Why is the total energy of an electron in an atom negative?
Negative energy levels in the diagram for a hydrogen atom
I have 16GB RAM. Do I need 32GB swap? I read many places that the rule of thumb for swap space is to double the amount of physical RAM. However, 32 GB does seem a LOT. Do I need that much? Do I need it at all with this high amount of physical RAM? Notify us when the question has been edited after posting an answer Show that $\left| \frac{1}{x}-\frac{1}{z} \right| \le \left|\frac{1}{x} - \frac{1}{y} \right| + \left|\frac{1}{y} - \frac{1}{z} \right|$. I'm trying to prove that  $$ \left| \frac{1}{x}-\frac{1}{y} \right| $$  is a metric on $X = \mathbb R^+$.  The properties of non-negativity and symmetry are obvious (do I actually even need to prove them?), but the tricky one is, as usual the inequality property, i.e.  $$ d(x,z) \le d(x,y) + d(y,z) \text{ for any } z \in X. $$  So, ok, I have that  $$ d(x,z) \le d(x,y) + d(y,z) \\ \iff \left| \frac{1}{x}-\frac{1}{z} \right| \le \left|\frac{1}{x} - \frac{1}{y} \right| + \left|\frac{1}{y} - \frac{1}{z} \right| \\ \iff \left| \frac{1}{x} \right|- \left| \frac{1}{z} \right| \le \left| \frac{1}{x}-\frac{1}{z} \right| \le \left|\frac{1}{x} - \frac{1}{y} \right| + \left|\frac{1}{y} - \frac{1}{z} \right| $$  ...and that's about as far as I've gotten–which is to say, not very far. I also tried multiplying both sides by $ \left| \frac{1}{x}-\frac{1}{z} \right| $, but that wasn't very fruitful. If someone could point me in the right direction, it'd be greatly appreciated. Thanks. How do you get the logical xor of two variables in Python? How to install "vanilla" TeXLive on Debian or Ubuntu? Debian and Ubuntu have very good texlive packaging. But I'd like to use tlmgr to have a more fine-grained control about upgrades and which packages I want to be installed.   So how do I install "vanilla" TeXLive on Debian/Ubuntu?   Also how do I make dpkg aware that I have TeXLive installed? How do I time a method's execution in Java? Random point uniform on a sphere How to colorize an image as in the example? I have seen such an effect being used in many places however I haven't figured out how to replicate it via Photoshop.  I have tried desaturating the image, placing it under a layer of solid colour and then lowering the opacity of the colour. I have also tried a couple of blend modes, to no avail. The effect is nowhere near the same.  Can someone please tell me the effect used on these images? How does the Ranger's Extra Attack feature work with Two-Weapon Fighting? I think I know the answer to this one but I'm double checking.  At the 5th level the Ranger gets Extra Attack, which lets them attack a second time during the Attack action. If they are fighting with 2 weapons, the second weapon normally uses their bonus action.  Does this mean when using the Extra Attack and using two weapons, a ranger can attack twice during the Attack action, and still make 1 extra attack as a bonus action, for a total of 3 attacks? Number of $n^2\times n^2$ permutation matrices with a 1 in each $n\times n$ subgrid I found the following question in a paper I was trying to solve:  The following figure shows a $3^2 \times 3^2$ grid divided into $3^2$ subgrids of size $3 \times 3$. This grid has $81$ cells, $9$ in each subgrid.    Now consider an $n^2 \times n^2$ grid divided into $n^2$ subgrids of size $n \times n$. Find the number of ways in which you can select $n^2$ cells from this grid such that there is exactly one cell coming from each subgrid, one from each row and one from each column.  My try:  Since we have $n^2$ rows, $n^2$ columns and $n^2$ subgrids in total, we have to choose one and only one cell from each of them. Let's choose them one at a time. We can choose the first cell in $n^4$ many ways. Then, we'll have to avoid that subgrid, that column and that row that we've chosen the first one from when choosing the second cell. So, we have $n^4-n^2-2n(n-1)$ choices. We can continue this to get the total number of possible ways. But, I think there's a hole. Say, we've chosen the first cell from the subgrid of the up-left corner and the second from the subgrid just right to it so that it doesn't violate any rules. Then, when finding the number of ways we can choose the third cell, we would have substracted some of the cells twice. I think you get it. Please, if anyone can help me solving this problem, it'd be greatly appreciated. Order of elements is lcm-closed in abelian groups How can I prove that if $G$ is an Abelian group with elements $a$ and $b$ with orders $m$ and $n$, respectively, then $G$ contains an element whose order is the least common multiple of $m$ and $n$?  It's an exercise from Hungerford's book, but it's not homework. I could not solve it when I take the course on groups and I think it should be easy. There is a hint that says to first prove when $m$ and $n$ are coprimes. I did this part. But I have no idea how to solve the general case.  Thanks. How to play a YouTube clip in background/minimised? Is there any other way to do a clean install withot formatting /home? I think my Ubuntu is a bit broken from messing around with it.  Unfortunately during installation, I did not make a separate partition for /home. Now I need to re-install but want to keep my music and films.  Any ideas... Q. Find all pairs $(p,n)$ of positive integers where $p$ is prime and $p^3-p=n^7-n^3$.   Rewriting the given equation as $p(p+1)(p-1)=n^3(n^2+1)(n+1)(n-1)$, we see that $p$ must divide one of the factors $n,n+1,n-1,n^2+1$ on the $\text{r.h.s}$.   Now, the $\text{l.h.s}$ is an increasing function of $p$ for $p\ge1$. This implies that for any given $n\ge1$, there is exactly one real $p$ for which $\text{l.h.s}=\text{r.h.s}$. For $p=n^2$, we get $\text{l.h.s}=n^6-n^2&lt;n^7-n^3=\text{r.h.s}.$ This means that either $p&gt;n^2$ or $p&lt;n^2$ must hold.  