sorting in multi-dimensional array
How to Sort Multi-dimensional Array by Value?
I want to do something like String.Format("[{0}, {1}, {2}]", 1, 2, 3) which returns:  [1, 2, 3]   How do I do this in Python? Prove if $|a_{n+1}|/|a_n| \leq |b_{n+1}|/|{b_n}|$ and $\sum |b_n|$ is convergent, then $\sum |a_n|$ is convergent Axiom of Choice: Family of non-empty sets mutually disjoint or not? I have noticed that the family of non-empty sets referred to in statements of the Axiom of Choice is sometimes required to be mutually disjoint, and sometimes not. Why is that? What do you call a person who "shuts down" ideas? My primary router has trouble extending wifi coverage to some bedrooms in the house so I am planning to add a second router to extend the wifi network using the wired Ethernet ports in one of the areas of poor coverage. Basic layout -   The ISP's router is in my family room downstairs (fiber connection) and the internet connects to the fiber via a box in an upstairs closet.   I purchased and installed an unmanaged Ethernet switch/splitter. In the upstairs closet where all of the Ethernet wires meet, the family room Ethernet cable was plugged into the ISP's box. I disconnected this wire and replaced it with a CAT5 cable running from the ISP's box in the closet to the Ethernet switch's port 1.  I think connected the family room Ethernet cable to port 2 and did the same in order for all the remaining Ethernet cables to various bedrooms.   So far, the switch is working fine.  I have wifi working with internet coming off of the router in the living room so I believe the switch is installed correctly otherwise that wifi should not have internet coming off of the living room jack.   Here is where things go awry - I took my second wifi router and connected the WAN/Yellow port to the Ethernet wall jack in another bedroom.  I then connected a lap top to one of the Ethernet ports on this router and found "no internet" after connecting.  On a whim, I changed the connection so that the router was connected to the Ethernet wall jack in one of the white ports but with the same results.   I don't think this counts as cascading and I don't really need a new network since I do not share the connection with any non-family members.  The router has been reset to factory settings to clear out any old instructions - but I still can't get it to connect to the internet or access the router to change the settings even - it's a netgear and the routerlogin.net will not let me access the router even when I am on the working WIFI for internet and have a cable running from the laptop to the netgear router.   I could use some help - just trying to extend the wifi range and it seems like it should be much less complicated. Tried following instructions in several other posts and they did not work for me - because I cannot access the router to change any settings. Why are so many important verbs irregular? Every vector space has a basis Prove that every vector space has a basis. I am going to use Zorn's lemma for this also here is a necessary definition regarding totally ordered subsets: one element will be contained in the other.   proof: A basis for $V$ is a maximally linearly independent set. Let $X$ be equal to all linearly independent subsets of $V$. On, $X$ define $A\leq B$ means $A\subset B$. Let $S$ be a totally ordered subset of $X$, let $T = \bigcup_{A_{s}\in S} A_s$ then $A_s\subset T$ for all $s\in S$. Does $T\in S$? i.e. is $T$ a linearly independent set?. Let $$O = \alpha_1 v_1 + \alpha_2 v_2 + \ldots + \alpha_n v_n \ \ \text{where} \ \ v_i\in T = \bigcup_{A_{s}\in S} A_s$$ Then there exists $S_i$ such that for all $i\in A_{s_{i}}$ all $i$ $A_{s_{1}},A_{s_{2}},\ldots, A_{s_{n}}$ (these will have one contained in the other then we would take the larger). So, there exists $A_{s_{k}}$ such that $A_{s_{k}}\supset A_{s_{i}}$ all $i$ just compare two at a time. Therefore, $V_i \in A_{s_{k}}$ all $i$, but $A_{s_{k}}$ is a linearly independent set. So, $\alpha_i = 0$ all $i$. Hence, $T\in X$ so $T$ is an upper bound for $S$. Hence every totally ordered subset of $X$ has an upper bound, hence by Zorn's lemma, there exists a maximal element and this would be the basis.  I am not sure if my proof is right, I apologize if it is a bit messy. If it is right, any suggestions on cleaning it up would be greatly appreciated. Don't display pending edit counter if you've exhausted your suggested edit votes I'd like to know if there is a way to run  program/shell script without typing full path:  /opt/idea/bin/idea.sh Prove homotopic attaching maps give homotopy equivalent spaces by attaching a cell Prove: If $f,g:S^{n-1} \to X$ are homotopic maps, then $X\sqcup_fD^n$ and $X\sqcup_gD^n$ are homotopy equivalent.   