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 $[0,1]$.
122266
How to prove that the $\{$ fractional part of $n\alpha\mid n \in \mathbb{N}$ $\}$ is dense in $[0,1]$ for an irrational number $\alpha$.   NOTICE that $n$ is in $\mathbb{N}$ Also notice that this is not a duplicate of the mentioned question as it does not carry a correct answer and the partially correct answer in the said question is given for integer multiple case, NOT for $n \in \mathbb{N}$
120607
This is a proof from a book on number theory I'm reading. I'm having a hard time following. I think there's a variable here that means two different things at two different times...  Theorem:  If n is a positive integer and p is a prime, then $p^e \vert\vert n!$, where $e=\lfloor\frac{n}{p}\rfloor + \lfloor\frac{n}{p^2}\rfloor + \dots + \lfloor\frac{n}{p^r}\rfloor$ and r is determined by n by the inequality $p^r \le n &lt;p^{r+1}$  Proof:  For a given integer k, the multiples of $p^k$ that do not exceed  n are $p^k, 2p^k, \cdots, qp^k$ , where q is the largest integer such that $qp^k \le n$. But this says that q is the largest integer not exceeding $n/p^k$, so that  $q= \lfloor n/p^k \rfloor$. Thus, $\lfloor n/p^k\rfloor$ gives the number of positive multiples of $p^k$ that do not exceed n.     Interlude:      I understand the proof so far. We've simply chosen some power of p   and shown that $\lfloor n/p^k \rfloor$ is the largest integer that we can multiply by $p^k$ and stay under n. Here's where I start to get confused.   Now, if $1 \le m \le n$, then $m=qp^k$ with $(q,p) = 1, 0 \le k \le r$, and m contributes precisely k to the total exponent e with which p appears in the canonical representation of n!     Interlude 2:      How exactly does $m=qp^k$ follow from $1 \le m \le n$? Also, this q seems to mean something other than $\lfloor n/p^k\rfloor$ since we seem to be considering the set of all numbers less than or equal to n and greater than or equal to 1 that have the form $m=qp^k$?   Moreover, m is counted precisely k times by the sum  $\lfloor\frac{n}{p}\rfloor + \lfloor\frac{n}{p^2}\rfloor + \dots + \lfloor\frac{n}{p^r}\rfloor$,   once as a multiple of p, once as a multiple of $p^2$, ... , once as a multiple of $p^k$, and no more. Of course, if $k=0$, then m is not counted in the sum. Therefore, the sum above accounts exactly for the contribution of each m between 1 and n to the exponent e as claimed.     Interlude 3:      What exactly is meant here by counted? I really don't understand the last part of this proof at all...
89403
Is it true that ${O}_n({\mathbb R}) \cap {\mathbb Q}^n$ is dense in ${\cal O}_n({\mathbb R})$ for any $n\geq 2$ ? It obviously suffices to consider the density of ${SO}_n({\mathbb R}) \cap {\mathbb Q}^n$ in $SO_n({\mathbb R})$.  This is easy for $n=2$ : consider the matrices of the form  $$ \frac{1}{p^2+q^2}\left(\begin{array}{cc} p &amp; -q \\ q &amp; p \end{array}\right) $$  where $p,q$ are integers.
9279
Let's suppose that $|z|&lt;1$ and $|w|=1$. Show that the modulus of $\displaystyle \frac{z-w}{1-\bar{z}w}$ is always $1$. Some hint.
30025
I am not able to prove that this set is dense in $\mathbb{R}$. Will be pleased if you help in a easiest way. $A=\{a+b\alpha: a,b\in \mathbb{Z}\}$ where $\alpha$ is a fixed irrational number.
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 $[-x]=-[x]-1$
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 $c=b$.
32601
Problem  One of my best friends asked me to think about the following problem:     Suppose a sequence $\{a_n\}_{n=1}^\infty$ satisfies $\lim_{n\to\infty}a_{\lfloor\alpha^n\rfloor}=0$ for each $\alpha&gt;1$. Is it true that $\lim_{n\to\infty}a_n=0$?   He told me that the preceding proposition (if true) implies Kolmogorov's strong law of large numbers in probability theory. I don't know how, but it's irrelevant.   Thoughts  Suppose $E\subseteq(1,+\infty)$ is countable (for example, $E=(1,+\infty)\cap\mathbb Q$) and $\lim_{n\to\infty}a_{\lfloor\alpha^n\rfloor}=0$ for each $\alpha\in E$, we cannot conclude that $\lim_{n\to\infty}a_n=0$. In fact, we can choose an infinite set $S\subseteq\mathbb Z_{&gt;0}$ such that $S\cap\{\lfloor\alpha^n\rfloor\colon n\in\mathbb Z_{&gt;0}\}$ is finite for each $\alpha\in E$ as follows:  Suppose $E=\{\alpha_1,\alpha_2,\dotsc\}$. We choose $S$ inductively. Suppose $T_0=\mathbb Z_{&gt;0}$. Given $T_{n-1}$, we set $s_n=\min T_{n-1}$ and $T_n=(T_{n-1}\setminus\{\lfloor\alpha_n^k\rfloor\colon k\in\mathbb Z_{&gt;0}\})\setminus\{s_n\}$. By a density argument, it's easy to see that $T_n$ are infinite therefore the process doesn't terminate. Let $S=\{s_n\colon n\in\mathbb Z_{&gt;0}\}$. It's apparent that $\#(S\cap\{\lfloor\alpha_n^m\rfloor\colon m\in\mathbb Z_{&gt;0}\})\le n$.  Given $S$, we set $a_n=1/n$ if $n\not\in S$, and $a_n=1$ if $n\in S$, then $\lim_{n\to\infty}a_{\lfloor\alpha^n\rfloor}=0$ for each $\alpha\in E$, but $\lim_{n\to\infty}a_n$ doesn't exist.  Here we choose $S$ by a diagonal process, therefore we cannot mimic the construction when $E$ is uncountable. In fact, the falsehood of the original statement is equivalent to the existence of $S$, therefore it's essential combinatorial, or related to some topological structure of $\mathbb Z_{&gt;0}$ (say, compactness or Baire category, etc). I have no idea on the general case. Any idea? Thanks!
