$||x|-|y|| \leq |x-y|$  when $(x,y \in R^k)$  In Principles of MA(Rudin), the author said one sees easily that      $||x|-|y|| \leq |x-y|$  when $(x,y \in R^k)$ (p.88, Rudin)   from the triangle inequality. But I'm not sure how to use the triangle inequality to show this. Can you help me show this?
I'm trying to show that $\lvert \lVert x \rVert - \lVert y \rVert \rvert \overset{\heartsuit}{\leq} \lVert x-y \rVert$. A hint would be nice.
Which side of wave-particle duality to choose in a given situation Why is quadratic integer ring defined in that way? Quadratic integer ring $\mathcal{O}$ is defined by \begin{equation}    \mathcal{O}=\begin{cases}       \mathbb{Z}[\sqrt{D}] &amp; \text{if}\ D=2,3\ \pmod 4\\       \mathbb{Z}\left[\frac{1+\sqrt{D}}{2}\right]\  &amp; \text{if}\ D=1\pmod 4    \end{cases} \end{equation} where $D$ is square-free.   I understand $\mathbb{Z}\left[\frac{1+\sqrt{D}}{2}\right]$ is not closed under multiplication if $D=2,3\pmod 4$. But still, isn't it more natural to define $\mathcal{O}=\mathbb{Z}[\sqrt{D}]$ for all square-free $D$? (in that case, it really seems like 'quadratic integer') I wonder what is the motivation of this definition. What is the black thing on the circuit What is the black thing on the below circuit?   Is it used to hide the IC (or other components) that is used on the circuit?   What material is it? Can I get cheaper plane tickets by using a VPN? In a Youtube video by ADVChina, the speaker is recommending using a VPN in China for various reasons, and to sign up using his referral code.  He says there are various practical uses including "You could also go and get cheaper plane tickets by hopping around different countries." and shows a screenshot of a flight website called KAYAK, with the VPN software changing his country to Italy.  Is it possible to get cheaper plane tickets by using a VPN in this way? I've never heard of KAYAK, but if I use something like Skyscanner, is it common for tickets to be priced differently depending on which country the user's IP is from?  Is it because of different taxes in different countries? So could it be illegal to get discounts like this? Or is there some pricing strategy from airlines to charge different amounts depending on the country of purchase? Proof relation between Levi-Civita symbol and Kronecker deltas in Group Theory In order to prove the following identity: $$\sum_{k}\epsilon_{ijk}\epsilon_{lmk}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$$ Instead of checking this by brute force, Landau writes thr product of Levi-Civita symbols as: $$\epsilon_{ijk}\epsilon_{lmn}=\det\left|    \begin{array}{cccc}       \delta_{il}   &amp; \delta_{im}  &amp; \delta_{in}   \\       \delta_{jl}   &amp; \delta_{jm}  &amp; \delta_{jn}   \\       \delta_{kl}   &amp; \delta_{km}  &amp; \delta_{kn}       \end{array} \right| $$  The proof that the equalty holds is quite straightforward if you consider what values the indices can take. But I've been told that there's a much more profound and elegant demonstration based on the representation of the symmetric group.  Does anybody know this approach based on group theory? Only half of the display works, can I set to display only in that region? I have a Macbook with a cracked screen, only the top left half of the screen works, is there anyway to reduce the screen resolution, keep the DPI the same, and thus have the entire OS visible in the top left section of the screen which works?  The below photo demonstrates what I mean, with parts of the creen totally unusable, it would make sense to be able to display the whole screen in the remaining usable part. Get a resource using getResource() We were asked to prove the following:  Let $ A $ be an $n \times n$ matrix with $n$ distinct real eigenvalues. If $AB=BA$, show that $B$ is diagonalizable.   It was suggested I show that an eigenvector of $A$ is also an eigenvector of $B$. I am both having trouble doing this and failing to see how I would complete the proof after. Any help would be appreciated.  Thanks Proof for the funky trace derivative : $d (\operatorname{trace} (ABA'C))$? Given three matrices $A$, $B$ and $C$ such that $ABA^T C$ is a square matrix, the derivative of the trace with respect to $A$ is:  $$ \nabla_A \operatorname{trace}( ABA^{T}C )  = CAB + C^T AB^T $$  There is a proof . However, I was not able to understand the second line of the proof. it says something like this:  $$ \nabla_A \operatorname{trace}( ABA^T C )= \nabla_\bullet \operatorname{trace}(f(\bullet)A^{T}C) + \nabla_\bullet \operatorname{trace}(f(A) \bullet^T C) $$  Where  $f(A) = AB$  I tried searching the proof or any hint using google, but I couldn't fine anything. Could anyone please help me with this second line. I cannot sleep. I am obsessed with this proof. The Challenge  I have been trying to write a filter that applies (prepends a string) to (a) post titles of a certain post type and (b) to that same post type WordPress admin backend for use by an auto social sharing plugin.  I've gotten really close.  I've achieved: 1.  prepending only to custom post_type post titles 2.  prepending to custom_post type titles AND all nav menu items 3.  prepending to ALL titles (post, page, nav menu, and backend auto-sharing plugin)  The final, 3rd attempt got me very hopeful, but yet still, no matter the function, conditional statement, or combination I use, I can't get custom_post title AND backend social sharing plugin, but not nav menu items.  My Code - what I've tried  I've tried different conditionals, like is_admin() and !is_nav_menu with all nav menu IDs.  I've also tried comparing the ID of the post to the ID of the menu.  