I have the name of a computer on the network, and I need to know how to get the ip address of said computer from a batch file? Thanks.
I need to know the ip address of another machine in a lan network.  What is the Windows command for getting ip address from machine name and vice versa?
Let A and B be sets. Show that A is a subset of B if and only if for any set C, one has A union C is a subset of B union C. Can you verify my proof if it is right?  Let $A$ and $B$ be sets.  (a) Show that $A$ is a subset of $B$ if and only if for any set $C$, one has $A$ union $C$ is a subset of $B$ union $C$. (b) Show that $A$ is a subset of $B$ if and only if for any set $C$, one has $A$ intersect $C$ is a subset of $B$ intersect $C$.  Edited Solution :  (a)  (=>) Given A is a subset of B, this implies A U B = B. Given a set C, A U B U C = B U C. Since A is a subset of B, this implies A U C is a subset of B U C.      Solution 1 :(&lt;=) Given A U C is a subset of B U C , For A U C , x is a member of A U C . For B U C, x is a member of B U C. Suppose x is a member of A but not a member of B, meaning A is not a subset of B. Therefore , A U C is not a subset of B U C. Hence, A must be a subset of B.      Solution 2 :(&lt;=) Given A U C is a subset of B U C, For A U C, x is a member of A U C and for B U C, x is a member of B U C. Suppose C is a disjoint set. Since A U C is a subset of B U C, it follows that A is a subset of B.  (b) (=>) Since A is a subset of B , it follows that A U B = B. [A intersect C] is a subset of A and  [B intersect C ] is a subset of B by definition of intersection. Let x be member of A and member of C. Since [ A intersect C ] is a subset of A, x must be in A. Let x be a member of B and member of C. Since [ B intersect C ] is a subset of B, x must be in B. Therefore, A intersect C is a subset of B intersect C.  (&lt;=) Since A intersect C is a subset of B intersect C, suppose we let Set C be a subset of B and a subset of A as well. C is a subset of A implies A intersect C is equals to C. C is a subset of B implies B intersect C is equals to C as well. Since A intersect C is equals to B intersect C, A is a subset of B. Bypass BIOS password set by faulty Toshiba firmware on Satellite A55-S1065 laptop? How to tell if a  tag failed to load I'm dynamically adding &lt;script&gt; tags to a page's &lt;head&gt;, and I'd like to be able to tell whether the loading failed in some way -- a 404, a script error in the loaded script, whatever.  In Firefox, this works:  var script_tag = document.createElement('script'); script_tag.setAttribute('type', 'text/javascript'); script_tag.setAttribute('src', 'http://fail.org/nonexistant.js'); script_tag.onerror = function() { alert("Loading failed!"); } document.getElementsByTagName('head')[0].appendChild(script_tag);   However, this doesn't work in IE or Safari.  Does anyone know of a way to make this work in browsers other than Firefox?  (I don't think a solution that requires placing special code within the .js files is a good one. It's inelegant and inflexible.) Cancellation Law for Direct Sums - What is wrong with this argument? Notation for intervals I have frequently encountered both $\langle a,b \rangle$ and $[a,b]$ as notation for closed . I have mostly encountered $(a,b)$ for open intervals, but I have also seen $]a,b[$. I recall someone calling the notation with $[a,b]$ and $]a,b[$ as French notation.   What are the origins of the two notations?  Is the name French notation correct? Are they used frequently in France? Or were they perhaps prevalent in French mathematical community at some point? (In this  Bourbaki is mentioned in connection with the notation $]a,b[$.)      Since several answerers have already mentioned that they have never seen $\langle a,b \rangle$ to be used for closed intervals, I have tried to look for some occurrences for this. The best I can come up with is the article on , where these notations are called Czech notation and French notation. Using $(a,b)$ and $[a,b]$ is called English notation in that article. (I am from Central Europe, too, so it is perhaps not that surprising that I have seen this notation in lectures.) I also tried to google for  or . Surprisingly, this lead me to a  where this notation is used for open interval. Bootable USB of Ubuntu Minimal CD I want to burn Ubuntu Minimal CD ISO to an old, 1 GB USB thumbdrive. Regular ISO doesn't fit. I already tried Startup Disk Creator and Unetbootin but both failed. I tried installing GRUB and pointing it to boot like this:  insmod part_msdos linux /linux initrd /initrd.gz   but it doesn't work. I also tried dd but I ended up with not bootable drive.  Rufus worked on Windows with the same thumbdrive but I no longer use Windows.  What else can I do? Installing Ubuntu after Windows 10 How can I install Ubuntu on Windows 10 as dual boot? I have a DVD with Ubuntu and when I tried to install Ubuntu I had a warning that the hard disk will be erased. How can I install Ubuntu and keep my Windows 10? Thanks Remove GRUB and keep a Linux system On a crop sensor camera, e.g. Canon EOS 7D, what would be the best kit to travel with around Europe for a few weeks? My shooting style will focus on landscape, architecture, and general travel shots (of note would be my lack of interest for portraits).     