Show that if $n^2$ is even then $n$ is even, for $n$ an integer.
Proving $a^2$ is even $\implies$ $a$ is even by contradiction
Some images shown outside of preview rectange Some images shown outside of preview rectange before uploading, eg: Algorithm for offsetting a latitude/longitude by some amount of meters Need to understand parallel query execution error Today we experienced a degradation in performance on our production sql server.  Durring the time this occurred we logged several "The query processor could not start the necessary thread resources for parallel query execution" errors.  The reading that I've done suggests that this has to do with how many CPUs to use when executing a complex query.  However when I checked during the outage our CPU Utilization was only at 7%.  Is there something else this could be referring too that I haven't come across yet?  Is this a likely culprit of the performance degradation or am I chasing a red herring?  My sp_configure values for this are as follows:  name                                minimum maximum config_value run_value cost threshold for parallelism      0       32767   5            5 Python Flask shutdown event handler Moderncv adding date of birth to personal information Redshifting of Light and the expansion of the universe So I have learned in class that light can get red-shifted as it travels through space. As I understand it, space itself expands and stretches out the wavelength of the light. This results in the light having a lower frequency which equates to lowering its energy.  My question is, where does the energy of the light go? Energy must go somewhere!   Does the energy the light had before go into the mechanism that's expanding the space? I'm imagining that light is being stretched out when its being red-shifted. So would this mean that the energy is still there and that it is just spread out over more space? Kinetic energy is simple, right?! from the equation it seems as though if you know the speed and mass you'll know the work. Let's try to prove it. let's assume we know the acceleration and distance for now. $$w=mda$$ So the distance is $$d= 1/2 *at^2$$ Then  $$t = \sqrt{2d/a}$$ So $a*t$, which is the speed, is  $$a*\sqrt{2d/a} = a* \sqrt{2d}/\sqrt{a} = \sqrt{2da} = \sqrt{2w/m}$$ So regardless of the distance or acceleration, if you know the work and the mass, you can calculate the speed which is  $$v=\sqrt{2w/m}$$  What if you know the speed and the mass? You can easily calculate the work. $$v = \sqrt{2w/m}$$ Then  $$v^2 = 2w/m$$  That would mean  $$m*1/2*v^2=W $$ Rearrange it and  $$W= 1/2 *m *v^2$$ would be the equation.  What if you have an initial speed and a new speed and want calculate the amount of work done after an initial speed to get the new speed? You'll just subtract the total work from the initial speeds work.  $$1/2mv^2 - 1/2ms^2$$  where s is  the initial speed and v is the new speed. However if you do it the other way which is  $$d= 1/2at^2 + vt$$ and solve it the answer would be $$1/2ms^2  + 1/2msv$$ Why aren't they equal? I've tried to construct examples of rings that match all except one of the properties in the definition of a Dedekind domain. (This is an old number theory qual question from Berkeleys MGSA website).  The only starting point that I can think of would be  $R:=K[X_1,X_2,\ldots]$  for $K$ a field. This is clearly integrally closed as any polynomial relation is contained in $K(X_1,\ldots,X_n)$ for some large enough $n$ and $K[X_1,\ldots,X_n]$ is integrally closed being a UFD.  The problem is then to get rid of enough ideals from this ring $R$, so that its dimension would be 1, but I can't seem to figure out what to do.  Anyone have any other ideas for rings that could work as a starting point? The polynomial ring with an infinite number of variables is pretty much the only example that I know of for how to construct a non-noetherian ring that's still a domain. Do Chinese citizens require a visa to go to Hong Kong for less than 7 days? What do you call a person who does/thinks the opposite of the crowd for just that reason? Such a person does not hold in high regard the decisions taken by a group (not a team) of people. She believes that groups tend to be anarchic and don't serve their purpose.  An example would be - for outings with friends, this person will not like planning it with the group since they will mess it up.  