I initially asked this question on MSE, but then thought it would appropiate to ask it here as well ( correct me if I'm wrong):   A well-known application of the MVT is to prove that any $f: \mathbb{R} \to \mathbb{R}$ with $f'= 0$ is constant. But of course, the MVT relies fundamentally on the properties of $\mathbb{R}$, and in particular does not hold in $\mathbb{Q}$. A standard counter-example is the function $f :\mathbb{Q} \to \mathbb{Q}$ defined by $ f(x)=0 $ if $ x &lt;\sqrt{2}$, and $f(x)=1$ if $ x &gt; \sqrt{2}$, for this function is locally constant. There are several ways of thinking about the difference between $\mathbb{R}$ and $\mathbb{Q}$ : $\mathbb{R}$ is connected while $\mathbb{Q}$ is totally disconnected, closed bounded intervals are compact in $\mathbb{R}$, but not in $\mathbb{Q}$, $\mathbb{R}$ satisfies the least upper bound axiom, $\mathbb{Q}$ does not.   Now, I wanted to see how far one could push this counterexample - is there a function $f :\mathbb{Q} \to \mathbb{Q}$ such that $f'=0$ and $f$ is strictly increasing? Clearly, any slight modification of the usual counterexample where $f$ is locally constant will not do. Also, one can show that if $f$ is uniformly differentiable with derivative zero, it must be constant, even in $\mathbb{Q}$, so I knew I was looking for non-uniformly differentiable but still differentiable $f$.   Eventually, I came up with a construction, albeit not a very explicit one: the idea is that, since $\mathbb{Q}$ is countable, one can enumerate it $q_{1}, q_{2}, q_{3}, \cdots$, and then define $f: \mathbb{Q} \to \mathbb{Q}$ inductively and recursively. If one could define $f$ in such a way that $ \forall \ q_{n} \in \mathbb{Q} \ \exists \ c_{n} \in \mathbb{Q} : \forall \ x \in \mathbb{Q} \ |f(x)-f(q_{n})| \leq c_{n}|x-q_{n}|^2$, and $f$ is strictly increasing, we would be done. The recursive definition essentially defines $f(q_{n+1})$, after having defined $f(q_{1}),f(q_{2}),f(q_{3}), \cdots , f(q_{n})$, in such a way that these inequalities hold and $f$ is strictly increasing, and at each step the new $c_{n}$ is chosen large enough so that these inequalities do not become incompatible at subsequent steps ( I am willing to provide more details of this construction if asked).  Note that, as $n \to \infty$, we must have $c_{n} \to \infty$, so that $f$ is not uniformly differentiable. Interestingly, the only key fact about $\mathbb{Q}$ used in the construction is that it is countable, so in a sense the construction provides a proof that $\mathbb{R}$ is uncountable, given the MVT.  While the construction works, it is not very simple nor explicit, which is perhaps to be expected, and the function must be evaluated point by point in the order given by the enumeration of $\mathbb{Q}$. So I was wondering if anyone else knows or can come up with a simpler construction of a function $ f : \mathbb{Q} \to \mathbb{Q}$ such that $f'=0$ and $f$ is strictly increasing. Any thoughts and ideas are welcome.
A well-known application of the MVT is to prove that any $f: \mathbb{R} \to \mathbb{R}$ with $f'= 0$ is constant. But of course, the MVT relies fundamentally on the properties of $\mathbb{R}$, and in particular does not hold in $\mathbb{Q}$. A standard counter-example is the function $f :\mathbb{Q} \to \mathbb{Q}$ defined by $ f(x)=0 $ if $ x &lt;\sqrt{2}$, and $f(x)=1$ if $ x &gt; \sqrt{2}$, for this function is locally constant. There are several ways of thinking about the difference between $\mathbb{R}$ and $\mathbb{Q}$ : $\mathbb{R}$ is connected while $\mathbb{Q}$ is totally disconnected, closed bounded intervals are compact in $\mathbb{R}$, but not in $\mathbb{Q}$, $\mathbb{R}$ satisfies the least upper bound axiom, $\mathbb{Q}$ does not.   Now, I wanted to see how far one could push this counterexample - is there a function $f :\mathbb{Q} \to \mathbb{Q}$ such that $f'=0$ and $f$ is strictly increasing? Clearly, any slight modification of the usual counterexample where $f$ is locally constant will not do. Also, one can show that if $f$ is uniformly differentiable with derivative zero, it must be constant, even in $\mathbb{Q}$, so I knew I was looking for non-uniformly differentiable but still differentiable $f$.   Eventually, I came up with a construction, albeit not a very explicit one: the idea is that, since $\mathbb{Q}$ is countable, one can enumerate it $q_{1}, q_{2}, q_{3}, \cdots$, and then define $f: \mathbb{Q} \to \mathbb{Q}$ inductively and recursively. If one could define $f$ in such a way that $ \forall \ q_{n} \in \mathbb{Q} \ \exists \ c_{n} \in \mathbb{Q} : \forall \ x \in \mathbb{Q} \ |f(x)-f(q_{n})| \leq c_{n}|x-q_{n}|^2$, and $f$ is strictly increasing, we would be done. The recursive definition essentially defines $f(q_{n+1})$, after having defined $f(q_{1}),f(q_{2}),f(q_{3}), \cdots , f(q_{n})$, in such a way that these inequalities hold and $f$ is strictly increasing, and at each step the new $c_{n}$ is chosen large enough so that these inequalities do not become incompatible at subsequent steps ( I am willing to provide more details of this construction if asked).  