I have an HP Pavilion dv6 Notebook with an AMD AE-35200M APU with Radeon HD Graphics 1.60 GHz. I has 8 gigs of ram and a 64-bit operating system.   I have only one storage device and I already have windows 7 installed. I am attempting to run Ubuntu on my computer without having to uninstall windows. The steps I took so far are as follows: I went to the official Ubuntu downloader and burned an installation disk. I then ran the disk to begin installation. A prompt window came up asking for a name and password along with a few other things that I left at default. It began installing and made it to the part mentioned above and then stopped.  Everything goes well until the install got to the "expanding" section. It has been on this screen for five hours, and the progress bar has not moved.  I have retried this twice with no success. So, can I assume something is wrong? How can I fix this?
I need my pre-installed version of Windows 7 (or any other version of Windows), how could I install Ubuntu without erasing it?
Failed review audit - for me quite clearly off-topic, can it be removed from the audit set? $3\times3$ linear system organization One of us is wrong, aren't we? I have just learned from what I consider a reliable source, that the following sentence is correct:     One of us is wrong, aren't we?   I would never in my life have written this, but I am assured that   As I realize comments don't live forever I will quote the relevant parts:     oerkelens : So you  would really write One of us is wrong, aren't we? I guess by analogy you would not bat an eyelid at One of these balls is blue, aren't they?   Matt Эллен : yes, "Then one of us is wrong, aren't we?" is exactly how it would be written. Same for the balls.   Could someone please enlighten me how it is possible that the number in a question tag supposedly has to be in discordance with the subject of the main clause?  I admit that I am not the youngest any more, and my school days are long gone, but back in the days, I was taught that  1. a verb and its subject concord in number 2. a question tag concords with the subject of the main clause  I was given examples like:     It is warm today, isn't it?   We will be on time, won't we?   Mary is pretty, isn't she?   John isn't the brightest, is he?   Some people may notice that the subject of the main clause seems to determine every time the subject of the question tag. When the main clause subject is singular, so are verb and subject in the question tag.  I was under the impression that a) this made sense and b) this would be a general rule.  I fully realize that grammar does not have to make sense, so a) is immaterial. As for b), today I learned I was wrong :)  So when (and possibly why?) do we form question tags that are different in number from their main clause?   To exemplify, also the other example sentence that I used and was corrected on:     One of these balls is blue, isn't it? (so this is wrong)   One of these balls is blue, aren't they? (and this is correct)   As an afterthought, does this strange grammatical number mix-up only appear in question-tags, or should I always refer to singular subjects in the plural if certain conditions are met? And what are those conditions?  Which versions are correct?     One of the cars is broken, aren't they? They (the car(s?) that is(are?) broken) should be fixed.   One of the cars is broken, aren't they? It (the car that is broken) should be fixed.   One of us must be wrong, mustn't we? We (the one(s?) that is (are?) wrong) should make amends.   One of us must be wrong, mustn't we? He (the one that is wrong) should make amends.   (In the last one, they could of course be used as the singular they, but that would avoid the issue...) How can I bypass Activation Lock? I bought a used iPhone 4.  The previous owner re-set the phone, but did not remove the device from their iCloud.  It is asking me for their iCloud information when I try to set-up the phone.  I don't have contact with the previous owner anymore.  