Assuming $p&gt;n^2$, it follows that the prime $p$ cannot divide any of $n,n+1,n-1$. So $p$ must divide $n^2+1$ and hence $p=n^2+1\quad (\because p&gt;n^2)$.  Substituting the value of $p$ in the given equation we get, $n^2+2=n^3-n\implies n^3-n^2-n=2$. As the factor $n$ on the $\text{l.h.s}$ must divide $2$, the above equation has a unique integer solution $n=2$.  Finally, we get $(5,2)$ as the solution to the given equation.      But how do I conclude this is the only solution possible? Also, why does'nt $p&lt;n^2$ (the case which I ignored) hold? As a bonus question, I would like to ask for any alternative/elegant solution (possibly using congruence relations) to the problem. Is it possible for an x86 processor to match an ARM processor in terms of performance per watt? From my personal experience with my tablet, and from the benchmarks and articles I've read, it always seems ARM processors, as seen in virtually all mobile devices, deliver incredible performance for the amount of power they consume.  It really seems like ARM is inherently more power-efficient than x86.  To me, it looks like that this is because x86 is a  while ARM is a . However, Intel Atom processors deliver very good performance relative to their power consumption as well.  In addition, having studied computer architecture, it is my understanding that in modern processors, the complex x86  is internally translated into simpler  that use an internal instruction set that is closer to the simpler ARM or MIPS ISAs than x86 proper.  As such, the ISA as presented to the programmer is little more than an interface to issue commands to the processor, rather than a representation of the actual low-level operations the chip performs.  Then again, implementing this translation layer requires additional silicon space on the chip...  That said, assuming that they are implemented using the same semiconductor process, is ARM inherently more efficient than x86? With a properly designed , is it possible for an x86 processor to deliver the same performance per watt as an ARM processor?  Remember, I'm looking for a well-written technical answer in the spirit of Stack Exchange and not mere speculation. No transaction fee ETF trades - what's the catch? I just noticed TD Ameritrade has program where you can trade from a list of ETFs with zero fees.  Sounds like it could save me a bit of money, but my cynical side wonders how they make money off such a program...  Is it a loss leader? (They entice me to have an account there, and then make money off the other trades I make?)  Or do they simply want to grow the size of their asset base -- what is the mechanism by which they make money on the funds that I have in my account?  Edit: in case it wasn't clear in the original question above, this is an ongoing program for free ETF transactions -- not a one-time deal for some number of free trades when you open an account. I have had an account there for years and I'm eligible to participate -- it's not a program explicitly designed just to gain new accounts. Does GTA V Online Network Merge Xbox 360 and PS3 players? How to adjust the gap between figure and caption within the same subfloat My question is:  How to adjust the gap between figure and caption within the same subfloat  Not the gap between two subfloat.  For the following example:  \begin{figure}     \centering     \subfloat[A]{\includegraphics[ scale=0.5]{images/a.jpg}}     \vspace{-10pt}     \subfloat[B]{\includegraphics[scale=0.5]{images/b.jpg}}     \vspace{-10pt}     \caption{}\label{fig:online_test} \end{figure}   I got the following figure:    The sub-caption of (a) is too close to the sub-figure B. I hope the caption (a) to stay closer to sub-figure A. How can I adjust the space between the caption and figure within a subfloat?  Thank you all for helping me and I would prefer a solution for subfloat, instead of using other packages. But I would also appreciate a solution of using other packages.   I know my question looks similar to . But please make it clear. I'm asking for how to adjust the gap between figure and caption within a subfloat, not the gap between two subfloat. That is, the gap between the caption "a" and the figure "A"  Thank you. Make an object invisible to a specific camera? (Cycles) I'm trying to set up multiple cameras for a scene set in a room. Is there a way to link object visibility to a specific camera? Ideally, a wall or ceiling would not render if it were in the way of the specified camera.  I've tried to get around this by keyframing Restrict Rendering, but that gets difficult to keep track of. I've also tried adjusting the clipping plane, but that clips too much and would also need to be keyframed if the camera moves.   Thanks! Trouble with Vakil's FOAG exercise 11.3.C I'm having trouble with the exercise in the title, even with part (a), which asks to prove that if $X$ is a closed subset of $\mathbb{P}^n_k$ of dimension at least 1 and $H$ is a non-empty hypersurface, then $X$ and $H$ meet.   Following the hint, I've done the following: By assumption, $X$ is given by $V(I)$ for some homogenous prime ideal $I\subset k[x_0,\dots x_n]=:A$ and $H$ by $V(f)$ for some homogenous polynomial $f\in A$. We wish to prove that there is a homogenous prime ideal $\mathfrak{p}$ (that's not the irrelevant one) containing both $I$ and $f$.  Consider the affine cone $\mathrm{Spec}\ A/I$ of $X$ in $\mathrm{Spec}\ A$. Being homogenous, both $I$ and $f$ are contained in the ideal $(x_0,\dots x_n)$. By Krull's principal ideal theorem, every minimal prime ideal containing $f\mod I$ has codimension one. $X$ having dimension at least one implies that there are homogenous prime prime ideals $\mathfrak{q}\supsetneq \mathfrak{r}\supset I$. This implies that $(x_o,\dots x_n)$ has codimension at least two. Thus there are prime ideals containing $f\mod I$ that are contained in $(x_o,\dots x_n)$. However, I'd need to show that there are homogenous such prime ideals, right? I'd rather get hints than full solutions.