I have a PostGIS database where the data is in an unknown or incorrect coordinate system.  Is there a way given some known points in the data to translate all the data to a specific coordinate system.
I have a Shapefile but its coordinate system is Unknown, and there is no *.prj file.   How can I identify it now?
What does "may they rot there" mean? Terence Tao, Analysis I, Ex. 5.5.2: Entry point needed Terence Tao, Analysis I, 3e, Exercise 5.5.2:     Let $E$ be a non-empty subset of $\mathbb{R}$, let $n \ge 1$ be an   integer, and let $L &lt; K$ be integers. Suppose that $K/n$ is an upper   bound for $E$, but that $L/n$ is not an upper bound for $E$. Without   using Theorem 5.5.9, show that there exists an integer $L &lt; m \le K$   such that $m/n$ is an upper bound for $E$, but that $(m-1)/n$ is not   an upper bound for $E$. (Hint: prove by contradiction, and use   induction. It may also help to draw a picture of the situation.)    does not do the trick for me, since I'm looking for the proof by contradiction, equipped only with the set of propositions and theorems covered so far.  What I'm actually looking for is an entry point to the proof, since I find it quite hard to negate the assumption.  My attempt is to induct over $n$. Assume I have been able to prove the statement for $n = 1$, is the following a valid negation of the assumption?  For all integers in the range $(L, K]$, at least one of the following statements holds:   $(m-1)/(n+1)$ is not an upper bound and $m/(n+1)$ is not an upper bound. $(m-1)/(n+1)$ is an upper bound and $m/(n+1)$ is not an upper bound. $(m-1)/(n+1)$ is an upper bound and $m/(n+1)$ is an upper bound.   If I were able to prove that there is an integer in the given range such that none of the above holds, am I done?       Theorem 5.5.9 (Existence of least upper bound). Let $E$ be a non-empty subset of $\mathbb{R}$. If $E$ has an upper bound, (i.e.,   $E$ has some upper bound $M$), then it must have exactly one least   upper bound.     I went through another loop considering  of Steve Kass, but I ended up wondering why the proof needs induction at all:  Assume for all integers $m$ where $L &lt; m \le K$ we have that $m/n$ and $(m-1)/n$ are upper bounds of $E$. Then $(m-1)/n = L/n$ is also an upper bound. But $L/n$ is no upper bound, by assumption. Assume $m/n$ and $(m-1)/n$ are no upper bounds. Then $m/n = K/n$ is also no upper bound. But $K/n$ is an upper bound, by assumption.   Hence, there is an integer in the range $(L, K]$ s.t. $m/n$ is an upper bound for $E$, and $(m-1)/n$ is not. I have a netbook that I want to install Ubuntu onto. I do have a good Mac computer that I can make a bootable USB drive from, but I'm confused on how to go about making it. The axiom of extensionality is:  ${\forall}a({\forall}b({\forall}c(((c{\in}a){\iff}(c{\in}b)){\iff}(a=b))))$  The definition of $(P{\iff}Q)$ is:  ${\lnot}((P{\implies}Q){\implies}({\lnot}(Q{\implies}P)))$  or, in other words  $(P{\land}Q){\lor}({\lnot}P{\land}{\lnot}Q)$  In the case of the statement $((c{\in}a){\iff}(c{\in}b))$, this leads us to two options: $((c{\in}a){\land}(c{\in}b))$ and $({\lnot}(c{\in}a){\land}{\lnot}(c{\in}b))$. So far nothing weird. Now, let's see all the possible results of the axiom.  1.$((c{\in}a){\land}(c{\in}b)){\land}(a=b)$ 2.${\lnot}((c{\in}a){\land}(c{\in}b)){\land}{\lnot}(a=b)$ 3.$({\lnot}(c{\in}a){\land}{\lnot}(c{\in}b)){\land}(a=b)$ 4.${\lnot}({\lnot}(c{\in}a){\land}{\lnot}(c{\in}b)){\land}{\lnot}(a=b)$  Coincidentally, they are also all possible statements regarding the statements $((c{\in}a){\land}(c{\in}b))$ and $(a=b)$: both parts are true, the first part is true and the other is false, the second part is true and the first is false and neither is true. The axiom seems not to state anything new, because rephrased it's just produces a list of all possible options regarding the two parts of the statement. Worse, it doesn't seem to say anything about equality, since it allows for $a=b$ with both $(c{\in}a){\land}(c{\in}b)$ and ${\lnot}((c{\in}a){\land}(c{\in}b))$. So, what is the point of this axiom? And in what way does it say anything about equality? A general question asked from time to time on stack exchange is question of trigger bulkification.  Therefore I have 2 questions:   Which best practices are you using? What is the best way to handle updates of more than 200 records? (In case of more than 200 records the trigger is executed more than once.) Why is my variable local in one 'while read' loop, but not in another seemingly similar loop? Ubuntu installer can't find any disk on Dell XPS 13 9350 I am trying to install Ubuntu 14.04 / 15.10 on my Dell XPS 13 9350 (November 2015).  