I just got a new computer (windows 8.1) and I have two drives.  The main C drive is an SSD and the D drive is a regular drive.  I want to move some of the less important data to the secondary drive, such as Downloads.  I know I can do it using MKLINK but I can't remove the Downloads folder as I get an Access is denied error, and I think that it may be a special folder anyway (a folder that may have some 'magic' attached to it).  Is there a clean way of doing this?  Does it have 'magic' attached to it?
My machine has a SSD (fairly small) and a Hard disk (large). I want to have windows and a couple of programs on the SSD but I want to have the documents folders (such as my documents and my downloads) on the HD. I also want the default install directory pointing at the larger HD. Can anyone advise or point me to a relevent guide? Thanks.
org.apache.tomcat.util.bcel.classfile.ClassFormatException: Invalid byte tag in constant pool: 15 sudo: unable to resolve host   I think I need to edit the /etc/hosts file to make an entry for 127.0.1.1 PC name.  But how can I do that if I can't access sudo? It seems to have gone since I updated to 13.10.  I have tried nano, which still doesn't work. In the midst of attempting to prove that for a commutative ring $A$ with identity, and an ideal $I$ of $A$, $I = \operatorname{rad}(I)$, where $\operatorname{rad}(I) = \{x: x^m \in I, m &gt;0\}$, implies that $I$ is an intersection of prime ideals, I've run into a little bit of confusion.   An answer to another question on this site, , indicates a strategy that I am attempting to use. I know that if $x$ is an element of $A\setminus I$, then so too is $x^m$ for any positive integer $m$. I use this fact to consider a subset $X = \{x^m : m &gt; 0\}$, which is closed under multiplication, and use Zorn's lemma to find an ideal of $A$ containing $I$ maximal with respect to the constraint that said ideal, $J$, is disjoint from X.  I want to prove $J$ is prime by contradiction. I suppose that $J$ were not prime; this also means that $J$ is not maximal in $A$. Then there would exist two elements, $f$ and $g$, of $A$ so that $fg$ is in $J$, but neither $f$ nor $g$ is in $J$. Following the prompting of the aforementioned answer, I know that $J + \langle a\rangle$ and $J + \langle b\rangle$ would be ideals strictly containing $J$, and since J is the maximal ideal disjoint from $X$, both $J +  \langle a\rangle$ and $J + \langle b\rangle$ contain at least some elements of $X$ (I am tempted to say all of them, but I'm not entirely sure that's true, in case for example $\langle a\rangle$ started containing powers of $x$ at $x^2$, in which case it isn't necessarily true that $\langle a\rangle$ contains $x$ itself), and so both $\langle a\rangle$ and $\langle b\rangle$ must themselves contain elements of $X$.  And this is where I am stuck. I feel like there is something I should know about multiplicative subsets of rings that would help me arrive at the desired contradiction, but I can't remember what it might be. The new top bar layout is fine, but the black color is very visually jarring/intrusive.  Is using a lighter color an available option? Metric is continuous function Let $X$ be a metric space , $d$ is the metric , show that $d$ is a continuous function from $X\times X$ to $R$.  I think the definition is all we need , but I just don't know where to start , can anyone help me. We want to show that if $a,b\in G$ where $G$ is a finite Abelian group, we have  $\operatorname{LCM}(|a|,|b|) = |ab|$ given that $ab \neq e$.   How I approached this question was by saying let $\operatorname{LCM}(|a|,|b|) = L$. Then if we show that $L$ divides $|ab|$ and $|ab|$ divides $L$ then $|ab| = L$. Showing that $|ab|$ divides $L$ was fine. But then I am having trouble with the second part that is showing $L$ divides $|ab|$. For this, so far I have:  Consider $(ab)^{|ab|} = e = a^{|ab|}b^{|ab|} $  since we know $a \neq b^{-1}$ by assumption then we can conclude that $|a|$ divides $|ab|$ and $|b|$ divides $|ab|$. We know that $L = \frac{|a||b|}{\operatorname{gcd}(|a|,|b|)}$ from here can we conclude that $L$ divides $|ab|$? Is it the airport security at the country you flying from? I highly doubt it because I have been to countries where it's so corrupted that you buy the airport officials for the price of a dinner.   Is it the aircrew' responsibility? Again I don't think they care much - unless there is severe punishments - in corrupted third world countries.   I have been personally in a situation when an officer at a certain airport insisted on examining my British passport before he allowed me to move to the gate because I refused to slip him a $100 note. I was travelling to the UK yet I have another passport which is valid and it is issued by the country I was flying from, which means he had no reason whatsoever - nor I think he cared - to examine my British passport to check if it's valid.  Edit to summarize it, I want to expose as little as possible of information about my dual nationality. For that I need to know whom I am obligated to show my passport(s) Comparing two strings in Bash I have the following if block in my bash script:  if [ ${PACKAGENAME} -eq kakadu-v6_4-00902C ]; then   echo "successfully entered if block!!" fi   The script execution is not entering my if block even though $PACKAGENAME is equal to kakadu-v6_4-00902C. What am I doing wrong? file /dev/* prints the description of a bunch of files as "block special (M/N)" and "character special (M/N)", where M and N are numbers. For example:  $ file /dev/null /dev/null: character special (1/3)   man file doesn't seem to document these, but refers vaguely to man stat, which seems to refer to these as major and minor device types. apropos 'device type' finds nothing. So what do these numbers mean? is incompatible with the enumerate and itemize environments. Is there a typographical reason not to wrap a list around a figure or is it a technical limitation of LaTeX? If it is a technical limitation, is there any work (new packages or changes to LaTeX) currently being done to eliminate the limitation in the future?  