Reflexive, symmetric or non transitive relations? I'm currently doing work on discrete mathematics in my free time and am having some difficulties with understanding some questions pertaining to Relations and Functions. To be specific, I'm stuck on the following practice question:      Let A = {a, b, c, d}. Give an example of a relation R on A^2 which is     reflexive, symmetric and not transitive. Explain.   I don't really understand how to do reflexive etc., how would I find the resulting relation? Any help is appreciated! Thanks.
Give an example of a relation R on $A^2$ which is reflexive, symmetric, and not transitive I am just looking for some clarification on this exercise:  Let $A = \{a,b,c,d\}$. Give an example of a relation $R$ on $A^2$ which is reflexive, symmetric, and not transitive.  I understand that if I was only dealing with $A$ and and not $A^2$ I could give an answer similar to:  $R = \{(a,a),(a,b),(a,d),(b,a),(b,b),(c,c),(d,a),(d,d)\}$  Here $R$ is reflexive, symmetric, and not transitive.  What is confusing me is the fact that we have $A^2$. I am not sure what I am missing here. I am looking through my book to see if I can figure this out, but if someone could help shed some light on this it would be appreciated.  Thanks,  Tony
From where do the permanent magnets get energy from? I have a doubt about permanent magnets. If a magnet is permanent it can attract some materials permanently.  Attracting something involves energy. If a permanent magnet can do this forever, from where does this energy come from? How can it not run out of energy? Doesn't it contradict with the laws of energy? Cold fusion is being mentioned a lot lately because of some new setup that apparently works. This is an unverified claim.  See for example:         ( of that last link in the Wayback Machine, given frequent  from that page.)   While we should give the scientific community time to evaluate the set up and eventually replicate the results, there is undoubtedly some skepticism that cold fusion would work at all, because the claim is quite extraordinary.  In the past, after Fleischmann and Pons announced their cold fusion results, in perfectly good faith, they were proven wrong by subsequent experiments.  What are the experimental realities that make Fleischmann and Pons style cold fusions experiments easy to get wrong?  Would the same risks apply to this new set up? I am given two charged particles of same charge at a distance of $r$. They initially apply force $F$.  Now an infinite dielectric (of dielectric constant $4$) of width $\frac{r}{2}$ is introduced between the particles. What will be the new force?   I find this problem confusing because I have only been told about forces when its either fully dielectric or fully vaccum given by Coulomb's law. How do we get forces when only partial space is dielectric? Number of automorphisms of a direct product of two cyclic $p$-groups What do the bars between keyframes mean? Is there a way to show the console in a Windows application?  I want to do something like this:  static class Program {     [STAThread]     static void Main(string[] args) {         bool consoleMode = Boolean.Parse(args[0]);          if (consoleMode) {             Console.WriteLine("consolemode started");             // ...         } else {             Application.EnableVisualStyles();             Application.SetCompatibleTextRenderingDefault(false);             Application.Run(new Form1());         }     } } Converting a string to a tuple in python What is the speed of sound in space? Is it possible to add new predictions/autofill in Overleaf? For example, when I use the command (in Overleaf/ShareLaTex):  \begin{figure}[H]     \centering     \includegraphics{some_graphic}     \caption{some_caption}     \label{fig:some_label} \end{figure}   the \includegraphics{} comes without the [] parameter to change the size of the image, and \begin{figure} without the [H] option by default.   Is it possible to renew the command to make it default? By that I mean that when I type the \begin{figure} command it adds the [H] and [h] as well as [width=\textwidth] in the \includegraphics command, rather than making a global change in the preamble e.g. Is it true that $aH = bH$ iff $ab^{-1} \in H$ Let $H$ be a subgroup of a group $G$. The I know that for $a,b\in G$ we have $aH = bH$ if and only if $a^{-1}b \in H$.  My question is if it is also true that $aH = bH$ if and only if $ab^{-1} \in H$? Does it matter where the $-1$ goes?  I am guessing that this is not true since we keep putting the $-1$ on the left element, but I am also thinking that it might be true because they look like the same.  If $G$ is abelian, then it doesn't matter, so a possible counter example would have to involve a non-abelian group. If $H$ is a normal subgroup, then it should also be true because then $a^{-1}bH = Ha^{-1}b$. How to create a new object instance from a Type How should I transform non-negative data including zeros? If I have highly skewed positive data I often take logs. But what should I do with highly skewed non-negative data that include zeros? I have seen two transformations used:   $\log(x+1)$ which has the neat feature that 0 maps to 0. $\log(x+c)$ where c is either estimated or set to be some very small positive value.   Are there any other approaches? Are there any good reasons to prefer one approach over the others? In-universe, a lightsaber is basically a controlled, fixed-length plasma stream that can be used as a sword. As such, the blade wouldn't have any mass (or infinitesimal mass if you want to get technical about E=mc2).  Anyone familiar with fencing or swordplay knows that the weight of the handle and blade, and thus the overall balance of the weapon, is critically important while fighting. A sword that is too blade-heavy is unwieldy; it is difficult to start, stop, and change the path of the blade as it's swung, which is vital to outmaneuvering your opponent in a duel. Conversely, a sword with too light a blade (equivalently, too heavy a hilt) is too easy to swing around. it's hard to "feel" where the blade is, and thus to make the minute changes needed to block a blow or land one. It can also be just as exhausting, in a different way, to wield such a sword; because most of the mass is behind your hand, it takes muscle power to point the blade forward instead of upward, while a balanced weapon would point directly forward from the user's hands. You also get no mechanical advantage while swinging it.  