Therefore, this person has convinced herself that when more people are involved with something, it will be worse. She distrusts crowds instinctively basically. Transaction log reported full but is not Is there a word that describes when you are unable to remember the word you want to say? Can the expression $\sqrt{n} + \sqrt{m}$ be rational if neither $n,m \in \mathbb{N}$ are perfect squares? It doesn't seem likely, the only way that could happen is if for example $\sqrt{m} = a-\sqrt{n}, \ \ a \in \mathbb{Q}$, which I don't think is possible, but how to show it? My dad's old TV! When I switch it off after watching and get my hands closer, my hands' hair gets attached to the screen. And they stick for sometimes! Now, I know how a cathode ray oscilloscope works. It is just the clever use of Hall - Effect . But why does my hair get stuck to the screen? What is the cause? Note, I checked this for my LED TV but it didn't work (I couldn't perceive anything) unknowingly. What's the magic in my old TV?? What exactly is the Box-Jenkins method for ARIMA processes? The  page says that Box-Jenkins is a method of fitting an ARIMA model to a time series. Now, if I want to fit an ARIMA model to a time series, I will open up SAS, call proc ARIMA, supply the parameters $p,d,q$ and SAS will give me AR and MA coefficients. Now, I can try different combinations of $p,d,q$ and SAS will give me a set of coefficients in each case. I select the set with the lowest Akaike information criterion.  My question is: where in the above procedure did I use Box-Jenkins? Am I supposed to use Box-Jenkins to come up with initial estimates of $p,d,q$? Or did SAS use it internally somehow? $ \tan 1^\circ \cdot \tan 2^\circ \cdot \tan 3^\circ \cdots \tan 89^\circ$ How can I find the following product using elementary trigonometry?     $$ \tan 1^\circ \cdot \tan 2^\circ \cdot \tan 3^\circ \cdots \tan 89^\circ.$$   I have tried using a substitution, but nothing has worked. Address is invalid on local machine [on windows 8 only] how can i forward requests to my own IP address to a local IP address in my network (possibly using named)? My setup  I have a linux server running in my basement. I currently have my router looking to that server for DNS lookup, so I can use named to spoof a bunch of domain names to point to my local server's local address, 192.168.0.111. This works great.  I also am running Apache and PHP on this server, and have forwarded port 80 on my router to this machine, so my public IP xx.xxx.x.xx is sent to the basement server.  This is a great development set up for me, because I can develop websites in my local network and pull them up on numerous devices with a *.mylocal.net address. I can also quickly share a website on my public IP with others that are outside of my network.  I have a static IP, so that makes this even easier.  The problem  Some applications need to have an absolute URL. My spoofed DNS doesn't work outside my network, and neither does my internal IP address.  I need to be able to access the sites at the same IP that they are accessed publicly, which is my public IP: xx.xxx.x.xx    For some reason, I can't access my own public IP address from within my own network.  My router's public IP is xx.xxx.x.xx, and when I hit xx.xxx.x.xx from outside my network I see exactly what I want -- my development website. But, when I hit xx.xxx.x.xx from inside my network, the connection times out.    I assume my ISP has a firewall that blocks requests from xx.xxx.x.xx to xx.xxx.x.xx where the origin and the destination are the same IP (or maybe this is impossible on any network?)  So how can I set up my network to forward the internal request to my public IP to the local IP of my basement server?   I would like to do this within the local DNS server I set up, or in my router, so the change can be applied to all devices in my network, and not just the basement server or the windows tablet/laptop, etc individually. Can spells with movable areas of effect be moved out of sight or behind obstacles? With spells like Moonbeam or Flaming Sphere, how should we treat the movement of their areas-of-effect when it comes to passing them behind a thick pillar or around a blind corner?  Should we treat any movement of them as if they are being 're-cast' (and so requiring a clear unobstructed view)? Or could a mage presumably will them somewhere out-of-sight? I am relatively fresh to the DI party and am struggling to get my head around how exactly to use Dependency Injection. I understand that I can pass requirements as part of a service, but what about in my own class?  Say I have the below class where I want the language manager service to be injected:  class Foo implements ContainerInjectionInterface {    protected $languageManager;     public function __construct(LanguageManagerInterface $languageManager) {     $this-&gt;languageManager = $languageManager;   }    public static function create(ContainerInterface $container) {     return new static(       $container-&gt;get('language_manager')     );   } }   When I call new Foo() I am required to pass an instance of LanguageManagerInterface. That makes sense in terms of a normal constructor method, but that would require me to instantiate the class like the below:  $languageManager = \Drupal::service('language_manager'); $foo = new Foo($languageManager)   That just doesn't feel right to me.  Is there something I am missing, or would this class always have to be a service to take advantage of dependency injection?