Don’t do nothing stupid
How is double negation interpreted in English?
I've seen many posts regarding a combinatorial proof of the following question. But for a non-combinatorial proof would the following method work? Also... is this the easiest way to arrive at a proof? It seems to be rather verbose.  Show the formula $k {n\choose k} = n {n-1 \choose k-1}$ is true for all integers $n,k$ with $0\le k \le n$.  My answer:  Observe that  $$\begin{align*} {n\choose k}&amp;= \frac {n}{k} {n-1 \choose k-1}\\ &amp;= \frac {n}{k}\left[ {n-2 \choose k-2} + {n-2 \choose k-1}\right]\\ &amp;= \frac {n}{k}\left[ \frac{(n-2)!}{(k-2)!(n-2-(k-2))!} + \frac {(n-2)!}{(k-1)!(n-2-(k-1))!}\right]\\ &amp;= \frac {n}{k}\left[ \frac{(n-2)!}{(k-2)!(n-k)!} + \frac {(n-2)!}{(k-1)!(n-k-1)!}\right]\\ &amp;= \frac{(n-1)!}{(k-1)!(n-k)!} + \frac {(n-1)!}{(k)!(n-k-1)!}\\ &amp;= \frac{(n-1)!}{(k-1)!(n-1-(k-1))!} + \frac {(n-1)!}{(k)!(n-1-k)!}\\ &amp;={n-1 \choose k-1} + {n-1 \choose k}={n \choose k}. \end{align*}$$ If the velocity is a relative quantity, will it make inconsistent equations when applying it to the conservation of energy equations?  For example:  In the train moving at $V$ relative to ground, there is an object moving at $v$ relative to the frame in the same direction the frame moves. Observer on the ground calculates the object kinetic energy as $\frac{1}{2}m(v+V)^2$. However, another observer on the frame calculates the energy as $\frac{1}{2}mv^2$. When each of these equations is plugged into the conservation of energy, they will result in 2 distinct results (I think). $$\int_{-\infty}^{\infty} \frac{\cos(αx)}{(x^2+1)(x^2+4)} \mathrm dx. $$  I am not sure how to solve this question. Can anyone help me to approach this problem. Thanks. If $M$ is an $R$-module and $I\subseteq\mathrm{Ann}(M)$ an ideal, then $M$ has a structure of $R/I$-module Let $M$ be an $R$-module and let $I$ be an ideal of $R$ such that $I$ is a subset of $\mathrm{Ann}(M)$. Define a product of an element of $R/I$ by an element of $M$ as follows: $$(r+I)\cdot m=rm.$$ (a) Show that this product is well defined. (b) Show that this product, together with the sum of elements of $M$, turns $M$ into an $R/I$-module.  My answer for (a): Let $r+I,s+I \in R/I$ such that $r+I=s+I$. Then $(r+I)m=(s+I)m$, for all $m \in M$. Then $rm=sm \implies r=s$  (since the inverse of $m$ is in $M$), so the product is well defined.  Is my answer to (a) correct?  How can I use Ann(M) to answer (b)? Value of $\lim_{n\to \infty}\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}$ I remember that a couple of years ago a friend showed me and some other people the following expression:  $$\lim_{n\to \infty}\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}.$$  As shown below, I can prove that this limit exists by the monotone convergence theorem. I also remember that my friend gave a very dubious "proof" that the value of the limit is $\frac{1}{e-1}$. I cannot remember the details of the proof, but I am fairly certain that it made the common error of treating $n$ as a variable in some places at some times and as a constant in other places at other times. Nevertheless, numerical analysis suggests that the value my friend gave was correct, even if his methods were flawed. My question is then:  What is the value of this limit and how do we prove it rigorously?  (Also, for bonus points, What might my friend's original proof have been and what exactly was his error, if any?)  I give my convergence proof below in two parts. In both parts, I define the sequence $a_n$ by $a_n=\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}$ for all integers $n\ge 2$. First, I prove that $a_n$ is bounded above by $1$. Second, I prove that $a_n$ is increasing.   (1) The sequence $a_n$ satisfies $a_n&lt;1$ for all $n\ge 2$.  Note that $a_n&lt;1$ is equivalent to $1^n+2^n+\cdots+(n-1)^n&lt;n^n$. I prove this second statement by induction. Observe that $1^2=1&lt;4=2^2$. Now suppose that $1^n+2^n+\cdots+(n-1)^n&lt;n^n$ for some integer $n\ge 2$. Then  $$1^{n+1}+2^{n+1}+\cdots+(n-1)^{n+1}+n^{n+1}\le(n-1)(1^n+2^n+\cdots+(n-1)^n)+n^{n+1}&lt;(n-1)n^n+n^{n+1}&lt;(n+1)n^n+n^{n+1}\le n^{n+1}+(n+1)n^n+\binom{n+1}{2}n^{n-1}+\cdots+1=(n+1)^{n+1}.$$  (2) The sequence $a_n$ is increasing for all $n\ge 2$.  We must first prove the following preliminary proposition. (I'm not sure if "lemma" is appropriate for this.)  (2a) For all integers $n\ge 2$ and $2\le k\le n$, $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.  We observe that $k^2-1\le kn$, so upon division by $k(k^2-1)$, we get $\frac{1}{k}\le\frac{n}{k^2-1}$. By , we may find:  $$\frac{k+1}{k}\le 1+\frac{n}{k^2-1}\le\left(1+\frac{1}{k^2-1}\right)^n=\left(\frac{k^2}{k^2-1}\right)^n.