Finally, I've also tried adding the filter only to loop_start and that seems promising if combined with other statements.  I will clean up my code for your review but I leave it this way for now in hopes it helps to see what I've tried and maybe where I went wrong with any of those methods.  // , $id = NULL function append_album_review_to_title( $title ) {     global $post;     global $dont_apply_title_filter;     $text = 'Album Review: ';     $prepended_title = $text . $title; /*     $nav_menus_obj = wp_get_nav_menus();     $nav_menu_ids = '';     foreach( $nav_menus_obj as $nav_menu ) {         $nav_menu_ids .= $nav_menu-&gt;term_id . ',';     }     rtrim($nav_menu_ids, ','); */ //  if ( get_post_type( $post-&gt;ID ) == 'album_review' &amp;&amp; in_the_loop() ){ //  if ( get_post_type( $post-&gt;ID ) == 'album_review' &amp;&amp; $id == get_the_ID() ){ //  if ( !is_nav_menu( '32057,32058,35135,32054,32056' ) ) { //  if ( get_post_type( $post-&gt;ID ) == 'album_review' &amp;&amp; is_nav_menu( $id ) ) {     if ( get_post_type( $post-&gt;ID ) == 'album_review' &amp;&amp; !$dont_apply_title_filter ) {         //print_r( $nav_menu_ids );         //print_r( is_nav_menu( $nav_menu_ids ) );         return $prepended_title; /*     } elseif ( get_post_type( $post-&gt;ID ) == 'album_review'  ) {         return $title; */     } else {         return $title;     }; } add_filter('the_title', 'append_album_review_to_title'); //add_action('save_post', 'custom_post_type_title', 100); /* function set_custom_title() {     add_filter( 'the_title', 'append_album_review_to_title', 10, 2 ); } add_action( 'loop_start', 'set_custom_title' ); */ What do you call someone who solves puzzles? What is a term or name for someone who is very adept at solving puzzles or situations that require though processing and logic.   I ran across , however this only deals with crossword puzzles. I'm looking for something that could be applied on a broader sense. How do I prove that a graph is a topological manifold I'm having problems with that:  Prove that M is a topological manifold.  $f:U \to \mathbb R^k$, U $\subset \mathbb R^{n}$ open, f continuous  M = $\{\left(x,y\right)\in \mathbb R^{n+k} \mid x \in U, y=f(x) \}$  Can anyone help me please? I have a sample composed by 2500 count data values. I've plotted in R the corresponding histogram and ecdf. I've run the One-Sample Kolmogorov-Smirnov test to check if the distribution is either exponential or poisson, but I got negative results.  Is it possible that the One-Sample Kolmogorov-Smirnov test, it's not the correct test to work with this count data.  What is the best way to identify a distribution for this kind of data? Find all $n\in\mathbb N^+$ such that the sum of the digits of $5^n$ is equal to $2^n$.   I've solved this, but I think the proof is a bit weird and I wonder if there's a better one.    My proof:  We can see that $n\neq 1$, $n\neq 2$, $n=3$, $n\neq 4$ and $n\neq 5$.  Let $n&gt;5$. Notice that $5^6$ has $5$ digits, while $\frac{2^6}{9}=7\frac{1}{9}$, so $5^6$ has to have at least $8$ digits in order for the sum of the digits to be able to at least reach $2^6$.  Everytime $1$ is added to the exponent of $5$, the amount of digits $5^n$ gets is at most $1$, while $\frac{2^{n+1}}{9}-\frac{2^n}{9}=\frac{2^n}{9}&gt;3$, so the lowest amount of digits it has to have to reach $2^n$ becomes higher by at least $4$.  When $n=6$, the amount of digits $5^n$ has is already lower than the amount it has to have in order to reach $2^n$, and I've just proved that the amount it has to have in order to reach $2^n$ grows faster than the amount of digits $5^n$ has, therefore $5^n$ will never be able to reach $2^n$ when $n&gt;5$.  Hence the only solution is $n=3$. Is there a CSS selector for elements containing certain text? I applied to a program that I had been striving for in my undergrad, at a top school, but I've since felt overwhelmed with the thought of going this year. I think I should have waited until I was absolutely sure I could handle grad school emotionally and financially. I am so anxious now, thinking I've ruined my opportunity with this school if I don't go.   I am seriously considering withdrawing my application, but I'm very worried about how it might hurt me in the future when I reapply.  Otherwise I may have to decline if I am accepted and reapply. The story:   Question posted on Programmers (let's call it ), Same question posted on Code Review (let's call it ), I migrate Q1 to Code Review, Someone notices that Q1 is a duplicate of Q2 and votes to close, I notice the duplicate notice and the auto-generated comment and also try to vote to close, but can't. No error message, nothing, it just doesn't work.   And then I notice that Q2 no longer exists ("voluntarily removed by its author"). My guess is the OP also noticed the duplicate vote, realized how messy cross posts are and decided to be a good citizen and delete one of the two questions. He picked the wrong one though, and now we're left with a duplicate notice pointing to a non existent question and an unresponsive dialog.  It's a mess ;) A weird and rare mess, but still a mess. What can we do about it? How to include python script inside a bash script Finder: keep sort-by per-directory Some directories make sense to be sorted by date modified, while others make more sense to be kept sorted alphabetically.   How can I make it so that finder will apply a specified sort by selection on a per-directory basis?   If no such feature exists (In which case I hope somebody at APPL is paying attention), I suppose using/remapping the keyboard shortcuts (Defaults being Cmd+Ctrl+5 for Modified and Cmd+Ctrl+1 for Alphabetical by Name) is the next best thing. How many Avatars have there been?