I am currently learning about Radioactivity and Nuclear Physics. Why are nuclides with an even number of protons and neutrons more stable? I have read that nuclides with an even number of protons or neutrons or both, are more stable than those with an odd number of nucleons. Is this something that is based on theoretical reasoning? Or just something taken from observation? I understand that nuclides with more neutrons would be more stable due to more nuclear forces to act against the electrostatic repulsion between protons (to an extent), but I don't know why odd/even counts matter. Is there more stability due to possible greater symmetry if the nuclide has an even number of nucleons? Related:
In basic chemistry, we are taught that an atom has roughly the same number of neutrons and number of protons, this doesn't seems to hold for larger atoms, but it is always roughly proportional (i.e. you seldom find an atom with 100 protons but 1 neutron, that just does not happen). Why is that?
Does there exist a bijective $f:\mathbb{N} \to \mathbb{N}$ such that $\sum f(n)/n^2$ converges? We know that $\displaystyle\zeta(2)=\sum\limits_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ and it converges.        Does there exists a bijective map $f:\mathbb{N} \to \mathbb{N}$ such that the sum $$\sum\limits_{n=1}^{\infty} \frac{f(n)}{n^2}$$ converges.      If our $s=2$ was not fixed, then can we have a function such that $\displaystyle \zeta(s)=\sum\limits_{n=1}^{\infty} \frac{f(n)}{n^s}$ converges Bluetooth from the command line in 12.04? I've been trying to pair up my bluetooth keyboard with my computer after reinstalling from a standard ubuntu 12.04 to a minimal install.  In the minimal install I have no gui, so I've been trying to use the various command line tools available, but I can't figure out how the pairing is supposed to go. Pairing when I had a gui worked flawlessly.  I've asked for help here:   No one has replied but a lot of details of my situation is available there.  How is one supposed to pair bluetooth devices from the command line? I have a table tabName with the following rows:  id(PK) .. Created(datetime)     Resolved(datetime) Status(new/resolved/open) 1          2015-05-10 00:00:00   1970-01-01 ..       open 2          2015-05-10 ..         1970-01-01 ..       new 3          2015-05-10            2015-05-12          resolved 4          2015-05-11 ..         1970-01-01 ..       open 5          2015-05-11 ..         1970-01-01 ..       new 6          2015-05-11            2015-05-11          resolved 7          2015-05-12 ..         1970-01-01 ..       open 8          2015-05-12 ..         1970-01-01 ..       new 9          2015-05-12            2015-05-12          resolved   NOTE:  1970-01-01 .. is default time for all the tickets which are still open.  I have to frame a query to get the count of tickets still open, date wise. Say like on 10 May I have 3 tickets opened. Noting that for id = 3, resolved date is not 10 but it's closed on 12 May.  I framed a query like  SELECT DATE_FORMAT(Created, '%d-%m') AS createdd, COUNT(1) FROM tabName WHERE DATE_FORMAT(Resolved, '%d-%m') &gt; DATE_FORMAT(Created, '%d-%m')       AND YEAR(Created) = '2015' GROUP BY createdd;   The above query gets all the tickets whose Resolved date is greater than the date which is considered for the count. This may count newer tickets as well. And also if the older tickets which are not resolved yet is not counted.  Is there any other way to frame the required query? I have been juggling with this query from last 3 days. Not able to find out still.  Updated: As suggested by Abhik, i have tried in sqlfiddle site.  Here is what i created:  CREATE TABLE test    (idint,createddatetime,resolveddatetime, status varchar(15), PRIMARY KEY (id));    INSERT INTO test       (id,'created','resolved',status`) VALUES       (1,'2015-05-10 00:00:00','1970-01-01 00:00:00','open'),    (2,'2015-05-10 00:00:00','1970-01-01 00:00:00','new'),    (3,'2015-05-10 00:00:00','2015-05-12 00:00:00','resolved'),    (4,'2015-05-11 00:00:00','1970-01-01 00:00:00','open'),    (5,'2015-05-11 00:00:00','1970-01-01 00:00:00','new'),    (6,'2015-05-11 00:00:00','2015-05-11 00:00:00','resolved'),    (7,'2015-05-12 00:00:00','1970-01-01 00:00:00','open'),    (8,'2015-05-12 00:00:00','1970-01-01 00:00:00','new'),    (9,'2015-05-12 00:00:00','1970-01-01 00:00:00','open')    ;   `   When