I created a bootable USB disk which boots fine but the installer and the file manager do not detect the hard drive.  What is the problem? How can I solve it? Use Cauchy product to find a power series represenitation of   $$1 \over {(1-x)^3}$$ which is valid in the interval $(-1,1)$.   Is it right to use the product of $1 \over {1-x}$ and $1 \over {(1-x)^2}$ if I know thier expantion (HOW).. Is there another way? Who is the Maker? C-3PO frequently speaks of “the Maker.” Is this a true droid religion, or is this a religious reverence by droids for their specific creators (in this case, Anakin, even though C-3PO has forgotten who he is)? Is this even a droid thing or is it just C-3PO? What else do we know about this “Maker”-worship? Limit with inverse trignometry We have to find the value of $$\lim_{n\to \infty}\sum_{k=2}^n\arccos\left(\frac{1+\sqrt{(k-1)k(k+1)(k+2)}}{k(k+1)}\right)$$  I tried it a lot but could not find any start.   Please help me in this . A limit question (JEE $2014$) The following is a JEE (A national level entrance test) question:  Find the largest value of the non-negative integer ( a ) for which:  $$ \displaystyle \lim_{x \to 1} \left( \dfrac{-ax + \sin(x-1) + a} { x + \sin(x-1) -1 } \right)^{\dfrac{1-x}{1-\sqrt x} } = \dfrac 1 4 $$  On solving this, we get $ a = 0 $ or $ a = 2 $. So we take the answer as $ 2 $. But the official answer key says that the answer is $ 0 $. The reason given (not in the official key - they simple contain the answer. I saw the reason in an "unofficial" solution to the question paper) is that if $ a = 2 $, then the term $\displaystyle \dfrac{-ax + \sin(x-1) + a} { x + \sin(x-1) -1 } $ tends to a negative value ( $-0.5$ ). So $ a = 0$.  BUT:  By putting $a = 2$ in the limit, and inputting that to wolframalpha (), the answer turns out to be $0.25$. So, by putting $ a = 2$ also, we get the same value of the limit - $0.25$.  How to solve this dispute? Also, could you please check with Mathematica? Unfortunately, I don't have access to it!  (This is very important as the change in the answer key will affect the rankings in a dramatic way) Prove $(1-a)(1-b)(1-c)(1-d)\geq abcd$ if $a^2+b^2+c^2+d^2=1$ Let $a,b,c,d\geq0$, $a^2+b^2+c^2+d^2=1$ Prove $\displaystyle (1-a)(1-b)(1-c)(1-d)\geq abcd$ I mutiplied both with $\displaystyle (1+a)(1+b)(1+c)(1+d)$ to use $1-a^2=b^2+c^2+d^2$ and try using the cauchy-schwarz and holder but it is doesn't work. How to remove a repository? How to not count blank pages in frontmatter of two-sided document I'm using terminal-ide  which is basically a terminal for android with all sorts of utilities. One is able to create bash scripts for example. Is there a way to create an icon on the home screen that will launch one of these scripts? How much energy does a photon need to form a black hole? I was wondering if it is possible for a single photon to form a black hole if it has a small enough wavelength.If so, what would this wavelength be? I came across this question because I am reading about loop quantum gravity, and have heard that it is 'untestable'. I know that the loops making up spacetime in loop quantum gravity are thought to have a size in the order of $10^{-35}$m, and I was wondering whether making a photon with a wavelength small enough to investigate this would form a black hole.  I am aware of the two posts here, but neither seem to really answer my question because they refer to several photons:  and .  Thank you. In Photoshop I've taken Magic Wand as far as I can, I think. Now I'm left with this grayish edge around the coffee saucer. How do I remove it? I've tried altering tolerance in Magic Wand but don't seem able to get it. Any suggestions please? Close and flag links do nothing on mobile site Limit point compactness implies compactness. I have a question about the following counterexample, why  $U_n = \{n\} \times Y$ has not finite collection covering $X$?   Is it because if you take a finite set $S\subset U_n$ then it's impossible to have $X\subset \bigcup S$ since X is infinite?  Counterexample:  Let $Y$ consist of two points; give $Y$ the topology consisting of $Y$ and the empty set. Then the space $X = \mathbb{Z}_{+} \times Y$ is limit point compact, for every nonempty subset of $X$ has a limit point. It is not compact, for the covering of $X$ by the open set $U_n = \{n\} \times Y$ has not finite collection covering $X.$ Is there a way to detect if a browser window is not currently active?