I lost 2 reputation points because a question was removed. I didn't asked that question. So these are my guesses:  1.-I gave an answer to that question. 2.-I edited that question. 3.-I upvoted that question.    But I can't remember. Its not the only one. I have got this like 3 times total. The reputation help page doesn't say anything about it   What is the reason? and someone should totally add that to that help page :)
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I've heard that in the study of finite fields, and other concepts related to finite fields, mathematicians have found a sort of gap: there are various results and things that seem like they correspond to a field with one element, $\mathbf{F}_1$, even though there is no such field. Of course, there is an object which seems similar to a field with one element: the zero ring, $\mathbf{Z}_1$, defined as having one element, $0$, obeying the equations $$0 + 0 = 0 - 0 = 0 \cdot 0 = 1 = 0.$$ The zero ring seems to behave like a field in every respect, except for the fact that it fails to satisfy $0 \ne 1$. (But the importance of the axiom $0 \ne 1$ isn't clear to me.) However, in any case, there seems to be consensus that the zero ring is not the field with one element. People say that the zero ring &quot;does not behave like a finite field&quot; () or that it &quot;does not have the features that mathematicians need&quot; (). I'm not familiar enough with algebraic geometry to understand the properties that $\mathbf{F}_1$ is expected to have. Is there an elementary, undergraduate-level explanation of  why it seems like a field with one element &quot;ought to exist&quot; in the first place, what properties we expect it to have and why, and how the zero ring fails to satisfy these properties?  Or do I need to study algebraic geometry if I want to have an understanding of any of this? (A side question: do we have a good answer to the question of whether the single element of $\mathbf{F}_1$ ought to be $0$, $1$, both, or neither?) I am trying to get from a particular point more informations about the surroundings (which are also points). So I wrote this:  SELECT *    FROM osm_point    WHERE ST_DWithin(way,(ST_SetSRID(ST_MakePoint(12.2385578688105, 51.8435615931295), 4326, 1000);   But there is an error     ERROR: FEHLER:  LINE 3: ...T_MakePoint(12.2385578688105,   51.8435615931295), 4326, 1000) ^ SQL state: 42601 Character: 133   What is wrong? unicode-math broken Prove $x^3 + 3 =4y(y+1)$ has no integer solutions Proof by contradiction:  1. Suppose  $x^3 + 3 =4y(y+1)$ has an integer solution   Then  $x^3 + 3 =4y^2+4y$ Then  $x^3 + 3 + 4 = 4y^2 + 4y + 4$ Then $x^3 + 7  = (2y + 2)^2$   Not sure how to simplify it further... I was trying to prove this inequality using induction, but couldn't do.  Question: Suppose $a$ and $b$ are real numbers with $0 &lt; b &lt; a$. Prove that if $n$ is a positive integer, then:  $$a^n-b^n \leq na^{n-1}(a-b)$$ Lie Algebra of $SL_n(\mathbb H)$ The Lie algebra of $SL_n(\mathbb C)$ are the matrices where the trace is $0$. But what is the Lie algebra of $SL_n(\mathbb H)$ where $\mathbb H$ is the quaternions? Out of five needed members for my doctoral thesis committee, four are fixed. For the fifth member, I approached Mr.A. He has no real connection to my research or topic but I contacted him because he was just as good a fit as anybody else in the department to be the fifth member. I have never worked with him or taken any classes but we did talk a lot in the first few years where he would just ask about my progress by the copy machine, etc. We don't even have a "rapport". The relationship is absolutely neutral, nothing positive/friendly, and nothing negative. I asked him to be on my committee and he said yes. I said I'll get back to you with dates, etc. He is a senior tenured professor.  