i run  ` select created_date, @open:= @open+open_ticket-resolved as open from(  select   date_format(created,'%d-%m') as created_date ,  sum( case when status='open' then 1 else 0 end )+  sum( case when status='new' then 1 else 0 end )+  sum( case when status='resolved' and ( date(resolved) > date(created) or  date(resolved) = date(created)) then 1 else 0 end ) as open_ticket,  sum(case when status = 'resolved' and date(resolved) = date(created) then 1   else 0 end) as resolved  from test  where  YEAR(Created) = '2015'  group by created_date )x,  (select @open:=0)r order by created_date ; `  I got following output  created_date   open 10-05   3 11-05   5 12-05   8  Here count of still open tickets for 12-05 must be 7. The above query includes ticket id "3" , which is closed on 12-05. So is there any way to exclude this? If i am able to get this right, then i will have full fledged query. What Does it Really Mean to Have Different Kinds of Infinities? Can someone explain to me how there can be different kinds of infinities?  I was reading "" by  and came across the concept of countable and uncountable infinities, but they're only words to me.  Any help would be appreciated. Is there a way to open a raster that is stored in ESRI V10.0 Geodatabase in QGIS V2.4?    I can open the vector layer, but there does not seems to be a way to open a raster. Determine all $f: \mathbb{R} \rightarrow \mathbb{R}$ such that, for all $x, y \in \mathbb{R}$, $$f((x + 1) f(y)) = y (f(x) + 1).$$    Source: Vuong Lam Huy, on a Facebook group. Is there any way to deserialize JSON into an object using JSON.deserialize if some of the property names in the JSON are reserved words in apex? I want to do something like this:  string jsonString = '{"currency" : "ABC"}'; public class JSONResult {     public string currency; } JSONResult res = (JSONResult) JSON.deserialize(jsonString, JSONResult.class); system.debug(res);   but of course I get an error that the identifier name is reserved. Getting No module name exifread error from QGIS Photo2Shape plugin? When you use the allowframebreaks option in beamer then you get additional frames if the content is too high for one frame. The vertical spacing gets disturbed when you use a list (itemize, enumerate, description), for example, see the following animation. I used the t option (for top alignment) for the frame.     \documentclass[aspectratio=169]{beamer}   \begin{document}  \begin{frame}[t,allowframebreaks] \frametitle{Frame Title} \setlength{\mylength}{\baselineskip} \begin{itemize} \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \end{itemize}  \end{frame}  \begin{frame}[t,allowframebreaks] \frametitle{Frame Title}  Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\ Text\\  \end{frame}  \end{document}    I thought of a solution, when I read  question where they explain the different length in lists. The main idea is to check if the current frame is not the first frame of a frame continuation and add the needed vertical space. I thought that \topsep is the needed length but apparently, it is not quite right. It ist almost perfect but it still "dances" a little bit. Which additional length do I need? Here's the animation and the used code.     \documentclass[aspectratio=169]{beamer}  \setbeamertemplate{frametitle continuation}{%     (\insertcontinuationcount)%     \ifnum\insertcontinuationcount&gt;1%     \vspace*{\topsep}%     \else%     %     \fi% }  \begin{document}  \begin{frame}[t,allowframebreaks] \frametitle{Frame Title}  \begin{itemize} \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \item Test \end{itemize}  \end{frame}  \end{document}   Related        Definition of \itemize in beamer  In beamerbaselocalstructure.sty I found  \renewcommand{\itemize}[1][]{%   \beamer@ifempty{#1}{}{\def\beamer@defaultospec{#1}}%   \ifnum \@itemdepth &gt;2\relax\@toodeep\else     \advance\@itemdepth\@ne     \beamer@computepref\@itemdepth% sets \beameritemnestingprefix     \usebeamerfont{itemize/enumerate \beameritemnestingprefix body}%     \usebeamercolor[fg]{itemize/enumerate \beameritemnestingprefix body}%     \usebeamertemplate{itemize/enumerate \beameritemnestingprefix body begin}%     \list       {\usebeamertemplate{itemize \beameritemnestingprefix item}}       {\def\makelabel##1{%           {%             \hss\llap{{%                 \usebeamerfont*{itemize \beameritemnestingprefix item}%                 \usebeamercolor[fg]{itemize \beameritemnestingprefix item}##1}}%           }%         }%       }   \fi%   \beamer@cramped%   \raggedright%   \beamer@firstlineitemizeunskip% }   But I do not understand where the different kinds of spaces are defined that cause the "dancing".  "Evil" Stuff in beamer  Motivated by the comment of @cfr I searched the beamer documentation for "evil" stuff -- this is what I have found (beamer guide version 3.41):   Do not use subsubsections, they are evil. Never ever use the evil option shrink.  Using this [again shrink] option is very evil. The use of this [allowframebreaks] option is evil. Command line move file with long directory path without repeating directory path Is there any way of executing a mv command without typing the full path in the second argument?  Example:  mv my/file/that/has/a/really/long/path/foo.bar some_magic_operator_that_means_the_same_directory/foo.baz Ubuntu has been crashing on me recently. I think its because it runs out of memory so I ran the free -m command and found that my memory usage was really high. So then, i ran top to find the culprit, but the displayed processes were using less than 1.5% of memory. How do I know which program is making ubuntu crash/run out of memory? Below is the output:  shafee@shafee-pc:~$ free -m              total       used       free     shared    buffers     cached Mem:          3824       3714        110          0        978       1954 -/+ buffers/cache:        780       3044 Swap:           99          0         99  shafee@shafee-pc:~$ top top - 02:12:14 up  1:24,  2 users,  load average: 0.16, 0.42, 1.49 Tasks: 182 total,   1 running, 181 sleeping,   0 stopped,   0 zombie Cpu(s):  2.9%us,  1.9%sy,  0.3%ni, 79.3%id, 15.5%wa,  0.0%hi,  0.2%si,  0.0%st Mem:   3916708k total,  3803848k used,   112860k free,  1002308k buffers Swap:   102396k total,        0k used,   102396k free,  2001852k cached    PID USER      PR  NI  VIRT  RES  SHR S %CPU %MEM    TIME+  COMMAND              4200 root      20   0  289m  53m  38m S    2  1.4   1:06.45 Xorg                 5590 shafee    20   0 19348 1368  956 R    2  0.0   0:00.01 top                     1 root      20   0 24124 2136 1264 S    0  0.1   0:02.05 init                    2 root      20   0     0    0    0 S    0  0.0   0:00.00 kthreadd                3 root      20   0     0    0    0 S    0  0.0   0:24.23 ksoftirqd/0             6 root      RT   0     0    0    0 S    0  0.0   0:00.00 migration/0             7 root      RT   0     0    0    0 S    0  0.0   0:00.00 migration/1             9 root      20   0     0    0    0 S    0  0.0   0:00.11 ksoftirqd/1            11 root      RT   0     0    0    0 S    0  0.0   0:00.00 migration/2            13 root      20   0     0    0    0 S    0  0.0   0:03.89 ksoftirqd/2            14 root      RT   0     0    0    0 S    0  0.0   0:00.00 migration/3            16 root      20   0     0    0    0 S    0  0.0   0:00.16 ksoftirqd/3            17 root       0 -20     0    0    0 S    0  0.0   0:00.00 cpuset                 18 root       0 -20     0    0    0 S    0  0.0   0:00.00 khelper                19 root       0 -20     0    0    0 S    0  0.0   0:00.00 netns                  21 root      20   0     0    0    0 S    0  0.0   0:00.01 sync_supers         shafee@shafee-pc:~$   Or is it normal to have 110mb of free memory and is my issue related to something else? How to get Lookup ID from Search API with REST I'm trying to get the ID of a lookup field when using search api. My query looks something like this:  /_api/Search/query?querytext='ListId:12345678-1234-1234-1234-123456789012'&amp;selectproperties='Title,Company'   Where Company is the lookup field, in a specific list.  The result looks like this:  &lt;d:element m:type="SP.KeyValue"&gt;   &lt;d:Key&gt;Company&lt;/d:Key&gt;   &lt;d:Value&gt;Some company&lt;/d:Value&gt;   &lt;d:ValueType&gt;Edm.String&lt;/d:ValueType&gt; &lt;/d:element&gt;   I have created a managed property which looks like this: $\varepsilon$-number countability without choice Let $\alpha\mapsto\varepsilon_\alpha$ be the enumeration of the $\varepsilon$-numbers--that is, those $\alpha$ such that $\omega^\alpha=\alpha$--by the ordinals.  If we know that countable unions of countable sets are countable, (slightly weaker than countable choice), then we can show fairly easily that $\varepsilon_\alpha$ is countable iff $\alpha$ is countable.  