Note that, as $n \to \infty$, we must have $c_{n} \to \infty$, so that $f$ is not uniformly differentiable. Interestingly, the only key fact about $\mathbb{Q}$ used in the construction is that it is countable, so in a sense the construction provides a proof that $\mathbb{R}$ is uncountable, given the MVT.  While the construction works, it is not very simple nor explicit, which is perhaps to be expected, and the function must be evaluated point by point in the order given by the enumeration of $\mathbb{Q}$. So I was wondering if anyone else knows or can come up with a simpler construction of a function $ f : \mathbb{Q} \to \mathbb{Q}$ such that $f'=0$ and $f$ is strictly increasing. Any thoughts and ideas are welcome.  Edit: I've asked the same question on MO:
I need to find node voltage at A. This seems like a trivial question, but when I set up the resistive voltage divider equations they are not equal to each other. Am I doing something wrong?  My equations are \$V_A = V_1(R3/(R3+R1)) = V_2(R3/(R3+R2))\$      I tried solving this problem with an LTSpice simulation and with the superposition method, but my answers did not match (although very close to each other). Is this because the simulator uses an approximation? Can someone compare results? Cannot change lock screen background on Ubuntu 18.04 We had a quiz in class today where we had to break the ciphertext with the key given, but not the algorithm. Suffice to say that I wasn't able to decrypt it within the allotted time of 12 mins and will probably get a 0% score on the quiz.  So, I was just wondering if there are some kind of standard techniques that are followed when decrypting a simple substitution ciphertext. How to ask a smart question The title of  of  is, "Is it posible [sic]?"  This doesn't provide any information at all. It should be mandatory for everyone asking questions on the Internet to read , which says:     Help your community learn from the question. When you field a good question, ask yourself "How would the relevant documentation or FAQ have to change so that nobody has to answer this again?" I know someone or the other has asked about disabling ⌘Q, but I like the shortcut, and I use it often. However, I use an email client called Airmail that unfortunately does not keep a service (for notifications) running in the background when it is quitted. However, if I press the close button a service runs in the background, allowing me to receive notifications.  But since I have a habit of pressing ⌘Q all the time to close apps, I was looking into a way I could disable it for only Airmail.  Anyone know of a what to do this? When using sudo powers and calling gedit the top level menu with File Edit View Search Tools Documents Help is missing.  If I create user ID root and sign in once and call gedit will it create the necessary user configuration files that might fix this?  Will this also make the annoying error messages go away whenever I use sudo, gksu or pkexec as a regular user with elevated privileges togedit ?  Will there be other benefits with Nautilus and other Gnome derivative Ubuntu applications?  NOTE: by sudo powers you can google pkexec gedit 3.5k hits, gksu gedit 40k hits or sudo gedit 500k hits. I'm in the minority using the first method but I believe it will become the standard in Ubuntu 17.04.    June 12, 2017 Update Here are a list of errors you get when using pkexec gedit:  (gedit:13003): Gtk-WARNING **: Calling Inhibit failed: GDBus.Error:org.freedesktop.DBus.Error.ServiceUnknown: The name org.gnome.SessionManager was not provided by any .service files  ** (gedit:13003): WARNING **: Set document metadata failed: Setting attribute metadata::gedit-spell-enabled not supported  ** (gedit:13003): WARNING **: Set document metadata failed: Setting attribute metadata::gedit-encoding not supported  ** (gedit:13003): WARNING **: Set document metadata failed: Setting attribute metadata::gedit-position not supported   On this  there is a commitment to fix these distracting messages as of June 5, 2017. No indication is given for when the fix will be upstream nor if the functionality will be implemented or the error messages will simply disappear.  Note the bug report was filed over a year ago and over that time it only effects 18 people. Getting inverse of a transposition key What is the best way to identify outliers in multivariate data? Suppose I have a large set of multivariate data with at least three variables. How can I find the outliers? Pairwise scatterplots won't work as it is possible for an outlier to exist in 3 dimensions that is not an outlier in any of the 2 dimensional subspaces.  