Then about a month later, I was looking at the CV of some of the new hired (junior) faculty in the department Mr.B (this is his first year) and I see that we have a collaborator in common. Mr.B has published multiple papers with him and Mr.B is working with the same code and the code managing team that we are working with. He is definitely a good person to have on my committee. I finally see Mr.B after a few more months (because of busy schedule/travel/etc.) and he agrees to be on my committee.  The question is, how can I tell Mr.A, as tactfully as possible, after about four months of being quiet that he is basically rejected from my committee because I didn't know about Mr.B but now that I know, Mr.B is a much better match than he is? That I don't need him but I do appreciate his offer? I don't want to antagonize Mr.A or burn any bridges. Should this be in email or in person? Should I explain or just keep it terse? Should I keep it vague that I don't need him and thanks for his time? Or should I tell him about Mr.B and why he is a better match?  FYI, this was before anything official/paperwork was done so there is no problem with that. Are used Schengen/UK visas acceptable for Turkish e-visa How do I update Custom Label's value using apex? Which one would be a better option metadata api or tooling api? Please provide example The help pages in R assume I know what those numbers mean, but I don't. I'm trying to really intuitively understand every number here. I will just post the output and comment on what I found out. There might (will) be mistakes, as I'll just write what I assume. Mainly I'd like to know what the t-value in the coefficients mean, and why they print the residual standard error.   Call: lm(formula = iris$Sepal.Width ~ iris$Petal.Width)  Residuals:      Min       1Q   Median       3Q      Max  -1.09907 -0.23626 -0.01064  0.23345  1.17532    This is a 5-point-summary of the residuals (their mean is always 0, right?). The numbers can be used (I'm guessing here) to quickly see if there are any big outliers. Also you can already see it here if the residuals are far from normally distributed (they should be normally distributed).  Coefficients:                  Estimate Std. Error t value Pr(&gt;|t|)     (Intercept)       3.30843    0.06210  53.278  &lt; 2e-16 *** iris$Petal.Width -0.20936    0.04374  -4.786 4.07e-06 *** --- Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1    Estimates $\hat{\beta_i}$, computed by least squares regression. Also, the standard error is $\sigma_{\beta_i}$. I'd like to know how this is calculated.  I have no idea where the t-value and the corresponding p-value come from. I know $\hat{\beta}$ should be normal distributed, but how is the t-value calculated?  Residual standard error: 0.407 on 148 degrees of freedom   $\sqrt{ \frac{1}{n-p} \epsilon^T\epsilon }$, I guess. But why do we calculate that, and what does it tell us?  Multiple R-squared: 0.134,  Adjusted R-squared: 0.1282    $ R^2 = \frac{s_\hat{y}^2}{s_y^2} $, which is $ \frac{\sum_{i=1}^n (\hat{y_i}-\bar{y})^2}{\sum_{i=1}^n (y_i-\bar{y})^2} $. The ratio is close to 1 if the points lie on a straight line, and 0 if they are random.  What is the adjusted R-squared?  F-statistic: 22.91 on 1 and 148 DF,  p-value: 4.073e-06    F and p for the whole model, not only for single $\beta_i$s as previous. The F value is $ \frac{s^2_{\hat{y}}}{\sum\epsilon_i} $. The bigger it grows, the more unlikely it is that the $\beta$'s do not have any effect at all. Negative Mass and gravitation I just updated from Ubuntu 10.10 to 11.04 and have an issue regarding the screen resolution. I have Intel integrated gfx chip and my monitor supports resolutions larger than 1024x768. (in 10.10 I've been using 1280x1024) But as soon as I upgraded, I'm stuck with 1024x768 resolution and seems I can't change it.  running xrandr In terminal yields the following results,   Screen 0: minimum 320 x 200, current 1024 x 768, maximum 4096 x 4096 LVDS1 connected 1024x768+0+0 (normal left inverted right x axis y axis) 0mm x 0mm    1280x800       58.1 +    1024x768       60.0*     800x600        60.3     56.2      