Troube DualBooting Ubuntu 14.04 after install on windows 7 disk
How can I install Windows after I've installed Ubuntu?
Here is my proof, I would appreciate it if someone could critique it for me:  To prove this statement true, we must proof that the two conditional statements ("If $\mathcal{P}(A)⊆ \mathcal{P}(B)$, then $A⊆B$," and, If $A⊆B$, then $\mathcal{P}(A)⊆ \mathcal{P}(B)$) are true.    Contrapositive of the first statement: If $A \nsubseteq B$, then $\mathcal{P}(A) \nsubseteq \mathcal{P}(B)$  If $A \nsubseteq B$, then there must be some element in $A$, call it $x$, that is not in $B$: $x \in A$, and $x \notin B$. Since $x \in A$, then $\{x\} \in \mathcal{P}(A)$; moreover, since $x \notin B$, then $\{x\} \notin \mathcal{P}(B)$, which proves that, if $A \nsubseteq B$, then $\mathcal{P}(A) \nsubseteq \mathcal{P}(B)$. By proving the contrapositive true, the original proposition must be true.    To prove the second statement true, I would implement nearly the same argument, so that isn't necessary to write. So, does this proof seem correct? Also, was the contrapositive necessary? Or is there another way to prove the initial statement? Is $\exists x(x=a)$ a valid formula in first order logic with equality? Under , is $\exists x(x=a)$ true in every interpretation in first order logic (FOL) with equality? If so, could $\exists x(x=a)$ be considered a tautology? This question was motivated by  which was closed for lacking clarity. Here I have attempted to add that clarity myself so that the same question can be useful to others who are wondering a similar thing. This is a self answer question. notify-send not working under ssh Maybe this is answered somewhere else, but I didn't see it.  I am running Ubuntu 14.04. When I SSH into my machine, for example:  ssh &lt;user&gt;@&lt;machineip&gt; notify-send "Hello"   I don't see anything on the monitor where I am logged into the machine. If I prefix notify-send with DISPLAY=:0.0 or DISPLAY=:0 nothing different happens. I just never see any notification on the current session.  Is there some trick/switch to getting this working?  In case this isn't clear, allow me to reiterate: From Computer A, I SSH into Computer B. Within the SSH session, I wish to execute notify-send to run on Computer B. I expect a growl-type notification to appear on the monitor of Computer B. Sequence uniform convergence but the derivatives are not. Give a sequence $(f_n)_{n\in \mathbb{N}}$ of differentiable functions which uniformly converge to $0$, but for which the seqeunce $(f_n')_{n\in \mathbb{N}}$ of the derivatives isn't even pointwise convergent.    I found this one in my textbook marked as "Fun things to solve" and although it may be fun, it's kind of hard to do.   To be honest I already failed at the first hurdle. I couldn't even find a sequence of functions which uniformly converges to $0$.   Is there a certain way of dealing with this kind of problem? Because it seems kind of hard for me to come up with sequences without a mathematical of way of doing so (or maybe there is a mathematical way I just don't know yet). Areas versus volumes of revolution: why does the area require approximation by a cone? Suppose we rotate the graph of $y = f(x)$ about the $x$-axis from $a$ to $b$.  Then (using the disk method) the volume is $$\int_a^b \pi f(x)^2 dx$$ since we approximate a little piece as a cylinder.  However, if we want to find the surface area, then we approximate it as part of a cone and the formula is $$\int_a^b 2\pi f(x)\sqrt{1+f'(x)^2} dx.$$  But if approximated it by a circle with thickness $dx$ we would get $$\int_a^b 2\pi f(x) dx.$$  So my question is how come for volume we can make the cruder approximation of a disk but for surface area we can't. Probability that two random numbers are coprime is $\frac{6}{\pi^2}$ This is a really natural question for which I know a stunning solution. So I admit I have a solution, however I would like to see if anybody will come up with something different. The question is     What is the probability that two numbers randomly chosen are coprime?   More formally, calculate the limit as $n\to\infty$ of the probability that two randomly chosen numbers, both less than $n$ are coprime. Simple approximation of Inverse Incomplete gamma function What is the complexity of finding SHA-1 collisions? Given the , what is the complexity of finding a SHA-1 collision?  Marc Steven's  is still detailing about 260. Does the previous complexity still hold? 