$$  A little multiplication and we arrive at $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.  We may now first apply this to see that $\left(\frac{n-1}{n}\right)^n\le\left(\frac{n}{n+1}\right)^{n+1}$. Then we suppose that for some integer $2\le k\le n$, we have $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$. Then:  $$\left(\frac{k-1}{n}\right)^n=\left(\frac{k}{n}\right)^n\left(\frac{k-1}{k}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}\left(\frac{k}{k+1}\right)^{n+1}=\left(\frac{k}{n+1}\right)^{n+1}.$$  By backwards (finite) induction from $n$, we have that $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$ for all integers $1\le k\le n$, so:  $$a_n=\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\cdots+\left(\frac{n-1}{n}\right)^n\le\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}&lt;\left(\frac{1}{n+1}\right)^{n+1}+\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}=a_{n+1}.$$  (In fact, this proves that $a_n$ is strictly increasing.) By the monotone convergence theorem, $a_n$ converges.  I should note that I am not especially well-practiced in proving these sorts of inequalities, so I may have given a significantly more complicated proof than necessary. If this is the case, feel free to explain in a comment or in your answer. I'd love to get a better grip on these inequalities in addition to finding out what the limit is.  Thanks! Are the rationals a closed or open set in $\mathbb{R}$? My Ubuntu 12.04 was normally updated through the Update Manager, something about gtk. Then I rebooted and now all that shows after the splash screen is a blue screen that only shows the cursor. I'm reading Galois Theory by Steven H. Weintraub (second edition), and finding that I'm at least somewhat short on the prerequisites. However the following proof looks wrong to me - am I misunderstanding something, or is it actually an incorrect proof?  Lemma 2.2.3. Let $F$ be a field and $R$ an integral domain that is a finite-dimensional $F$-vector space. Then $R$ is a field. Proof. We need to show that any nonzero $r \in R$ has an inverse. Consider $\{1, r, r^2, \cdots\}$. This is an infinite set of elements of $R$, and by hypothesis $R$ is finite dimensional as an $F$-vector space, so this set is linearly dependent. Hence $\sum_{i=0}^n{c_i r^i} = 0$ for some $n$ and some $c_i \in F$ not all zero.  It then goes on to show, given the above, that we can derive an inverse for $r$. However, if I consider examples like $r = 2 \in Q[\sqrt{2}]$, $r = \sqrt{2} \in Q[\sqrt{2}]$ or $r = 2 \in Q[X]/{&lt;X^2&gt;}$, the set $\{1, r, r^2, ...\}$ doesn't look linearly dependent to me. I do believe the lemma is true (and might even be able to prove it), but this does not look like a correct proof to me. Am I missing something? [Edit] Well yes, I am. Somehow I had managed to discount the possibility of any $c_i$ being negative, despite repeatedly looking at each fragment of the quoted text in an attempt to find what I might be misunderstanding. How to calculate the difference between two dates using PHP? Look at  edit. I didn't really expect that I would able to suggest adding  tag, but I could!  How is it possible? (seems that usual users can add mod-only tags without moderator help!)  Edit  Oded removed  back :-( Is Ubuntu compatible with my hardware? Statements equivalent to the Axiom of Choice The Axiom of Choice reads:     The product of a collection of non-empty sets is non-empty.   As you know well, this axiom is equivalent to many other statements. A few examples (probably the most known) are the following:   Zorn's lemma: Let $(P,\ge)$ be a poset in which every chain has an upper bound. Then $P$ has a maximal element.   This is used in a number of different places to prove very powerful theorems. A couple of examples I can think of right now are: the proof that every ring has a maximal ideal (commutative algebra) and the  (functional analysis).   Well ordering theorem: Every set can be well-ordered (i.e. it admits a total order such that every subset has a least element). Tychonoff's theorem: Every product of compact topological spaces is compact.   This is used for example in the proof of  (again functional analysis).   gives various more. Among them:   Tarski's theorem: If $A$ is an infinite set, then there is a bijection $A\to A\times A$.   (As @Wicht dutifully commented, Tarski's theorem is the name given to the implication $|A|=A\times A \Rightarrow$ Axiom of Choice, while the other implication was known before the proof of this fact by Tarski.)   Every vector space has a basis.   My question is: what are other useful statements which are equivalent to the axiom of choice? Where are they used, and to prove what?  