I think it can be proved by showing they are both deformation retracts of $X\sqcup_H(D^n\times I)$ where $H$ is the homotopy between $f$ and $g$.  However, I have difficult in proving that the deformation retracts are continuous map. In fact, I have difficulty in representing a map in quotient spaces like $X\sqcup_fD^n$. I think a map from $X\sqcup_fD^n$ to $W$ can be represented by two maps: $m_1: X\to W$, $m_2: D^n\to W$, where for $x\in S^{n-1}$, $m_1\circ f(x)=m_2\circ i(x)$.  Then I construct the deformation retract this way: $m_1: X\to X$. For $x\in H(S^{n-1},t)$, $m_1(x)=H(S^{n-1},0)$, otherwise $m_1(x)=x$.  $m_2: D^n\times I\to D^n\times {0}$: $m_2((D^n,t))=(D^n,0)$.  It is easy to verify that $m_1$ and $m_2$ define a map from $X\sqcup_H(D^n\times I)$ to $X\sqcup_fD^n$. As long as this is a continous map, obvioulsy then we find a deformation retract. But it seems such a map is not continous? Does set difference distribute over set intersection? I am asked to prove that, if $A, B$ and $C$ are sets, then $$A-(B\cap C)=(A-B)\cap(A-C).$$  However, I think that either I have made an error in my working, or the wording of the problem contains a typographical error.    My working so far is as follows: $$\begin{align}x\in(A-(B\cap C)) &amp;\iff (x\in A)\wedge(x\not\in(B\cap C))\\ &amp;\iff(x\in A)\wedge((x\not\in B)\vee(x\not\in C))\\ &amp;\iff((x\in A)\wedge(x\not\in B))\vee((x\in A)\wedge(x\not\in C))\\ &amp;\iff(x\in(A-B))\vee(x\in(A-C))\\ &amp;\iff x\in((A-B)\cup(A-C)) \end{align}$$  First line to second line: by De Morgan's Law. Second line to third line: by distributivity of set intersection over set union. Third line to fourth line: by definitions of intersection and set difference.   If my working is correct, then I have shown that $$A-(B\cap C)=(A-B)\cup(A-C).$$ Am I correct? Where can I find gedit's backup files and how to open them? So, I just finished off the first part of an important task for school and today when I was going to continue, the document with my notes were gone. The only fragment I've got left of my file is the back up file, which doesn't show in the folder where I saved the document, but I can search for it.  Is there anyway which allows me to open the back up file, save it as a text file and continue my task? I don't really have the time for running restoring tools, which never has helped me in a satisfactory way anyway.  I wrote the document in gedit and my version of Ubuntu is 12.10. Is there a built-in checksum/hash utility on Windows 7? Capitalization for a bullet list Invite people who liked my first page to like another one in Facebook Usage of "shall we?" What does it mean and where would I use it? How can I get very vivid colors in night cityscape pictures? Why does the image not appear? I'm using IEEEtran to format an article, and I wanted to include an image. I included the graphicx and hyperref packages, and used the following code, like normal, to include an image.  \begin{figure}[!t] \centering \includegraphics[width=3in]{fetch} \caption{Pipeline: Fetch Instruction} \label{fig:fetch} \end{figure}   What appears, however, is this:    Fetch.png is a file in the directory, and if I remove it, pdflatex errors instead of producing an output. I'm using MiKTeX 2.9 on Windows 7. Any ideas? My wife wants a 4th baby, but I don't My wife wants a 4th baby, but I don't.  We have been talking about it for the last few days (talking strongly).   I have raised points such as:   We will get less time to spend with our current kids. The age gap between the first and the last will be too wide. It will cost more money. She will be out of work longer. We will need a bigger house. And we don't even have our own house yet (we are still renting). We will have to start again. Sleepless nights, nappies, feeding, new cot, car seat, etc etc.   She says:   She feels like she is missing something. She has always wanted a large family. Things will be the same regardless of having 3, or 4 kids. She agrees that money will be tighter, but wants to push through it.   Our current kids are 9 months, 2 years, and 4 years. We are young married couple (25), and I have a fairly decent job.  What would be the pros and cons of having another child?   How can we rationally talk to the other about having/or not having another child?  And what should we do if we cannot agree? How does the Comma Operator work