640x480        59.9   VGA1 connected 1024x768+0+0 (normal left inverted right x axis y axis) 344mm x 194mm    1366x768       59.9 +    1360x768       60.0      1024x768       75.1     72.0     70.1     60.0*     832x624        74.6      800x600        72.2     75.0     60.3     56.2      640x480        72.8     75.0     66.7     60.0      720x400        70.1     1280x1024_60.00 (0xce)  109.0MHz         h: width  1280 start 1368 end 1496 total 1712 skew    0 clock   63.7KHz         v: height 1024 start 1027 end 1034 total 1063           clock   59.9Hz   What maybe the problem? Is it a bug? What kind of steps I should take in order to get a higher resolution? (changing xorg.conf maybe?) Any insight is highly appreciated. Thanks in advance.   UPDATE    Screenshot after running xrandr --addmode VGA1 1360x768  As you can see, side bar is not completely visible and Ubuntu logo at the task bar is missing. Also when you open an application, the Task bar of the application (where it should go to the top panel) is missing as well.. I know if $\ i\ $  of the following form $\ 3x^2 + (6y-3)x - y\ $ or $\ 3x^2 + (6y-3)x + y - 1, \ \ x,y \in \mathbb{Z}^{+},i \in \mathbb Z_{\ge 0}$,  then $\ 12i + 5\ $ must be composite number, because:  $12(3x^2 + (6y-3)x + y - 1) + 5 = 36x^2 + (72y - 36)x + (12y - 7) = (6x + 12y - 7)(6x  + 1)$  How come if $\ i\ $ not of the following form $\ 3x^2 + (6y-3)x - y\ $ and $\ 3x^2 + (6y-3)x + y - 1\ $,  then $12i + 5$ must be prime? $\ x,y \in \mathbb{Z}^{+},i \in \mathbb Z_{\ge 0}$.  For example: $\ 5 = \ 3*1^2 + (6*1-3)*1 - 1\ $ ,when $\ x = y = 1\ $, as proved before,$\ 5*12+5\ $ must be composite number;  $\ 0, \ 1. \ 2.\ 3.\ 4\ $ can't of the following form $\ 3x^2 + (6y-3)x - y\ $ and $\ 3x^2 + (6y-3)x + y - 1\ $,  $\ 12*0 + 5 = 5\ $, $\ 5 $ is prime,$\ 12*1 + 5 = 17\ $, $\ 17 $ is prime, and so on, How to write permutations as product of disjoint cycles and transpositions $$\sigma=\begin{pmatrix} 1 &amp; 2 &amp;3 &amp; 4&amp; 5&amp; 6&amp;7 &amp;8 &amp;9 &amp;10 &amp; 11 \\ 4&amp;2&amp;9&amp;10&amp;6&amp;5&amp;11&amp;7&amp;8&amp;1&amp;3    \end{pmatrix}$$  (1) I am asked to write this permutation in $S_{11}$ as a product of disjoint cycles and also as a product of transpositions.   (2) Also find the order of the element. Is this permutation even or odd?  I think these are the disjoint cycles  $E_{1}=(1,4,10)$, $\;\operatorname{order}E_1 =3$  $E_{2}= (3,9,8,7,11),\;$ $\operatorname{order}E_{2} =5$  $E_{3}=(5,6),\;$ $\operatorname{order}E_{3} =2$  $S_{11}$=  $E_{1} \cdot E_{2}\cdot  E_{3}$  The order of $\operatorname{order}E_{3}$ is even so the order of the permutation is even. Why are they asking this? And what is the significance of it being even or odd?  Transpositions – I have read this a couple times but the one example in my textbook is rather unclear, I am not sure what this means?  I think the transposition for $E_{1}$ is $(1,4)(1,10)$ but I'm not sure what this means. LaTeX conditional expression I would like to be able to set a flag at the beginning of the main flag (say output) that tells if I want to produce a paper document or an electronic document.   Then I should check this flag to define or not empty pages, hypperref, oneside or twosides, etc.  How could I do that ? I want to put a link to my SO profile on my homepage. I remember a codesnippet i saw somewhere that would basically display my Username and my reputation. Where was that? Are continuous self-bijections of connected spaces homeomorphisms? I hope this doesn't turn out to be a silly question.  There are lots of nice examples of continuous bijections $X\to Y$ between topological spaces that are not homeomorphisms.  But in the examples I know, either $X$ and $Y$ are not homeomorphic to one another, or they are (homeomorphic) disconnected spaces.    My Question: Is there a connected topological space $X$ and a continuous bijection $X\to X$ that is not a homeomorphism?    For the record, my example of a continuous bijection $X\to X$ that is not a homeomorphism is the following.  