Obviously, out-of-universe the props were full "swords" that were then rotoscoped to paint the beam blade over the white wooden (later plastic) dowel that protruded from the saber. This gives the actual prop used by the actors a more realistic weight and balance, which can be seen in the duels (Darth Maul has to heave a bit to get that double-saber staff moving, and you can see the effort behind a full baseball swing of the sword by Luke or Vader in the original trilogy).  Was there any attempt in-universe to explain the weight and balance that an active lightsaber seemed to possess? There is a nice property of normal spaces, namely, closed disjoint subsets can be separated by continuous functions into $\mathbb R$.   Then you ask yourself, what about Hausdorff spaces?, are all Hausdorff spaces completely Hausdorff?   Well, if we choose our separating space $X$, that is, the space such that for any Hausdorff $Y$ and any distinct $a,b\in Y$ there is some continuous $f:Y\rightarrow X$ with $f(a)\neq f(b)$, to be completely regular and Hausdorff, this cannot happen by an example of E. Hewitt, Hewitt’s Condensed Corkscrew, call it $\mathbb H$; as any continuous function from $\mathbb H$ into $\mathbb R$ is constant, and any Hausdorff completely regular space $X$ can be embedded into a product of closed intervals $\prod_{i\in I}[0,1]$, and thus any continuous $f:\mathbb H\rightarrow X\subseteq \prod_{i\in I}[0,1]$ is constant, as each $\pi_i\circ f$ is constant, so our candidate to be the separating space $X$ cannot even be $[0,1]$.  My question is, does there exist a Hausdorff space $X$ such that for all Hausdorff spaces $Y$, and any distinct $a,b\in Y$ there exists a continuous $f:Y\to X$ such that $f(a)\neq f(b)$?  Thanks. I am trying to perform a multiple regression in R. However, my dependent variable has the following plot:    Here is a scatterplot matrix with all my variables (WAR is the dependent variable):    I know that I need to perform a transformation on this variable (and possibly the independent variables?) but I am not sure of the exact transformation required. Can someone point me in the right direction? I am happy to provide any additional information about the relationship between the independent and dependent variables.  The diagnostic graphics from my regression look as follows:    EDIT  After transforming the dependent and independent variables using Yeo-Johnson transformations, the diagnostic plots look like this:    If I use a GLM with a log-link, the diagnostic graphics are: All of my Firefox extensions were spontaneously and simultaneously disabled (converted to Legacy status) a few moments ago.  How can I re-enable them? Firefox 66.0.3 (64-bit) Plotting the function $f(x)=x^{1/3}$ defined for any real number $x$ gives us:   Since $f$ is a function, for any given $x$ value it maps to a single y value (and not more than one $y$ value, because that would mean it's not a function as it fails the vertical line test). This function also has a vertical tangent at $x=0$.   My question is: how can we have a function that also has a vertical tangent? To get a vertical tangent we need 2 vertical points, which means that we are not working with a "proper" function as it has multiple y values mapping to a single $x$. How is it possible for a "proper" function to have a vertical tangent?  As I understand, in the graph I pasted we cannot take the derivative of x=0 because the slope is vertical, hence we cannot see the instantaneous rate of change of x to y as the y value is not a value (or many values, which ever way you want to look at it). How is it possible to have a perfectly vertical slope on a function? In this case I can imagine a very steep curve at 0.... but vertical?!? I can't wrap my mind around it. How can we get a vertical slope on a non vertical function? JavaScript curry: what are the practical applications? I don’t think I’ve grokked currying yet. I understand what it does, and how to do it. I just can’t think of a situation I would use it.  Where are you using currying in JavaScript (or where are the main libraries using it)? DOM manipulation or general application development examples welcome.   mentions animation. Functions like slideUp, fadeIn take an element as an arguments and are normally a curried function returning the high order function with the default “animation function” built-in. Why is that better than just applying the higher-up function with some defaults?  Are there any drawbacks to using it?  As requested here are some good resources on JavaScript currying:    Crockford, Douglas (2008) JavaScript: The Good Parts  (Takes a detour into ML so skip the whole section from “A crash course in ML” and start again at “How to write curried JavaScript”)    (Mr. Resig on the money as per usual)    I’ll add more as they crop up in the comments.    So, according to the answers, currying and partial application in general are convenience techniques.  If you are frequently “refining” a high-level function by calling it with same configuration, you can curry (or use Resig’s partial) the higher-level function to create simple, concise helper methods. How to Store Fresh Herbs Is Canon 60D still a decent camera? I have a Canon 60D, bought it back in 2013. Use it quite a while, but in past two-three years, I lost interest in photography. I will be going on a long trip this summer and would like to take a decent camera with me. Is Canon 60D still a decent one or I will be missing on features and upgrades comparing to the latest versions?  In terms of lenses I own Tamron 24-70 2.8 And Canon 100 2.8 Macro.