Edit: To respond to some of the comments: I realize that there are "gymnastics" which can make this work. I do not have an example of the problem. Whenever I run into a problem with wrapfig because it is near an enumerate or itemize environment, I search for an easy solution, find none, and move the figure elsewhere. In the vast majority of cases, the cost of learning the complex solutions isn't worth the benefit of keeping the figure where I want it. I keep on seeing a fair number of questions (such as ) which get marked as answered within 5-10 minutes of being posted. Sometimes the questions like this are insanely popular and actually see some proper discussion, however often I see questions have few views and only 1 answer. The trouble with these questions is that very few people have scrutinised the answer for correctness, or possibly a better solution. I always make a point of not accepting my answers until a day or so has passed — would it not be a good idea to put a short (1 hour or so) time limit on accepting answers on new questions with few views or votes, or perhaps just an additional confirmation dialog, something along the lines of:  Your question is new and not many people have had a chance to look at it yet.  Are you sure you don't want to wait to see if anyone else has a better suggestion? I need to calculate the SQRT of  $x$ to $y$ decimal places. I'm dealing with $128$-bit precision, giving a limit of $28$ decimal places. Obviously, if $\,y &gt; 28$, the , which I'm currently using, becomes futile, as it simply doesn't offer $y$ decimal places.  My question to you is, can I calculate the Square Root digit-by-digit? For example, calculating the $n$-th digit, assuming I already have digit $\,n - 1$. I'm well aware this may be impossible, as I've not found any such method through Google yet, just checking here before I give up in search of arbitrary precision methods. A sum of series problem: $\frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \cdots + \frac{2008}{2006!+2007!+2008!}$ I have a question regarding the sum of this series: $$\frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \cdots + \frac{2008}{2006!+2007!+2008!}$$ My approach: I found that this sum is equal to: $$\sum_{n=3}^{2008}\frac{n}{(n-2)!+(n-1)!+(n)!}$$  I reduced it to : $$\sum_{n=3}^{2008}\frac{1}{n(n-2)!}$$  Please suggest how to proceed further. Does a gluon have mass or not? I was under the impression that gluons had no mass. Then Wikipedia throws this curveball:  Mass: 0 (theoretical value) &lt; 0.0002 eV/c2 (experimental limit)  Which one should a poor soul go by? If the theoretical value just the one that people &quot;thought&quot; was right, until they found out the truth, or is the experimental limit(?) an impossibility that still has not been explained? Or something third? Open more than one instance of Windows 10 settings How can I open more than one instance of the Windows settings?  It wasn't a problem with the old style, but with the new one I can't seem to open multiple settings simultaneously and side by side.  Example of the new settings window: Gerund preceded by a genitive? Is this sentence actually grammatical?     You know your having a rough day when kittens don't even make you smile.   The writer of this sentence may intend to mean you're instead of your but I'm just wondering if having a gerund preceded by a genitive is considered grammatical in this case.  If the sentence above is ungrammatical, why is this sentence grammatical?      He resents your being more popular than he is.   What's the difference between these two sentence that determines one to be grammatical and the other to be not?  This is how I analyze the two sentences and I can't see the difference:   "You know it when kittens don't even make you smile" for "it" = "your having a rough day" "He resents it" for "it" = "your being more popular than he is" Are proofs by contradiction really logical? While playing Phase Ten, is it legal to pick up the top card of the discard pile and then immediately discard that exact card? $f$ strictly increasing does not imply $f'>0$ We know that a function $f: [a,b] \to \mathbb{R}$ continuous on $[a,b]$ and differentiable on $(a,b)$, and if $f'&gt;0 \mbox{ on} (a,b)$ , f is strictly increasing on $[a,b]$. Is there any counterexample that shows the converse fails?   I have been trying to come up with simple examples but they all involve functions that are discontinuous or has derivative $f'=0$ which does not agree with the hypothesis hmmm Why won't Smart UV Project automatically unwrap a simple chair? I created a very simple chair mesh and I would like to unwrap it so that I can map a texture. I would like to have Smart UV Project unwrap it automatically, but whether I do it that way or try manually, I am unable to unwrap the model.   Is there something I'm missing?    I tried a few guides, but I'm not sure where I'm going wrong.  My File: