Prove the zero in the complex number system is unique.  I have an idea but i don't know how start this proof.   The idea:  I think start out with the assumption that there exist two different unities, say, $a$ and $b$, and then proceeds to show that this assumption leads to some contradiction.  But i'm a little stuck. Can someone help me?
We are supposed to prove that the zero vector is unique with the additive identity $x+O=x$ and then by creating another additive identity with $O'$, then $x+O'=x$.  We set them equal $x+O=x+O'$ and use the vector cancelling theorem to get $O=O'$, but how does this prove that the zero vector is unique?
Why disable Fast Boot on Windows 8 when having dual booting? Why is everybody keep mentioning to disable Fast Boot on Windows 8 if you have installed alongside Ubuntu? Is it something that is recommended only for UEFI machines or is it a suggestion for Legacy BIOS machines as well? Is it because it makes the Windows partition inaccessible from Linux or there is another more serious reason to disable it? Calculating Statistics Based on Expression in QGIS? Is it possible to calculate the sum of polygon areas based on whether or not it has a value in another column?   For example:  In column 2 I have the area of polygons. I want to calculate the area of all polygons that contain 'AG' in column 1 or all polygons that are IN column 1. But calculate them separately so I can get the total area of each type in column 1. In QGIS, I have been learning a few expressions to style my trail maps but still know little and am stumped on what to use to filter my symbol data. I am showing Parking locations on my map. I use a marker symbol as a background element, so I can do callouts. I created a data field and added values to display beside this symbol a number that references a list of parking addresses in the book's text. All good except if the field has no number to display, it is NULL, it does not display my marker symbol. I need to filter this, so it does show even when there is no value in the field. \overbrace and \underbrace with square bracket My current idea is to use the dual graph, and prove that that is a bipartite graph, and so proving the statement. But I can't seem to figure out how even degree for all verticies means that it is bipartite.  Any help is appreciated. Thanks! I'm trying to get through Spivak's Calculus on my own and even though I kinda understand induction I'm not so sure that's the case when it comes to complete induction. So I tried to do a starred problem which involves using it. But I'm not sure that my proof is valid, can someone check this for me?  So the problem is that I have to prove that     For every natural $n&gt;1$, if $n$ is not prime we can write it as a product of primes.   I'm starting with a base case. For $n=2$, $n$ is prime, so the assumption holds.  Let's assume that for some $n &gt; 1$ and also all numbers $p &lt;= n$ the assumption holds.  Then $n+1 = cd$. If $n+1$ is not prime, then $c &lt; n+1$ and $d &lt; n+1$, but as they are natural numbers, we can also write that $c &lt;= n$ and $d &lt;= n$. But we assumed that if $p &lt;= n$, and $p$ is not prime, we can write it as a product of primes. So if $c$ or $d$ are not primes, we can write them as a product of primes. That means that $n+1$ can be written as a product of primes if it's not prime, which completes the proof. Learning special relativity and working with Minkowski diagrams for a while now i am still trying to get my head around some oddity.   From my understanding time can often be seen as the dependent variable in those diagrams, so why put it to the vertical axis? What confuses me even more is the apparent convention to denote a point in space time $(t,x)$ even though $x$ is on the horizontal and $t$ is on the vertical axis, so the usual notation would be to refer to that vector as $(x, t)$.  Are there mostly historical reasons or why is it beneficial to use this notation and to violate usual conventions? Ubuntu 13.10 to 14.04 Lts does not upgrade, keeps showing error The update from Ubuntu 13.10 to 14.04 LTS does not work it keeps showing me an error as could not calculate the upgrade, even when I want it to get updated from the terminal it says the same error. I did not use the beta its the final Ubuntu 14.04 LTS.I have restarted my laptop for like 5-10 times no change. Disabling the software sources also does not help. I want to schedule my script for last Saturday of every month. I have come up with the below:  45 23 22-28 * *  test $(date +\%u) -eq 6 &amp;&amp; echo "4th Saturday"  &amp;&amp; sh  /folder/script.sh    Is this correct or I need to change it? I can't test it right now as it will be invoked only on last saturday. Please advise.  