Is there a way to remove the device from their iCloud so I can use a new one? Explanation of the binomial theorem and the associated Big O notation I've been watching a lot of pro replays ever since the SC2 beta came out, and I've noticed that almost always, when Protoss research upgrades, the shields are the last research chosen. Why is that? Seems to me that if anything, shields should be the most important upgrade, as they affect all Protoss units and buildings. Let $\mathbb S = \varnothing$.      Then from the definition: $ \bigcap \mathbb S = \left\{{x: \forall X \in \mathbb S: x \in X}\right\}$      Consider any $x \in \mathbb U$.      Then as $\mathbb S = \varnothing$, it follows that: $\forall X \in \mathbb S: x \in X$ from the definition of vacuous truth.      It follows directly that: $\bigcap \mathbb S = \left\{{x: x \in \mathbb U}\right\}$      That is: $\bigcap \mathbb S = \mathbb U$. ()   Proofwiki uses the above "proof" to "prove" that intersection of the empty set is the whole universe.   My question is, is the use of vacuous truth really allowed in axiomatic set theory, like ZFC? I don't see how the use of vacuous truth is justified.   The next problem I can think of is that we cannot really "define" the elements of empty set (to my knowledge, there is no element in empty set) so how can we then prove as the above proof did? This seems to contradict the use of vacuous truth.  And of course, there is issue of using the whole universe as a set, and I don't think this is allowed.... (Maybe proof above is using a different axiomatic set theory, as I am using ZF-minded thoughts...) Ordinal dependent variable with continuous independent variables One-hot vs dummy encoding in Scikit-learn $G$ is non abelian simple group of order $<100$ then $G\cong A_5$ Suppose $X$ and $Y$ are independent positive random variables with probability density functions $f_X$ and $f_Y$ respectively. Show that $Z=X/Y$ is absolutely continuous and find its probability density function.   $\textbf{My Thought:}$ In order to show that $Z$ is absolutely continuous, I need to show that it has a commulative distribution function. Then I can also differentiate the latter to obtain the probability density function of $Z$. I have the following calculation \begin{align*} P(X/Y\leq z)&amp;=P(X\ge zY,Y&lt;0)+P(X\le zY,Y&gt;0)\\ &amp;=\int_{-\infty}^{0}\left(\int_{yz}^{\infty}f_{X}(x)dx\right)f_{Y}(y)dy+\int_{0}^{\infty}\left(\int_{-\infty}^{yz}f_{X}(x)dx\right)f_{Y}(y)dy\\ &amp;=\int_{0}^{\infty}\left(\int_{0}^{yz}f_{X}(x)dx\right)f_{Y}(y)dy. \end{align*} Also, if $z&lt;0$, then we have that $F_Z(z)=0,$ since $X,Y$ are positive random variables. Therefore the random variable $Z$ is continuous, and we can differentiate it to obtain the probability distribution function of $Z$. We have that $f_Z(z)=0$ if $z\leq 0,$ and otherwise we have $$f_Z(z)=\int_{0}^{\infty}yf_{X}(yz)f_Y(y)dy.$$    Is my reasoning above correct?  Any feedback is much appreciated.  Thank you for your time. Visa for transit in Seoul I have a question about the visa needed to transit in Korea.  I'm an Indonesian passport holder and am traveling to the US. On the way back I have a transit of 24 hours at Incheon. If I want to go out and stay in Seoul, do I have to apply for a visa? Or can I get a transit visa on arrival? Convert a string/integer to superscript in C# Proof of Steinitz Theorem I want a source containing the proof of Steinitz Isomorphism Theorem stating:     For any Dedekind domain $R$ and any two nonzero ideals $I$ and $J$ of $R$ we have $I⊕J≅R⊕IJ$.   Thanks! I would like to draw an arrow like this using TikZ:    This is taken from . I didn't succeed in reproducing the arrow; is somewhat beyond my (extremely limited) TikZ abilities. I don't need all the workflow stuff. I simply want to draw such an arrow at a certain place in my text, e.g. -->  \documentclass{article} \usepackage{&lt;???&gt;} \usepackage{tikz} \usetikzlibrary{&lt;???&gt;} \begin{document} Donec vehicula augue eu neque. Pellentesque habitant morbi \tikz &lt;???code for the arrow???&gt; ; tristique senectus et netus \end{document}   PS: Some explanations in your solution would be much appreciated, otherwise it will be difficult to customize it. 'innerText' works in IE, but not in Firefox Show that $d_b(x,y)=\frac{d(x,y)}{1+d(x,y)}$ is a metric. where $(X,d)$ is a metric and $x,y \in X$.  