I am not thinking of a regression problem, but of true multivariate data. So answers involving robust regression or computing leverage are not helpful.  One possibility would be to compute the principal component scores and look for an outlier in the bivariate scatterplot of the first two scores. Would that be guaranteed to work? Are there better approaches? What I would like is a method to convert a double to a string which rounds using the half-up method - i.e. if the decimal to be rounded is 5, it always rounds up to the next number. This is the standard method of rounding most people expect in most situations.  I also would like only significant digits to be displayed - i.e. there should not be any trailing zeroes.  I know one method of doing this is to use the String.format method:  String.format("%.5g%n", 0.912385);   returns:  0.91239   which is great, however it always displays numbers with 5 decimal places even if they are not significant:   String.format("%.5g%n", 0.912300);   returns:  0.91230   Another method is to use the DecimalFormatter:  DecimalFormat df = new DecimalFormat("#.#####"); df.format(0.912385);   returns:  0.91238   However as you can see this uses half-even rounding. That is it will round down if the previous digit is even. What I'd like is this:  0.912385 -&gt; 0.91239 0.912300 -&gt; 0.9123   What is the best way to achieve this in Java? How to suppress certain .bib fields (e.g. year) only for certain entry types (e.g. inproceedings) Why are my wrists the first part of my body to start aching during a long ride? Most riders seem to say it's their groin, neck or lower back that starts hurting first after a decent distance has been ridden. For me it is my wrists.  I have no history of wrist pain (strains, breaks etc) so it can't be explained by that and I try to adjust my hand positions across my handlebars as often as I can.  My question has 2 parts:   1. Is there anything I can do to minimise pain in terms of riding position, gloves, exercises etc which will reduce soreness?  2. Why might this pain be occurring?  *I do all of my riding on paved roads on a road (Specialized Allez) bike. Parallel.ForEach Slower than ForEach I found a video on YouTube that shows someone using the command tasksel from the server's boot menu after a fresh install of Ubuntu Server. Simple enough it looked like, but when I go to the options I want to choose for installation it has no information on how to select them, and when I click Enter on one of them it takes me back to the command prompt.  So how do I make a selection of the items listed under tasksel?  I have tried Shift++ to add, but no luck. Anyone know?  I want to install Ubuntu Desktop, as Ubuntu GNOME GUI is not an option in the listing for Ubuntu 17.10 Server. Much effort has been expended on a famous unsolved problem about the .  Not surprisingly, it's called the Riemann hypothesis, which asserts:  $$ \zeta(s) = 0 \Rightarrow \operatorname{Re} s = \frac1 2 \text{ or } \operatorname{Im} s = 0 .$$  Now there are numerical methods for approximating $\zeta(s)$, but as I understand, no one knows any exact values except at even integers, which include the trivial zeroes for which $\operatorname{Im} s = 0 $.  So I've always wondered:  For the rest, how does one prove $\operatorname{Re} s = 1/2$ exactly?  (All I know that seems vaguely useful is the argument principle, which I'm not sure helps here, but I'd be happy to learn about other techniques that aren't too far advanced.)  Edit:  Looking over this again, I found out I missed the closed-form values at .  This doesn't affect the question but seemed worth correcting.  Thanks to the people who contributed. Why my differential equations become True? Is it possible to to take a screenshot of a webpage with JavaScript and then submit that back to the server?  I'm not so concerned with browser security issues. etc. as the implementation would be for . But is it possible? As the title, I was asked to show that $2^{44}-1$ is divisible by $89$.  My first attempt is to change the expression into $(2^{22}-1)(2^{22}+1)$.  Then further simplified it into $(2^{11}-1)(2^{11}+1)(2^{22}+1)$, I used my calculator and was able to show that $2^{11}-1$ is divisible by $89$ but then I don't know how to show it with modular arithmetic. I do think that it is quite similar to the form where we can use the Fermat's little theorem. $(\sqrt{2})^{88}-1$. (Though I do understand Flt can only be applied to integer.)  