I have the below for last sunday of every month but i can't understand much of it. The first part gives 24 which is sunday and after -eq gives 7 which i don't know what it means:  00 17  * * 0 [ $(cal -s | tail -2 | awk '{print $1}') -eq $(date | awk '{print $3}') ] &amp;&amp; /folder/script.sh   Can i modify the above to get last saturday?  With Romeo's help I was help to come up with the below answer:  00 17 * * 6 [ $(cal -s | tail -2 | awk '{print $7}') -eq $(date | awk '{print $3}') ] &amp;&amp; /folder/script.sh Removing intercept from GLM for multiple factorial predictors only works for first factor in model For each positive even integer $n$, set   $$P_n = \displaystyle {{n} \choose {\frac{n}{2}}}\frac{1}{2^n}.$$  Show that $\displaystyle \lim_{n \to \infty} P_n$ exists and determine its value.   Here's what I have so far: each $P_n$ can be thought of as the probability of tossing a fair coin $n$ times, and obtaining exactly $\frac{n}{2}$ heads. My expectation is that $P_n \to 1$, since in any large trial, one would expect to record just about as many heads as tails. But I'm not sure how to mathematically justify this hunch.  Hints or solutions are greatly appreciated. How to make Portable QGIS 3 Is it possible to make QGIS 3 portable with changed location for Qgis setup parameters folder.  For QGIS 2 it looked like this in qgis.bat file:     qgis --optionspath "X:\qgis" --configpath "X:\qgisini"   For QGIS 3 it doesn't work. Setup parameters are default placed in:      C:\Users\xxx\AppData\Roaming\QGIS I'm attempting to modify the Table of Contents in my document. Currently, chapter one in my document is listed as   1   Introduction   What I would like to see is  Chapter 1   Introduction   Seems like an easy enough thing to ask for, but can't seem to find a good way to do this. Any feedback would be most appreciative. As a new DM I'm still learning how to control the monsters in a battle. A situation came up last game and although it was fine and fun, I wonder if there are better strategies for a DM to avoid a bottleneck in a doorway.  Scenario: PCs were up against a door deciding whether they should go in. Hobgoblins on the other side of the door open it and an encounter ensues. However, the encounter is basically the PC at the door trading blows with the NPC at the door, while their respective allies in the back try to shoot over their friends' heads. Although you can move through an ally's space, you can't occupy an ally's square. This leads me to assume that you can't move in to an ally's square, attack from it, and move back out of the square (though please correct me if I'm wrong).  This scenario and understanding of the rules brings me to my question.  What tactics can a DM use in a combat scenario like the one above to make sure that the NPCs are all contributing to the battle? Especially if they have no ranged weapons, is there a way that they can all try to attack? Is there a way to save selection by faces I'm trying to find a way to save selection faces, tried Vertex Group, the issue with Vertex Group is that it adds the faces that are marked by 4 vertices that are added to the Vertex Group, there are some cases were this doesn't apply, the image explains what I mean. So is there a way to save my selected faces, so I don't have to go through the whole process of reselecting them each time I need to do so? If you check the image, I didn't assign the hatched faces to the Vertex Group, still when I select the Vertex Group, they get selected, I understand that they got selected because the four boundary vertices are assigned to the Vertex Group, I need to save only face selection, not vertex selection. Genetic linkage greater than 50 centimorgans Classically, the linkage between two loci can be measured in centimorgans (cM), which represents the percent chance that these two loci will recombine an odd number of times (generating a recombinant genotype).  