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) In a field that is the direct product of two fields, why isn't it a field? Suppose that we have the direct product of two fields, defined as: $\mathbb{F}^2 = \mathbb{F} \ \times\ \mathbb{F}$. Then, this is obviously not a field because we can take $(1,0)$, and this doesn't have a multiplicative inverse. However, I saw a mention to this today in a book, and the reasoning was:     "$\mathbb{F} \ \times\ \mathbb{F}$ has the zero-divisor $(1,0) \neq  (0,0)$, hence it is not a field".   (Lectures on Rings and Modules by Joachim Lambek, pg. 17).  I am not sure what they mean by zero-divisor here nor what $(1,0) \neq  (0,0)$ has anything to do with the direct product of fields not being a field. I always assumed there wasn't a multiplicative inverse, and that was it. Am I missing something deeper? In the cases of black holes that form from supernova and collapse of a massive star, I understand that in most of these cases, the star loses significant amounts of mass from the explosion. Presumably, after this point as the remaining mass becomes more dense as it further collapses unto itself, it eventually becomes a black hole with gravitational force greater than that of its parent star. But, if gravity is based on mass, how can the black hole have greater gravitational force than the star from which it is formed? What do the fields in ls -al output mean? The ls -al command shows the following output;  -rwxrw-r--    10    root   root 2048    Jan 13 07:11 afile.exe   What are all the fields in the preceding display? Script to convert filenames to lowercase depending on extension Extracting text from HTML file using Python Ideal of the twisted cubic The twisted cubic is the image of the morphism $\phi : \mathbb{P}^1 \to \mathbb{P}^3 , (x:y) \mapsto (x^3:x^2 y:x y^2:y^3)$, it is given by $X = V(ad-bc,b^2-ac,c^2-bd)$. Now I would like to compute $I(X)$, which equals by the Nullstellensatz the radical of the ideal $I := (ad-bc,b^2-ac,c^2-bd) \subseteq k[a,b,c,d]$. I think that that $I$ is already a radical ideal, even a prime ideal. Namely, I suspect that  $$\phi^* : Q:=k[a,b,c,d]/I \to k[s,t] , a \mapsto s^3, b \mapsto s^2 t , c \mapsto s t^2 , d \mapsto t^3$$ is an injection. If this is true: How can we prove that? I've already tried to find a $k$-basis of the quotient, but this turned out to be a big mess. Even the representation of the quotient as a monoid algebra doesn't seem to help. Another idea is the following: A formal manipulation of generators and relations implies $Q_a \cong k[a,b]_a$. Thus it suffices to prove that $Q \to Q_a$ is injective, i.e. that $a$ is not a zero divisor. Why do residuals in linear regression always sum to zero when an intercept is included? I'm taking a course on regression models and one of the properties provided for linear regression is that the residuals always sum to zero when an intercept is included.   Can someone provide a good explanation for why this is the case? What is the difference between "likelihood" and "probability"? The  claims that likelihood and probability are distinct concepts.     In non-technical parlance, "likelihood" is usually a synonym for "probability," but in statistical usage there is a clear distinction in perspective: the number that is the probability of some observed outcomes given a set of parameter values is regarded as the likelihood of the set of parameter values given the observed outcomes.    Can someone give a more down-to-earth description of what this means?  In addition, some examples of how "probability" and "likelihood" disagree would be nice. Is there a formal proof for the superposition theorem? Package babel Error: Unknown option `francais' As I was trying to validate our pcb design, I measured the impedance matching on the digital transmission lines, and got stuck with the oscilloscope not showing what I expected. We are using the PicoScope 4227, but I suspect that the strange measurement is not directly related to the specific oscilloscope but more to general on how to measure digital signals.  My setup is as follow: I measure the output-pin of an FPGA with both oscilloscope input A and B.  Input A (blue) is measured using the oscilloscope probe set to x10.  Input B (red) is measured using a 50-ohm coxial cable terminated at the scope input with 50-ohm.  As you can see, Input B seems correct, no reflection ect.  But Input A (the probe) has some overshoot, and ringing.  The probe should be calibrated and valid to 250MHz, so I don't get was is wrong. ?  I suspect that the small probe-clip adds some inductance, and therefore we see the ringing. But I really don't know.    EDIT After reading your answers, I walked got into our lab storage and started looking, and found a probe GND clip like the one showed in the answer below.   When I use that clip, I get the output below. Which is really a huge improvement.