I think I saw a result at some point that $\varepsilon_0$ is countable without relying on any choice principle (though I can't recall where I saw that). Is this true? More generally, for which $\alpha$ can we conclude that $\varepsilon_\alpha$ is countable in ZF?  Update: I have been able to prove (in ZF) that if $\varepsilon_\alpha$ is countable, then $|\alpha|&lt;\aleph_1$, so that's one direction. I'm still a bit uncertain about the answers I've got so far. Count below a comment is misleading Graph Theory: Tree has at least 2 vertices of degree 1 Let $\lambda(A)$ be the Lebesgue measure of $A$. There exists an interval $I$ such that $\lambda(E \cap I) / \lambda(I) < 1-\epsilon$ (Not mentioned in title but $\epsilon$ is a number greater than $0$ and $E$ a Lebesgue measurable subset of $\mathbb{R}^n$.) I know a question equivalent to this one has been asked (). But it was not answered. Also my thinking is a bit different. So I would love to get some help on this problem. So far I've got:  $$\frac{\lambda(E \cap I)}{\lambda(I)} &gt; 1- \epsilon \iff \frac{\lambda(I)-\lambda(E\cap I)}{\lambda(I)} &lt; \epsilon$$  And also, a theorem states that, for all $\epsilon$ there exists a closed set $F\subset E$ such that $\lambda(E\setminus F) &lt; \epsilon$ so it would sufice to prove  that there exists an interval $I$ such that $$\frac{\lambda(I)-\lambda(E\cap I)}{\lambda(I)}&lt;\lambda(E\setminus F).$$ I'm not sure if this helps me much but any help is appreciated. What would you do? I want to buy a home, but I'm not interested in using a realtor as a buyer's agent.  I know that in MLS systems, when a realtor lists a house they specify what commission a buyer's agent will be paid.  Since I am the one who is bringing a qualified buyer to the listing agent, I feel that I would should get the commission (even if it is in the form of a discount on the house).    Essentially, the seller would get the same amount of money at closing... the listing agent would get the same commission as they normally would for showing the house to someone with a buyer's agent... but I would pay less for the house.  Is there a tactful way to negotiate this with the listing realtor?  For example, should I make a side contract with the realtor to pay me the commission at closing?  Or perhaps I can specify in my offer that they should reduce their fees to the seller by the commission amount if the offer is accepted, in order to make my offer more attractive?  Are these types of deals ever done, and if so, is there a conventional way to make it happen?  Thanks Biometrika journal style file not getting along with TeXLive-2020 I cannot compile styleguide.tex in biometrika.zip (see ). Here is a MWE which leads to the same error message, though not sure yet that I've not tossed baby out with bath water. \documentclass{article} \RequirePackage[thmmarks]{ntheorem}  \makeatletter \def\arabic#1{{\rm\expandafter\@arabic\csname c@#1\endcsname}} \theoremnumbering{arabic} \newtheorem{definition}{Definition} \makeatother  \begin{document}  \begin{definition} This is a definition. \end{definition}  \end{document}  Here is the output of pdflatex: % pdflatex styleguide This is pdfTeX, Version 3.14159265-2.6-1.40.21 (TeX Live 2020) (preloaded format=pdflatex)  restricted \write18 enabled. entering extended mode (./styleguide.tex LaTeX2e &lt;2020-02-02&gt; patch level 5 L3 programming layer &lt;2020-03-06&gt; (/usr/local/texlive/2020/texmf-dist/tex/latex/base/article.cls Document Class: article 2019/12/20 v1.4l Standard LaTeX document class (/usr/local/texlive/2020/texmf-dist/tex/latex/base/size10.clo)) (./ntheorem.sty  Style `ntheorem', Version 1.24 &lt;2004/09/20&gt; ) (/usr/local/texlive/2020/texmf-dist/tex/latex/l3backend/l3backend-pdfmode.def ) (./styleguide.aux) ! Use of \@item doesn't match its definition. \update@series@target@value #1-&gt;\def \reserved@a {                                                   #1}\ifx \target@meta@famil... l.13 T       his is a definition. ? q OK, entering \batchmode Possible Duplicate:          Consider the field extension $\mathbb{Q}(\sqrt2)$. I want to show that $\sqrt5 \notin \mathbb{Q}(\sqrt2)$. If this were not the case, then we could write $\sqrt5 = a + b\sqrt2$ for $a,b\in\mathbb{Q}$. However, I do not see the contradiction here. Is there a better/easier way to prove this? "Bring" vs. "take" in American English