Due to independent assortment, markers on different chromosomes are expected to recombine 50% of the time. As I understand, if one conducts a classic experiment and counts various progeny, then deduces the apparent linkage, values of 50 cM or more are interpreted as "impossible to determine whether on separate chromosomes or just recombines very often". See (), with grammar corrected:     The final point that we need to make regards the maximum distance that we can measure. Because of the way in which the calculations are performed, we can never have more (than) 50% recombinant gametes. Therefore the (maximum) distance that two genes can be apart and still measure that distance is just less (than) 50 cM. If two genes are greater than 50 cM apart, then we can not determine if they reside on the same chromosome or are on different chromosomes.   Wikipedia gives an  and also remarks that (d is physical distance):     The probability of recombination is approximately d/100 for small values of d and approaches 50% as d goes to infinity.   However, what happens if the linkage is greater than 50 cM? It seems like classical experiments aside, such a situation is possible in reality.  For instance, , and . Let's take two genes on Chr.IV:    is at about +631 kb.  is at about +1370 kb.   The physical distance between these is about 734 kb. According to the cM/kbp value, the centimorgan distance is $734\text{ kb} \cdot 0.31\text{ cM}/\text{kb} = 228 \text{ cM}$ . How to interpret this 228?  If I mate a haploid yeast which is DNF2 TOM1 with another which is dnf2 tom1 (lowercase indicating minor allele rather than deletion), then sporulate them, what is the chance of getting spores with DNF2 tom1 or dnf2 TOM1 genotypes?  I realize that the cM/kbp value is only a simplification and that in practice linkage is a more complex phenomenon. Nevertheless, it seems plausible that several crossover events can happen on Chr.IV since it is so big. This includes the possibility of 1, 3, 5 and more crossovers which would create a hybrid (assuming they all happen between these two loci), as well as the possibility of 2, 4, etc crossovers which would create a non-hybrid spore (at least as far as our chosen markers are concerned). Confused About Trigonometric Substitution How do users navigate to Meta? How do you expect users to:   Know that a site's meta-site exists (and what it's for)? Find the hyperlink to the meta-site?   I know it exists, of course, and how to get there; I'm wondering how (or even whether) you expect new or naive users to do that.  I think there isn't a link to Meta from the main site, am I right (except one link in the "list of all 171 Stack Exchange sites" drop-down menu on the toolbar)?  Sometimes I tell people to post a question on Meta instead of on the main site -- I don't realise that they may not know that Meta exists, and that if they do (because they've been there before) they don't know how to get there (because they haven't yet investigated everything on the toolbar). Find roots of $3^x+x^3=17$ Find $x$ in the following equation: $$3^x+x^3=17$$ Creating node shapes I am currently learning how to use TikZ/PGF to make some figures for a project report, and so I have a question:     Is there is a practical way of making a particular object into a node?   The object I have in mind is shown in the Figure below. I want the object to consist in the big circle, the two circular arcs, the label and anchors at north, south, east and west. The dotted lines are simply there to indicate that I am going to have paths going out of these points, which is why I need the anchors there.     I include the source, in the case that it is useful to anyone.  \begin{tikzpicture} \draw[help lines, gray, dotted] (-2,-2) grid (2,2);  \begin{scope}[shift={ (0,0) } ]     \begin{scope}[rotate=90]         \draw[thick] (-1,0) arc[start angle=-90, delta angle=90, %                               x radius=1, y radius=1] -- (0,1) %NW                      (0,-1) arc[start angle=180, delta angle=-90, %                                    x radius=1, y radius=1] -- (1,0); %SE         \draw[gray] (0,0) circle (1);     \end{scope} % Anchor replacements     \node(N1)       at (0,1)    {};     \node(S1)       at (0,-1)   {};     \node(E1)       at (1,0)    {};     \node(W1)       at (-1,0)   {}; % Label     \node(Crossing number) at (0,0) {1}; \end{scope}  \path[thick, gray, dotted] [ out=90,   in=270]     (N1.center) edge (0,2) [ out=180,  in=0]       (W1.center) edge (-2,0) [ out=270,  in=90]      (S1.center) edge (0,-2) [ out=360,  in=180]     (E1.center) edge (2,0); \end{tikzpicture}   PS. I have had a look at the Manual, but I could not find a description of how to do this.