20989
I want to show that     For all irrational numbers $a \in \mathbb{R}$ there exists an $Q \in \mathbb{N}$ such that $|a-\frac{p}{q}| \lt \frac{1}{qQ}$ for $p,q \in \mathbb{Z}$ and $1 \le q \le Q$.   My attempt:  Let $a \in [0,1)$ and $Q \in \mathbb{N}$. Divide $[0,1)$ in $Q$ disjunct subintervals $I_i := [\frac{i}{Q}, \frac{i+1}{Q})$ for $i \in \{0,1,\ldots, Q-1\}$. Consider the fractional parts $\{na\} \in [0,1)$ of the number $na$ for $n \in \{0,1,\ldots, Q\}$. Since there are in total $Q+1$ such fractional parts but only $Q$ subintervals, by the Pidgeonhole Principle there must be at least one subinterval that contain at least two fractional parts. Let such two fractional parts be $\{na\}, \{ma\} \in I_i$. Wlog let $m \lt n$. Then  $$|\{na\} - \{ma\}| \lt \frac{1}{Q}$$  since those two fractional parts are in the same $I_i$ which has a width of $\frac{1}{Q}$. By the definition of fractional parts  $$\{na\} - \{ma\} = na - \lfloor na \rfloor - (ma - \lfloor ma \rfloor) = na - ma - \lfloor na \rfloor + \lfloor ma \rfloor = aq - p$$  for $q := n-m$ and $p := \lfloor na \rfloor - \lfloor ma \rfloor$. Also $p,q \in \mathbb{N}$, because $m \lt n$. It follows then   $$|aq - p| \lt \frac{1}{Q} \iff |a - \frac{p}{q}| \lt \frac{1}{qQ}$$  If $a \ge 1$ then we can write $a = a' + k$ for some $a' \in [0,1)$ and $k \in \mathbb{N}$. Then with $p' = p+qk \in \mathbb{N}$  $$|a - \frac{p}{q}| = |a' + k - \frac{p}{q}| = |a' - \frac{p+qk}{q}| = |a' - \frac{p'}{q}| \lt \frac{1}{qQ}$$  by the first case.  If $a \lt 0$ then the rational number must be negative too, so let $p \in \mathbb{Z}$ and $p \le 0$. Then with $a' := -a \gt 0$ and $p' := -p \ge 0$  $$|a - \frac{p}{q}| = |-(a' - \frac{p'}{q})| = |a' - \frac{p'}{q}| \lt \frac{1}{qQ}$$  by the first case again.  My question:  Is my proof correct? Is there anything which could be done better?  I'm still not really sure about the dependencies. $Q$ is really $Q_a$ so so only dependend on $a$?! And is $p,q$ really $p_a,q_a$ only dependent on $a$ too or also on $Q$?  Could I instead take numbers $na$ for $n$ from another subset than $\{0,1,\ldots, Q\}$ provided it has the same cardinality?  How can I guarantee that $p' \in \mathbb{N}$ because $k$ could be a negative integer. Also I think the final argument I gave, why considering  $a \in [0,1)$ is sufficient, is not really the essence.
89219
Could someone provide, please, a proof of the theorem below?  "Being $x$ and $b$ integers greater than $1$, which can not be represented as powers of the same basis (positive integer) and integer exponent, then the logarithm of $x$, in base $b$, is an irrational number."  Well... Assuming that the logarithm is a rational number $p/q$ ($p$ and $q$ are relatively prime integers), I know I can write $x^q = b^p$, but I can not conclude from this fact that $b$ and $x$ are powers of a same integer (and with integer exponent).
