if Higgs boson stopped existing, what would happen?
What happens to matter in a standard model with zero Higgs VEV?
How can I stop iPhone prompting me to setup Apple Pay in Settings? Don't show pending suggested edits to me after I run out of votes Why does logging in to Stack Overflow keep me signed in to my Gmail email after logging out and regardless of the Stay signed in checkbox?  Steps to reproduce:   Browse to  and confirm you are logged out Browse to  and confirm you are logged out Click login Click Google Enter Email/Password and confirm Stay signed in is unchecked Click Sign in Click logout (up top) Click Logout (down below) Browse to  and confirm you are still logged in Click Sign out &lt;- Shouldn't clicking Logout on Stack Overflow be doing the same thing as this? Browse to  and confirm you are logged out     Answers are making sense but my frustration is that clicking Logout (twice) on StackOverflow essentially does nothing. If I proceed to click login and Gmail I am immediately authenticated because I still have an active Gmail cookie. In order to effectively logout I need to click Logout twice on Stack Overflow and then manually browse to  and Sign out. Is it possible to get the information for a device tree using /sys of a running kernel? Commonly for arm systems, device trees supply hardware information to the kernel (Linux). These device trees exist as dts (device tree source) files that are compiled and loaded to the kernel. Problem is that I do not have access to such a dts file, not even to a dtb file.   I have access to /sys and /proc on the machine and I wanted to ask if that would allow me to "guess the correct values" to be used in a dts?  Also potential answer could highlight additionally the aspect if the answer to this question also depends on whether the device tree interface was used in the first place (i.e. a dtb was created and provided to the kernel) instead of some more hacking "we simply divert from vanilla and patch the kernel so as to solve the device information problem for our kernel only"-solution? Someone edited my message on StackOverflow, but it really bugs me out.  I'm not sure what's wrong with it:   As you see, the bigger the circle becomes, the more vertices I need for hiding the straight lines. As you see, the bigger the circle becomes, the more vertices I need to hide the straight lines.   The latter (corrected) sentence just bugs me out for some reason. It would sound better if there was a comma after the word "need", perhaps, but I guess that would be even worse in means of grammatical correctness.  Can someone explain why is the first sentence bad and the second sentence correct? And could there even be better solution to fix the first sentence? Came across the following exercise in Bartle's Elements of Real Analysis. This is the solution I came up with. Would be grateful if someone could verify it for me and maybe suggest better/alternate solutions.  I also looked up these related questions - , ,  - but was not happy with proofs given there. I seem to need some help understanding these. Any such help is appreciated.      Show that every sequence in $\Bbb R$ either has a monotone increasing sub-sequence or a monotone decreasing sub-sequence.   Let $(x_n)$ be a sequence in $\Bbb R$. Suppose $(x_n)$ is not bounded. Without loss of generality we may assume that $(x_n)$ is not bounded above. Therefore given any real number there is a member of the sequence which is greater. Let $x_{n_1}$ be any member of the sequence.   There is $x_{n_2} \gt \sup\{x_1, x_2, ..., x_{n_1} \}$. For $i \gt 1$ let $x_{n_i} = \{x_1, x_2, ..., x_{n_{i - 1}}\}$ then $(x_{n_k})$ forms a monotone subsequence of $(x_n)$.   Now suppose instead that $(x_n)$ is bounded. By the Bolzano-Weierstrass Theorem there is a subsequence $(y_n)$ of $(x_n)$ which converges to a limit $y$. Without loss of generality there are infinitely many distinct values in $(y_n)$ that are unequal to $y$. Let $y_{k1}$ be the first such element. Let $y_{k2}$ be any element in $\{ y' \in (y_n) \ \ | \ \ |y' - y | \lt |y - y_{k1}|  \}$.   For $i \gt 1$ let $y_{ki} \in \{ y' \in (y_n)  \ \ | \ \ |y' - y | \lt |y - y_{k \ i - 1}|  \}$. Such $y_{ki}$ exists for every $i \in  \Bbb N$ since $ \lim (y_n) = y $. Now let $(y_{kn})$ be the sub-sequence of $(y_n)$ thus formed. At least one of the two following sets must contain infinitely many elements.    $\{ y \in (y_{kn}) \ \ | \ \ y \gt x\}$ $\{ y \in (y_{kn}) \ \ | \ \ y \lt x\}$   The one which does forms a monotone subsequence. I added two torus objects, and when I click one of them, it also selects the other. They both share the same node thing. I want to delete one of them, but it keeps on deleting the other with it. I can't even undo placing the second torus without removing the other. Deleting comment removes "Add comment" link on Mobile If I delete a comment on the mobile version of the site (at least if it's the only comment on an answer), the "Add comment" link will vanish along with the comment, so I have to reload to leave another comment.     but got declined as , which is why I'm reposting this (as it may well be a new bug with the same symptoms).  I'm using Firefox 32.0.3 on Android. Suppose $\lim \limits_{n \to ∞} a_n=L$. Prove that $\lim\limits_{n \to ∞} \frac{a_1+a_2+\cdots+a_n}{n}=L$ Why isn't the eigenclass equivalent to self.class, when it looks so similar? MT7610U - Unable to install wifi driver What's the difference between $(stuff) and `stuff`? Are submodules of finitely generated free modules always free? I don't believe it's true that all submodules of a finitely generated free module are free just based on results from google. Is there a canonical simple example of a finitely generated free module with a submodule that is not free? Intuituive reason why Fermats last theorem holds I am unsure of whether it is normal, but to me, intuitively Fermats last theorem should not hold.  If anyone intuitively believed it to be correct, why?     Can someone explain so I understand somewhat why FLT holds without reading a massive and complex proof? How do I disable the alt-right click keyboard binding? What is the difference between a telephoto lens and a zoom lens? Looking at the , I notice a distinction between telephoto lens and zoom lens.  What is the difference between the two?  Why would I want one over the other?  I  but remain confused about this distinction.  Please talk to me like I'm stupid. :) I am hoping for a completely lay explanation.  In particular, I'd like to understand in what situations I'd use a telephoto lens, and in what (other) situations I'd use a zoom lens. What is a good asymptotic for $f_n = f_{n-1}+\ln(f_{n-1})$? I want the QED symbol to be 1 line lower at the end of proof How can I compile different .tex file on click on typeset in TeXShop? (MacTeX 2011) My problem is that I want to have different .tex files for the preamble, the presentation aroundings (my target is a beamer PDF file) and the content of the presentation.  Mostly I'll be working on the content file. I even think about partitioning this content .tex file into several chapter files for a better overview. My problem with this solution is that I have to open the presentation.tex file to typeset the whole beamer PDF for looking at the results of my previous changes on the content files. (My presentation.tex file includes the preamble and the content.)  Is there any way to get TeXShop to always typeset one particular file (my presentation.tex) although I'm currently working on a different one (e.g. the content.tex)?  A shortcut? A command in the .tex file? An option in TeXShop? Anything? Minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$ I had an example in the book given as follows:        Find the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$ .       Solution:  $~~~~~$$(\sqrt{2}+\sqrt{3})^2=5+2 \sqrt6$      $(\sqrt{2}+\sqrt{3})^4=49+20 \sqrt6$       Then $(\sqrt{2}+\sqrt{3})^4-10(\sqrt{2}+\sqrt{3})^2+1=0.$       Thus $a=\sqrt{2}+\sqrt{3}$ satisfies $f(x)=x^4-10x^2+1$ over $\mathbb Q$.       Let $p(x)$ be the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$.Then $(\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3})$ are also roots of $p(x).$   So degree of $p(x)$ is atleast $4$.      But $f(a)=0 $ and $f(x) \in \mathbb Q[x] \implies p(x)$ divides $f(x)$ .      So $f(x)$ is minimal polynomial of $\sqrt{2}+\sqrt{3}.$   But I can't get the step  how are $(\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3})$ also roots of $p(x).$   Kindly help with this.