They're obviously not opposed to messing with other races but do they place limits on their interference and/or interactions?  We have witnessed Q (John de Lancie) change or alter many things for his amusement, but he only goes so far and usually puts things back as they were when he gets bored or done with whatever he's up to. I have never seen them 'advance' a civilisation, though Q has saved one or two.  Do they have a rule for this?
Are there any rulesets or laws for Q or Q continuum? Anything, that would prevent them from doing... anything? Or at least limit them from doing something really bad?  In "" episode Q explains to  that they can do, have, get, become anything they want or appear anywhere they want, with just a snap of fingers. If that is true and if there are absolutely no borders or limits, they can't pass, then how can universe work in general? Or, how did it manage to exist so far?  I mean -- any Q could wipe out Borg, entire human population or even entire universe... with just a snap of fingers. What could prevent Q from doing that?
A command-line or batch cmd to concatenate multiple files I have 50 text files in one directory.  Is there a Windows command-line method to concatenate those files into a single file?  I am using .  I don't want to type the name of all files. Reset the password in Ubuntu / Linux Bash in Windows I have installed Linux Bash on Windows 10 (available from Microsoft), but I've forgotten my password.  I've searched for a solution but I only find suggestions that involve doing something before the system boots, like accessing recovery mode, but that will not work in this case as this Ubuntu is just an application within a Windows system and doesn't have its own boot process.  How can I reset my password? If $\sqrt[3]{a} + \sqrt[3]{b}$ is rational then prove $\sqrt[3]{a}$ and $\sqrt[3]{b}$ are rational Assume there exist some rationals $a, b$ such that $\sqrt[3]{a}, \sqrt[3]{b}$ are irrationals, but:  $$\sqrt[3]{a} + \sqrt[3]{b} = \frac{m}{n}$$  for some integers $m, n$  $$\implies \left(\sqrt[3]{a} + \sqrt[3]{b}\right)^3 = \frac{m^3}{n^3}$$  $$\implies a + b + 3 \cdot \sqrt[3]{ab}\left(\sqrt[3]{a} + \sqrt[3]{b}\right) = \frac{m^3}{n^3}$$  Since $a +b$, $\sqrt[3]{a} + \sqrt[3]{b}$ are rational, $\sqrt[3]{ab}$ must be rational as well.   For convenience let us say $\sqrt[3]{a} = p, \sqrt[3]{b} = q \implies pq$ is rational. This means, for all $i$, $p^iq^i$ is rational.  $$\implies (p + q)^2 = \frac{m^2}{n^2}$$  $$ = p^2 + q^2 + 2pq = \frac{m^2}{n^2}$$  Since $pq$ is rational, $2pq$ is rational and so is $p^2 + q^2$.  Assume, for some $i$ that $p^i + q^i$ and $p^{i-1} + q^{i-1}$ is rational.   $$\implies (p^i + q^i)(p + q) - pq(p^{i - 1} + q^{i - 1}) = p^{i+1} + q^{i+1}$$  is rational as well.  So for all $i$,  $$a^{\frac{i}{3}} + b^{\frac{i}{3}}$$  and   $$a^{\frac{i}{3}}b^{\frac{i}{3}}$$  are rational.  I know I'm really close to the answer, but it somehow just keeps slipping through my fingers. I have a Dell Studio 17. It's a few years old, and I wanted to upgrade the processor. It currently has an Intel Core 2 Duo T5750 @ 2.00GHz  My question is how can I find out which processors are compatible? What is the status of pdfsync? The pdfsync homepage states that:     At the end of this summer (2007), pdfsync is slowly migrating to pdftex itself. We are working at giving pdftex all the pdfsync ability and more...   The pdfsync README file discourages people from using pdfsync:     pdfsync uses extremely sensible code. You should not use pdfsync on final documents because it can change the layout rather significantly.   From your knowledge and experience, what is the actual status of pdfsync?   Is it safe to use pdfsync? Is there an alternative for pdfsync that is built into pdftex? Is there a different tool for pdflatex that offers DVI-like functionality (forward and backward search)? How do I specify a pointer to an overloaded function? Proving the two given vectors are orthogonal I am given the vectors $\mathbf{w}, \mathbf{v}, \mathbf{u}$ in $\mathbb{R}^n$ such that $\mathbf{u} \neq 0$ and $\mathbf{w} = \mathbf{v} - \frac{\mathbf{u}\,\bullet\,\mathbf{v}}{\|\mathbf{u}\|^2}\bullet\mathbf{u}$. I am asked to show that the vector $\mathbf{w}$ is orthogonal to $\mathbf{u}$.     So far, I have written out the definition of orthogonal: two vectors are orthogonal if and only if their dot product is zero.   So what we need to prove is $\mathbf{w}\bullet\mathbf{u} = 0$ where $\mathbf{w}\bullet\mathbf{u}$ is defined as $\mathbf{w}^T\bullet\mathbf{u}$.   However, I have been stuck on this problem for about an hour and haven't made any significant progress from here. How do we go about proving that the vectors are orthogonal? How do I check if someone accessed my computer? I need to check the list of files accessed in my computer (ex: yesterday on particular time). Is it possible?   Or   I need to check whether anybody accessed my computer (when I left it unlocked) or not for a particular time. This question has been asked before, at least twice by people who were trying to cheat on the PROMYS Europe 2020 Application, and once in a  because the OP showed no effort. I haven't been able to solve it, so I'm going to to ask it again.  PROMYS applications have closed, and I hope the effort shown below is enough to keep this question from being closed.     