Limit to around 1-3 lenses, and I prefer light ones.   Crop sensor camera chosen to save on weight over a full frame option. Question is open to all brands for general guidance, but I will be using a Canon body.   Cost is not a factor, I will rent the lenses I do not already own.   Quality is of paramount importance. I will not have a tripod.   I understand some may consider this subjective, but I do believe that for a crop sensor body, with weight as a consideration, and a limit to the amount of lenses, a true best kit can be determined. Following an online tutorial that uses an object named "icosahedron" from Add > Mesh > Regular Solids > Archimedean > Icosahedron. This object is not available in my version of Blender (2.74). Presenter asks me to include an add-on named "Add Mesh: Regular Solids", which again is not listed in my User Preferences dialog. What can I do about it? Asymptote: when white isn't white I'm trying to display a white 3D ball with Asymptote (3D PDF), but it always turns up gray:    import graph3; import solids;  defaultrender.merge=true; size(10cm,0);  currentprojection=orthographic(-Z);  //currentlight=Headlamp; //currentlight=light(-10,1,1); //currentlight=White; currentlight=Viewport;  draw(unitsphere,rgb(1,1,1));   I understand it's a lighting issue, and I've tried playing with its settings (the lines commented out in my source are some of my attempts), but I never manage to have white be white. If I understand correctly, I need a whiter diffuse component in my light, but I cannot see how to achieve that. And the documentation is not very clear…    This will be part of a molecular model, and the rendering I'm going for it something like this: I am currently wondering about the famous .  Where does it come from mathematically that when you point with your thumb in the direction of the current, your curved fingers will point in the direction of the $B$-field? In other words, which mathematical operation does it come from? Probably it has to do with some vector product or Stokes theorem application, but I am not quite sure about it. how to disable system apps without root? How to disable system apps without root?  My phone model is vivo x5 max+ and android 4.4.4  I have tried Debloater but it failed to connect (I am sure I've allowed USB debugging). Prove that at most $2k+1$ independent sets form the set of vertices of a graph, where $k$ is the maximum degree. Is it possible to have test error lower than training error?  I have a classification problem with 2000 samples, 500 of which are positives, 1500 are negatives. I split my data into 70% training data, 30% test data.  Run random forest with 200 estimators and cv=10. I did this several times and compared the recall and precision score and notice the scores for my test set is significantly better. Is this possible? Remove suggested apps option in iOS 10 As far as I can tell, suggested apps show themselves based on location.  There are two places where these suggested apps show themselves to me:   At the bottom left of the multitask screen when I double press the home button At the bottom left of the lock screen    article describes how to get rid of the small app icon appearing in the bottom left. However, this method is for iOS 8, and I am now on iOS 10, where I am unable to find iTunes and App Store > Suggested Apps.   How can I get rid of these (in my opinion) annoying little things? I know enough amount of set theory and logic to study grad-level math. However, I don't know more advanced set theory and logic, such as the ones on Kunen's or Shoenfield's texts. Although it's good to learn such advanced topics, I may not have enough time to do so, since I will probably put my emphasis on algebra, analysis or geometry as well as mathematical physics.   Although how much set theory and logic I should learn is totally up to what sort of analysis, algebra or geometry I will study, I want to know how much of them typical competitive algebraists/analysts/geometers know. If you are one of them, how much do you know? Can Blender render pngs with the background transparent? I would like to render an animation as a series of .png images with transparent backgrounds for use in a video. Is this possible? Why do we assume $a+b=b+a$ in a Ring with 1. Also, is it true with Rings without 1? I was wondering why do some people use redundant axioms in definitions?  If you just expand $(a+1)(b+1)=(a+1)b+a+1=ab+b+a+1$ $(a+1)(b+1)=a(b+1)+b+1=ab+a+b+1$. Hence, $ab+a+b+1=ab+b+a+1$, then cancel ab and 1. Then, you get it's commutative for free.   Why do we then assume it?   Also, is it even stronger. If we drop the condition that 1 is in R, then can we still deduce a+b=b+a? As clearly the argument just done falls apart if you haven't got 1 in R. Can there be two distinct, continuous functions that are equal at all rationals? Akhil showed that the  is the same as the continuum, using as a step the observation that continuous functions that agree at rational points must agree everywhere, since the rationals are dense in the reals.  This isn't an obvious step, so why is it true?