P.s.: By "useful statements" I mean either statements which are important theorems by themselves (as Tychonoff's theorem) or that are directly used in (simple versions of) the proof of important theorems (as Zorn's lemma). If Durmstrang and Beauxbatons were hidden from common knowledge, how did people from other schools go to them for the Triwizard tournament? So in GoF, they state that Beauxbatons and Durmstrang are hidden from view and don't have their locations public.     “Durmstrang’s another wizarding school?” said Harry      “Yes,” said Hermione sniffily, “and it’s got a horrible reputation.      According to An Appraisal of Magical Education in Europe, it puts a   lot of emphasis on the Dark Arts.”      “I think I’ve heard of it,” said Ron vaguely. “Where is it? What   country?”      “Well, nobody knows, do they?” said Hermione, raising her   eyebrows.      “Er — why not?” said Harry.      “There’s traditionally been a lot of rivalry between all the magic   schools. Durmstrang and Beauxbatons like to conceal their whereabouts   so nobody can steal their secrets,” said Hermione matter-offactly.   So how do students get to the other school for the Triwizard tournament? Analysis Question Involving Real Numbers Prove that sequence $ x_{(n+1)}= \frac {a}{1+x_n}$ is convergent to positive root of $x^2+x-a=0,$where $a >0$ and $x_1 >0$ Prove that sequence $$ x_{(n+1)}= \frac {a}{1+x_n}$$ is convergent to positive root of $x^2+x-a=0,$where $a &gt;0$ and $x_1 &gt;0$  we have $$x_{(n+2)}-x_n=\frac{-a(x_{(n+1)}-x_{(n-1)})}{(1+x_{(n-1)})(1+x_{(n+1)})}$$  Hence from this we see behaviour of even terms subsequence and odd terms subsequence  is opposite. how to proceed from there. any hint please Auto mount a disk drive in ubuntu 14.04 LTS (Bookmark directory not found) I'm using ubuntu 14.04 LTS. When I start my pc and click on any bookmark from the file manager it shows this error:   Unable to find the requested file. Please check the spelling and try again. Unhandled error message: Error when getting information for file '/media/angel/DOC/edu': No such file or directory   But when I open DOC drive and again clink on the same bookmark it works properly. Can anybody tell me why is this happening and how to solve this issue?   Edit: Output of lsblk is following: (ubuntu is installed in sda5)  NAME   MAJ:MIN RM   SIZE RO TYPE MOUNTPOINT sda      8:0    0 119.2G  0 disk  ├─sda1   8:1    0  91.3G  0 part  ├─sda2   8:2    0     1K  0 part  ├─sda5   8:5    0  24.1G  0 part / └─sda6   8:6    0   3.8G  0 part [SWAP] sdb      8:16   0 465.8G  0 disk  ├─sdb1   8:17   0  93.2G  0 part  ├─sdb2   8:18   0     1K  0 part  ├─sdb5   8:21   0  93.2G  0 part  ├─sdb6   8:22   0  89.3G  0 part  ├─sdb7   8:23   0  93.2G  0 part  ├─sdb8   8:24   0  93.2G  0 part  └─sdb9   8:25   0   3.8G  0 part  sr0     11:0    1  1024M  0 rom     After opening the disk drive the output of lsblk is bellow:   NAME   MAJ:MIN RM   SIZE RO TYPE MOUNTPOINT sda      8:0    0 119.2G  0 disk  ├─sda1   8:1    0  91.3G  0 part  ├─sda2   8:2    0     1K  0 part  ├─sda5   8:5    0  24.1G  0 part / └─sda6   8:6    0   3.8G  0 part [SWAP] sdb      8:16   0 465.8G  0 disk  ├─sdb1   8:17   0  93.2G  0 part  ├─sdb2   8:18   0     1K  0 part  ├─sdb5   8:21   0  93.2G  0 part  ├─sdb6   8:22   0  89.3G  0 part /media/angel/DOC ├─sdb7   8:23   0  93.2G  0 part  ├─sdb8   8:24   0  93.2G  0 part  └─sdb9   8:25   0   3.8G  0 part  sr0     11:0    1  1024M  0 rom   The output of sudo blkid gives the following information about sdb6:  /dev/sdb6: LABEL="DOC" UUID="04280A22280A12FA" TYPE="ntfs" How do I prompt for Yes/No/Cancel input in a Linux shell script? I want to pause input in a shell script, and prompt the user for choices. The standard Yes, No, or Cancel type question. How do I accomplish this in a typical bash prompt? Importing coordinate data from Excel into QGIS I have an Excel file containing data collected over many years as latitude and longitude coordinates for archaeological sites. How can I import it into QGIS as a new layer? Show that $\lim _{r \to 0} \|T_rf−f\|_{L_p} =0.$ I am having a hard time with the following real analysis qual problem.  Any help would be awesome.  Suppose that $f \in L^p(\mathbb{R})$, where $1\leq p&lt; + \infty$. Let $T_r(f)(t)=f(t−r)$. Show that $\lim_{r \to 0} \|T_rf−f\|_{L_p} =0$, that is $$ \lim_{r\to 0} \left( \int_{\mathbb R} |f(t-r) - f(t)|^p \mathrm d t\right)^{1/p} =0.$$ When I dip a paper towel in a cup of water the water gets drawn up due to . How is this reconciled with conservation of energy, as it would seem on the surface that the potential energy of the system has gone up with no work being done?