I know we need to show:   non-negativity: $d(x,y)\geq$ 0 $d(x,y)=0$ if and only if $x=y$ symmetry: $d(x,y)=d(y,x)$ $d(x,z)\leq d(x,y) + d(y,z)$   I think we need to consider the derivative of $f(t)=\frac{t}{1+t}$ however I am not sure what to do next. Nested Radical of Ramanujan I think I have sort of a proof of the following nested radical expression due to Ramanujan for $x\ge 0$.     $$\large x+1=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}}$$ for $ x\ge -1$   I just want to know if my proof is okay or there is a flaw, and if there is one I request to give some suggestions to eliminate them. Thank you. The proof is the following:  Proof: Let us define $$ a_n(x)=\underbrace{\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}}}_{n \ \mbox{terms}}$$ for $x\ge 0$ so that $$ a_1(x)=\sqrt{1+x},\ a_2(x)=\sqrt{1+x\sqrt{1+(x+1)}},\ a_3(x)=\cdots$$    and so on. Since $x\ge 0$ each one of the $a_n$ is defined (I am taking only the positive square root). Also, we note that $$a_{n+1}^2(x)=1+xa_{n}(x+1)$$  Now, we note that $\{a_n(x)\}$ is an increasing sequence and that $$a_n(x)&lt;x+1$$ $\forall n\ge 1$, this is easy to prove by induction as below:  For $n=1$, $a_1(x)=\sqrt{1+x}&lt;1+x$ since $x\ge 0\Rightarrow 1+x\ge 1$. SO it is true for $n=1$. Similarly, the truth can be proved for $n&gt;1$.  Then $a_n(x)$ converges to $$l(x)=\sup_{n}a_n(x)\le x+1$$ Now I make the following claim:  Claim: $l(x)=x+1\quad \forall i\ge 0$  Proof: Fix $x$. Let $l(x)&lt;x+1$. Then, $l(x)=x+1-\epsilon$ for some $\epsilon&gt;0$. Now, I claim that there must be a $n$ such that $$x+1-a_n(x)&lt;\epsilon$$, and if that is true then $$a_n(x)&gt;x+1-\epsilon=l(x)$$ which is a contradiction since $$l(x)=\sup_{n}a_n(x)$$ and then it implies that $$l(x)=x+1$$ To prove my claim it requires $$x+1-a_n(x) &lt;  \epsilon$$ Now, \begin{align} x+1-a_n(x) = &amp; x+1-\sqrt{1+xa_{n-1}(x+1)} \\ \ =&amp; \frac{(x+1)^2-({1+xa_{n-1}(x+1)})}{x+1+\sqrt{1+xa_{n-1}(x+1)} } \\ \ =&amp; x\frac{(x+1)+1-a_{n-1}(x+1)}{x+1+\sqrt{1+xa_{n-1}(x+1)} } \\ \ &lt;&amp; \frac{x}{x+2}((x+1)+1-a_{n-1}(x+1))\\ \ &lt;&amp; \frac{x}{x+2}\cdot\frac{x+1}{x+3}((x+2)+1-a_{n-2}(x+2))\\ \ &lt;&amp; \frac{x}{x+n-1}\cdot \frac{x+1}{x+n} ((x+n-1)+1-a_{1}(x+n-1))\\ \ &lt;&amp; \frac{x}{x+n-1}\cdot \frac{x+1}{x+n}(x+n-\sqrt{x+n})\\ \ &lt;&amp; \frac{x(x+1)}{x+n-1} \end{align}  Now, if one is able to find $n$ such that $$\frac{x(x+1)}{x+n-1}&lt;\epsilon \Rightarrow x&lt; \frac{-(1-\epsilon)+\sqrt{(1-\epsilon)^2+4\epsilon(n-1)}}{2}$$ then we're done.   Now from the upper bound it seems that there always exists some $n$ that satisfies this requirement. Hence the claim is proved. Mongoose Sub Document Woes Chrome address bar no longer shows protocol or www subdomain A couple of days ago I noticed that for all websites, chrome omits from the address bar the protocol and the subdomain if it is www. My co-workers report that they are not seeing this. Chrome version is 67.0.3396.87.  For example, if I type into the address bar https://www.google.com (without the quotes), Chrome immediately displays the chip "Secure" followed by google.com. See attached screenshots.      I have looked in vain for Chrome settings that affect this, plus the chrome://flags page and chrome://net-internals page.  I did not do anything consciously to make this change.  Any ideas?  EDIT: Suggestion from MrWhite that Ominbox UI Hide Stead-State URL Scheme and Trivial Subdomains had become enabled: good suggestion, except that the setting is still at “Default.” See screenshot:    However, as a shot in the dark, I am going to play with the setting to see what happens.  First I set it to “Disabled.” That restored the previous behavior of the address bar, which is to not to hide the protocol and “trivial subdomains.” Then I changed it back to “Default,” and it was hiding them again. One would suspect that somehow the default has been changed, except that co-workers are not reporting this issue.  Oh, well. I’ll just leave it as Disabled for now.  I would be ready to accept the answer from MrWhite, because it provides a workaround. Yet will leave it open for a bit longer, because it does not really answer why this behavior has changed, and why it has to be overridden.  EDIT 2: Chrome version is 67.0.3396.87.