proofs for $\alpha$ integer and for $\alpha=1/2$, but I wonder what happens with the other cases.
70466
Given $x\in\mathbb R_+$ and $m,n\in\mathbb Z_+$, is it true that $$\bigg\lfloor\frac{\lfloor \frac{x}{m}\rfloor}{n}\bigg\rfloor=\bigg\lfloor \frac{x}{mn}\bigg\rfloor?$$    Thanks for at least three different and convincing proofs! I'll use this result  trying to code the prime counting formula from Wikipedia $(2)$:  Given $m$, select $y$ so that  $\sqrt[3]{m}\le y\le\sqrt{m}$ and let $n=\pi(y)$. Then  $(1)\;$ $\pi(m)=\phi(m,n)+n-1-P_2(m,n)$, where  $(2)\;$ $\displaystyle \phi(m,n)=\phi(m,n-1)-\phi\Big(\frac{m}{p_n},n-1\Big)$,  $\;\phi(m,0)=\lfloor{m}\rfloor$ and  $(3)\;$ $\displaystyle P_2(m,n)=\sum_{y&lt;p\le\sqrt{m}} \Big(\pi\Big(\frac{m}{p}\Big)-\pi(p)+1\Big)$, $\;p$ is a prime.
123477
Mathematically, why is this true?  $$\left\lceil\frac{a}{b}\right\rceil= \left\lfloor\frac{a+b-1}{b}\right\rfloor$$  Assume $a$ and $b$ are positive integers.  Is this also true if $a$ and $b$ are real numbers?
4448
I'm doing some exercises on Apostol's calculus, on the floor function. Now, he doesn't give an explicit definition of $[x]$, so I'm going with this one:     DEFINITION Given  $x\in \Bbb R$, the integer part of $x$ is the unique $z\in \Bbb Z$ such that $$z\leq x &lt; z+1$$ and we denote it by $[x]$.   Now he asks to prove some basic things about it, such as: if $n\in \Bbb Z$, then $[x+n]=[x]+n$  So I proved it like this: Let $z=[x+n]$ and $z'=[x]$. Then we have that  $$z\leq x+n&lt;z+1$$  $$z'\leq x&lt;z'+1$$  Then $$z'+n\leq x+n&lt;z'+n+1$$  But since $z'$ is an integer, so is $z'+n$. Since $z$ is unique, it must be that $z'+n=z$.  However, this doesn't seem to get me anywhere to prove that  $$\left[ {2x} \right] = \left[ x \right] + \left[ {x + \frac{1}{2}} \right]$$  in and in general that   $$\left[ {nx} \right] = \sum\limits_{k = 0}^{n - 1} {\left[ {x + \frac{k}{n}} \right]} $$  Obviously one could do an informal proof thinking about "the carries", but that's not the idea, let alone how tedious it would be. Maybe there is some easier or clearer characterization of $[x]$ in terms of $x$ to work this out.  Another property is $$[-x]=\begin{cases}-[x]\text{ ; if }x\in \Bbb Z \cr-[x]-1 \text{ ; otherwise}\end{cases}$$  I argue: if $x\in\Bbb Z$, it is clear $[x]=x$. Then $-[x]=-x$, and $-[x]\in \Bbb Z$ so $[-[x]]=-[x]=[-x]$. For the other, I guess one could say:  $$n \leqslant x &lt; n + 1 \Rightarrow  - n - 1 &lt; x \leqslant -n$$  and since $x$ is not an integer, this should be the same as $$ - n - 1 \leqslant -x &lt; -n$$  $$ - n - 1 \leqslant -x &lt; (-n-1)+1$$  So $[-x]=-[x]-1$
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 !
50512
Let $n$ is a natural number. Prove that $$\left\lfloor\frac{n}{1}\right\rfloor+\left\lfloor\frac{n}{2}\right\rfloor+....+\left\lfloor\frac{n}{n}\right\rfloor+\left\lfloor{\sqrt{n}}\right\rfloor$$   is even.   I used induction and the inequality $$x-1&lt;\lfloor{x}\rfloor\le{x}$$ to prove it.  Is there any other, nicer way to do it?
110249
If $x\in\mathbb R$, solve $$4x^2-40\lfloor x\rfloor+51=0$$   where $\lfloor x\rfloor$ denotes the integer part of the number. $\lfloor x\rfloor\le x$ and $\lfloor x\rfloor=x-\{x\}$, where $\{x\}$ marks the fraction part of the number. $0\le\{x\}&lt;1$.   Not sure if all these denotions are conventions that always mean what I've mentioned. Also, I'm not too sure if $x\in\mathbb R$, since the problem doesn't mention it, but it seems quite obvious that it most likely is that way.
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 $[0,1]$.
112159
Let $n$ be the largest positive integer. Since $n ≥ 1$, multiplying both sides by $n$ implies that $n^2 ≥ n$. But since $n$ is the biggest positive integer, it is also true that $n^2 ≤ n$. It follows that $n^2 = n$. Dividing both sides by $n$ implies that $n = 1$.  The goal is to find the flaw in the reasoning of the proof rather than find a proof that proves it wrong. Here's where I think the problem is:  It was stated that $n^2 ≥ n$ after multiplying both sides of the inequality by $n$. But then because $n$ is the largest possible integer, $n^2 ≤ n$. This is where I think the flaw is, because if $n$ is the single largest integer, then we'd get $n^2 &lt; n$ rather than $n^2 ≤ n$, so writing the latter is incorrect.
75240
How do you prove this identity:  $$\int_0^{n^2}\lfloor\sqrt{t}\rfloor dt = \frac{1}{6}n(n-1)(4n+1)$$  I'd very much appreciate your help on this one!
73676
I'm trying to prove rigorously the following:  $\lfloor x/a/b \rfloor$ = $\lfloor \lfloor x/a \rfloor /b \rfloor$   for $a,b&gt;1$  So far I haven't gotten far. It's enough to prove this instead  $\lfloor z/c \rfloor$ = $\lfloor \lfloor z \rfloor /c \rfloor$   for $c&gt;1$  since we can just put $z=\lfloor x/a \rfloor$ and $c=b$.