Can someone tell me whether I can take square root in modular arithmetic as well? I am still pretty new to the modular arithmetic. Thank you very much. In Section 9 of Chapter 6, Lang constructrs for the first time a non-abelian extension as a splitting field $K$ of $X^n - a$ when the ground field $k$ does not contain primitive $n$-th roots of unity $\zeta_n$. He constructs an injective homomorphism of the Galois group $G(K/k)$ into the matrices of type   $$ M=\begin{pmatrix} 1 &amp; 0 \\ b &amp; d \end{pmatrix} \quad \hbox{with}\ b\in \mathbb{Z}/n\mathbb{Z} \quad \hbox{and} \quad d\in (\mathbb{Z}/n\mathbb{Z})^*,$$  where $(\mathbb{Z}/n\mathbb{Z})^*$ the multiplicative group of integers modulo n.  Afterwards, he states in Theorem 9.4 that if $n$ is an odd positive integer prime to the characterstic and $[k(\zeta_n):k]=\varphi(n)$ (i.e, the primitive roots of unity are not in the ground field), then the above homomorphism is, in fact, an isomorphism. Besides, the commutator group is $G(K/k(\zeta_n))$, so $k(\zeta_n)$ is the maximal abelian subextension of $K$.  Lang proves the theorem inductively over the number of prime factors in the decomposition of $n$ (i.e. if $n=9= 3\cdot 3$, then the proof "would take two steps" to complete). The induction start with $n$ prime is written clearly and I agree with it.   Now come the questions:   I do not understand Lang's argument to prove that the commutator subgroup is the subgroup $\begin{pmatrix}1 &amp; 0 \\ b &amp; 1\end{pmatrix}$, $b\in \mathbb{Z}/n\mathbb{Z}$. What is this projection on the diagonal and how does one see that the factor group of the commutator group is isomorphic to $(\mathbb{Z}/n\mathbb{Z})^*$?   By the way, direct computation of $x\, y\, x^{-1}\, y^{-1}$ for two arbitrary matrices $x=\begin{pmatrix} 1 &amp; 0 \\ b &amp; d \end{pmatrix}$, $y=\begin{pmatrix} 1 &amp; 0 \\ c &amp; e \end{pmatrix}$ yields a matrix $\begin{pmatrix} 1 &amp; 0 \\ b(1-e)+c(d-1) &amp; 1 \end{pmatrix}$. As $n$ is odd, $2 \in (\mathbb{Z}/n\mathbb{Z})^*$, so taking $e=1$ and $d=2$, we indeed reconstruct the above-mentioned group.   The second (inductive) part of the proof is a complete mystery to me with virtually every sentence being a Theorem which needs an extra proof. Can someone give me some hints?   2.a. Lang writes $n=p\,m$ and takes $\alpha$ to be the root of $X^n-a$. It is clear that $\beta=\alpha^p$ is a root of $X^m -a$. Now comes the argument "By induction we can apply the theorem to $X^m-a$". I agree that we can apply it here but the result is that $K=k(\beta,\zeta_m)$ is the splitting field of $X^m-a$. In the diagram below, there appears $k(\beta,\zeta_n)$ out of nowhere.   2.b. Now Lang proceeds to apply the theorem on $X^p-\beta$ over $k(\beta)$. Bergman's "Companion" reminds to check that $[k(\beta,\zeta_p):k(\beta)]=\varphi(p)$ before applying the theorem and gives a hint "$[k(\zeta_p):k]$ and $[k(\beta):k]$ are relatively prime". I disagree with this hint. Take $n=21= m p$ where $m=3$ and $p=7$. Then $[k(\zeta_p):k]=\varphi(p)=p-1=6$ but $[k(\beta):k]=m=3$ and both are not co-prime. Can we apply the theorem here at all? Or if so what is the argument? If Bergman's argument is valid, where is the fault in my logic?  2.c. Even assuming that we can apply the theorem in 2.b., then we can say something about $k(\alpha, \zeta_p)$ and not $k(\alpha,\zeta_n)$. How do we proceed to checking everything Lang states in the last section of the proof (basically, everything after the diagram)?  EDIT: I have found the essentially same question How do I disable a bottom panel with the list of open windows on GNOME 3? How do I disable "Workspace dock" (a bottom panel with the list of open windows and a workspace indicator-cum-switcher), the one that is over "Plank" in the screenshot below?    I'm using Ubuntu 17.10. Explanation of terminal objects using limits I understand terminal objects, but I falter when trying to achieve them using the approach of diagrams, the category of constant diagrams $\mathcal{C}^{\mathcal{J}}$ and limits.  I have the empty category 0, now I have a diagram $D: \mathcal{J} \to \mathcal{C}$. However this diagrams image should be empty, because there are no objects or morphisms to map. At this point, I'm confused, because I can't get to the universal property $\forall c \in \mathcal{C}. \exists! f: c \to T$ anymore, because I'm missing T. (EDIT: I get that this is how it is supposed to be)  But continuing with the category of constant diagrams, even if I have a constant diagram mapping everything to one object / morphism, there is nothing to map, so the functors image is still empty. So now I'm missing both $T$, and any kind of $c$.  EDIT: It seems to me, have I chosen 1, instead of 0, I would have come up with that is considered the definition of the terminal category. What am I missing?