Roughly, the idea is to find an ordered family of topologies $\tau_i$ ( $i\in \mathbb Z$) on a set $S$ and use the shift map to create a continuous bijection from $\coprod_{i\in \mathbb Z} (S, \tau_i)$ to itself.  Let $S = \mathbb{Z} \coprod \mathbb Z$.  The topology $\tau_i$ is as follows: if $i&lt;0$, then the left-hand copy of $\mathbb Z$ is topologized as the disjoint union of the discrete topology on $[-n, n]$ and the indiscrete topology on its complement, while the right-hand copy of $\mathbb Z$ is indiscrete.  The space $(S, \tau_0)$ is then indiscrete.  For $i&gt;0$, the left-hand copy of $\mathbb Z$ is indiscrete, while the right-hand copy is the disjoint union of the indiscrete topology on $[-n, n]$ with the discrete topology on its complement.  Now the map $\coprod_{i\in \mathbb Z} (S, \tau_i)\to \coprod_{i\in \mathbb Z} (S, \tau_i)$ sending $(S, \tau_i) \to (S, \tau_{i+1})$ by the identity map of $S$ is a continuous bijection, but not a homeomorphism. According to some grammars[1][2] and CMS...  "Interjections and vocative should be separated from the rest of the sentence by commas".  The vocative case [1] is used to indicate direct address (i.e., to show when you are talking to somebody or something directly). Nouns in the vocative case are set off using commas. CMS 6.38  As direct address (i.e. vocative case), the parenthetical element should be set off using commas (CMS 6.30). An absolute phrase is always treated as a parenthetical element, as is an interjection.  So,      Hello, world!   That is a very classical way to hello the world.  But, languages change. I am not sure, but today, I feel that it is very uncommon helloing someone with an interjection, providing a small pause and clarifying who is helloed. I have an impression that usually people use hello as a ("kind of") verb, not an interjection.   Question: Is that just a personal (bad) impression or some linguistic has already studied that phenomenon?     [1] - Nelson, Gerald C., Director Survey of English Usage Sidney Greenbaum, and Sidney Greenbaum. An Introduction to English Grammar. Routledge, 2013. [2] - Advanced Grammar in Use with Answers: A Self-Study Reference and Practice Book for Advanced Learners of English. 3 edition. Cambridge University Press, 2013. CMS - Chicago Manual of Style Optional argument after splat argument So, I'm running linux Mint (but I guess the same applies to Ubuntu) and as is common knowledge, their TeX distribution tends to be slightly outdated. Now I can't have none of that and decide to install it with the install-tl from the TeXlive homepage. Somehow, I cannot install it as a normal user (this I find normal as it is installing in the root directory), so I proceed to install TeXlive   sudo perl ./install-tl   and add the paths to .bashrc. All is well and my TeXlive installation works. Except for the texhash option. Running it as a normal user is no problem, except that I installed TeXlive as root, and so running texhash as normal user doesn't do anything (directory not writable. Skipping...). No problem I think, I'll just do it as root. But somehow texhash isn't installed for root.  Now, texhash is somehow connected to updating the directory tree, but I don't really know what it does. I use it on my ~/texmf/ folder to add my packages to a global path (I think), but when looking at info for this question I see in a comment for  question that this is apparently not needed, though I don't understand what I'm to do now.  So if I'm not adding new packages from CTAN or updating, I really don't need texhash and all is good?   To summarize: I'd like to know:    Why should I need texhash? Can I avoid installing as root on linux Mint/ubuntu ( they say you don't have to be root, that it's even discouraged)? Should I somehow install texhash for root?   I realize this might be a vague question at best, but I didn't find an answer I readily understood, so I'd appreciate any thought on this.