76017
I know and I have proved more than once that the set of irrational numbers ($\mathbb{I}$) is uncountable, but now I'm given to solve this problem:   Show that $|\mathbb{I}|=|\mathbb{R}|$,  How can I do that?  Do I need to assume the Continuum Hypothesis in order to that statement to be true?  Thanks
77194
$\newcommand{\R}{\mathbb R}$  Let $f: \R \to \R$ be an increasing function. $(a \leq b \Rightarrow f(a) \leq f(b))$. I want to prove that $f \in \mathcal B$, the Baire one functions.  I first tried to prove that $g(x) := \lfloor f(x)+x \rfloor$ is in $\mathcal B$. The points of discontinuity of $g$ are countable since $g$ is increasing. Now, let $\mathcal D$ be the points in $\R$ where $g$ is not continuous. If $z \in \mathcal D$ then I want to prove that $\exists \delta &gt; 0: (z-\delta,z+\delta) \cap \mathcal D = \{z\}$, i.e. $z$ is the only discontinuity in a certain neighborhood. Then it is not hard to prove that $g \in \mathcal B$ and that $f$ is the uniform limit of $\mathcal B$ functions and thus is also $\mathcal B$.  I already tried to prove this (existence of $\delta &gt; 0$) by contradiction. However, I could not see a nice way to get here a contradiction.
114439
Let $\alpha$ be an irrational real number and let $a_j$ be a sequence of rational numbers converging to $\alpha$. Suppose that each $a_j$ is a fraction expressed in lowest terms: $a_j = \alpha_j / \beta_j$. Prove that the $\beta_j$ tend to $\infty$  Attempt: AFSOC that $\beta_j$ does not tend to infinity, then it is bounded by some $M$. We can also find an interval $\alpha \in (k, k+1)$. Let $\epsilon &gt; 0$, $\exists N_0$ such that for all $N &gt; N_0$ we have $|a_N - \alpha|  &lt; \epsilon$. So $|\frac{p_N}{q_N} - \alpha| &lt; \epsilon$. We try to use the fact that $q$ is bounded but fail to derive a contradiction. I do not know how to use the fact that $\alpha$ is between two integers, although it might not be relevant at all.  Ideas?
29843
Let $\alpha = \frac{1+\sqrt{-19}}{2}$. Let $A = \mathbb Z[\alpha]$. Let's assume that we know that its invertibles are $\{1,-1\}$. During an exercise we proved that:     Lemma: If $(D,g)$ is a Euclidean domain such that its invertibles are $\{1,-1\}$, and $x$ is an element of minimal degree among the elements that are not invertible, then $D/(x)$ is isomorphic to $\mathbb Z/2\mathbb Z$ or $\mathbb Z/3\mathbb Z$.   Now the exercise asks:     Prove that $A$ is not a Euclidean Domain.   Everything hints to an argument by contradiction: let $(A, d)$ be a ED and $x$ an element of minimal degree among the non invertibles we'd like to show that $A/(x)$ is not isomorphic to $\mathbb Z/2\mathbb Z$ or $\mathbb Z/3\mathbb Z$.  How do we do that? My problem is that, since I don't know what this degree function looks like, I don't know how to choose this $x$!  I know that the elements of $A/(x)$ are of the form $a+(x)$, with $a$ of degree less than $x$ or zero. By minimality of $x$ this means that $a\in \{0, 1, -1\}$. Now I'm lost: how do we derive a contradiction from this?
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 !
13134
There's a problem statement:     For each $m \in \mathbb{N}$, we construct a sequence $m_0$, $m_1$, $m_2,\dots$ denoted $S_m$, recursively via $m_0=m$ and      $$m_{i+1} = m_i + \left\lfloor \sqrt{m_i} \right\rfloor$$      for all $i \ge 0$.   Here, $\lfloor x \rfloor$ is the floor of $x$, the greatest integer less than or equal to $x$.   Hence, we have $\left\lfloor \sqrt{10} \right\rfloor=3$ and $\left\lfloor \sqrt{29} \right\rfloor=5$.      Show that for each positive integer $m$, the sequence $S_m$ contains the square of some integer.   I'm pretty certain that this can be proved with induction. I am just not quite sure what to induct on. Examining examples shows that $S_m$ always results in a perfect square eventually, though I'm not sure how to prove it.
117048
Every finite subset of $\mathbb R$ contains its supremum (and its infimum)   Proof Let $A=\{a_1,...,a_n\}$ be a finite subset of $\mathbb{R}$. Since it is non-empty and it is bounded ($\max A$ is an upper bound), it has supremum, that is $\exists \sup A$ and by definition $\forall a \in A \;\, a \leq \sup A$. Let's suppose that $\sup A  \not\in A$ then, since $\max A \in A$ we have that $\max A &lt; \sup A$. But considering that $\mathbb Q$ is dense in $\mathbb R$ we can conclude that $\exists r \in \mathbb Q$ s.t. $\max A &lt; r &lt; \sup A$, but this is absurd since $r$ is an upper bound of $A$ and it is lower than the supremum. Necessarily, $\sup A\in A$.  Is there anything wrong? Is there any way to prove this without using density of $\mathbb Q$ or another property? Thanks in advance.