The Mathematical Forest is grown in a two-dimensional plane, where   trees can only grow on points with integer coordinates. To start with,   there are no trees at all. The foresters plant the first tree at   $(0,0)$. Each year, they carry out tree planting according to the   following rule. If there is a tree on the point (𝑚,𝑛) but there are   no trees on the points $(m+1,n)$ and $(m,n+1)$, then they can choose   to remove the tree on $(m,n)$ and plant new trees on the points   $(m,n+1)$ and $(m+1,n)$. For an integer $k\geq1$, the $k$th diagonal   consists of all points $(m,n)$ with $m+n=k−1$. Is it possible for the   foresters to arrange their planting so that eventually there are no   trees on the first $2$ diagonals? What about the first $3$ diagonals?    $4$ diagonals? Can you generalize?   It's easy to get to a position with no trees on the first two diagonals.  (It only takes $4$ plantings.)  I can prove it's impossible to reach a position with no trees on the first $4$ diagonals, and I believe it's impossible to reach a position with no trees on the first $3$ diagonals, but I can't prove it.  For the $k=4$ case, I used a pagoda function, as in a peg solitaire problem.  For $k=1,2,3,\dots$ define the potential of a tree on the $k$th diagonal as $\frac1{2^{k-1}}$, and the potential of the forest as the sum of the potentials of all the trees in it.  When a tree of potential $\frac1{2^{k-1}}$ is removed, it is replaced by $2$ trees of potential $\frac1{2^k}$ so the potential of the forest never changes.  Initially, the potential of the forest is $1$.  Since there are $k$ trees on diagonal $k$, the potential of all points beyond the fourth diagonal is $$\sum_{k=5}^\infty\frac k{2^{k-1}}=\frac34&lt;1,$$ so it is impossible that there are no trees on the first $4$ diagonals.  The potential of the fourth diagonal is $\frac12$, so this argument doesn't show that it's impossible to have no trees on the first $3$ diagonals.  I've done a lot of experimenting, though, and I believe that the statement is true.  (I even wrote a little computer game so I could experiment quickly.)  The best I've been able to do is to get down to one man on the third diagonal.   The black circles are trees eligible for removal, and the gray circles are other trees.  You can see that there is a "traffic jam" in front of the lone tree on diagonal $3$, and there seems to be no way to clear it.  I've been trying to come up with an argument that reflects this traffic jam, but I haven't been at all successful. We know that after $n$ removals there are $n+1$ trees, and I've been trying to prove somehow that they can't all be sufficiently far from the origin to allow all trees to get past diagonal $3$, but I haven't come close.   In the diagram above, if the tree at $(1,1)$ were moved to $(0,2)$, and the tree at $(1,2)$ were moved to $(2,0)$, then it would be possible to remove the tree at $(0,2)$, emptying the second diagonal.  In the fictitious position, we still have one tree on diagonal $2$ and two on diagonal $3$.   So, I think an argument along the lines I was trying cannot just be concerned with the distance from the origin; it must somehow take into account how the trees interfere with one another.  I haven't been able to do this.    I'd be grateful for solutions, hints, or counterexamples.  I wasn't really sure what tags to attach.  Please add any that seem appropriate. terminator: permission denied for sudo Prove $\sin(x)\tan(x) > x^2$ for $x \in ( \,0, \frac{\pi}{2}) \,$ Prove $\sin(x)\tan(x) &gt; x^2$ for $x \in ( \,0, \frac{\pi}{2}) \,$  So I did the following: Let $f(x) = \sin(x)\tan(x) - x^2$.  Then of course $f(0) = 0$ and I want to show that for $x \in ( \,0, \frac{\pi}{2}) \,$ $f'(x) &gt; 0$ but I don't really know where to go from here as I can't get anything reasonable done with the derivative.  I was thinking of approximating it using the following inequality $\sin(x) &lt; x &lt; \tan(x)$.   Is that a good direction? Let $m$ and $n$ be two integers. Prove that if $m^2 + n^2$ is divisible by $4$, then both $m$ and $n$ are even numbers.   I think I have to use the contrapositive to solve this. So I assume $\neg P\implies Q$ and I have to derive $\neg Q$? I know its easier to show odd numbers rather than even... but im still unsure how to do this practice problem. linear dependence proof using subsets Subdomain versus subdirectory I recently moved from having a main site like www.example.com with a subdomain associated site yyy.example.com, to moving everything off the subdomain to a subdirectory like www.example.com/yyy/, with the same hierarchy underneath the subdirectory, and now yyy.example.com is just a redirect page to the subdirectory, so yyy.example.com/abc/page-x is redirected to www.example.com/yyy/abc/page-x and so on.  The effect has been that the traffic increased by several times (the total page views are more than ten times at the present rate) what it was to the yyy subdomain. Previously almost all of the traffic to the yyy subdomain was from the www site. This is from monitoring the old yyy site for a year or more.  I'm curious to know if there's any research or results which would indicate whether this is universal (subdirectories beat subdomains) or "just me". The recipe that I used did not call for cornstarch or corn syrup. The result is that they are slightly wet. Can I do anything to dry them up, such as dusting them with corn flour?  I did dredge them in a mixture of confectioners sugar and corn flour; it helped a little, but I think I will find a recipe that has corn starch in it for next time! How do I change a close vote to a duplicate? Voted to close a question based on "lack of minimal understanding".  Later realized the question was a duplicate.  How to I change vote to mark as duplicate instead? Why is $2\pi i \neq 0?$ How do I brew an X potion? external variable in awk What is a comprehensive definition of Philosophy? Alternatively, is it impossible to define Philosophy? This is a pseudo- question, but it seems like it belongs here.