Consider this randomly chosen question opened in chrome, and later in the android app.        In Chrome, I can press and hold on any link to see the actual address before I even visit it. Whereas in the stack exchange android app, I'm not able to do so.        This is not an important issue, but I'd simply love to have such a preview feature on our app.
I think it would be good to have a way for the user to view the target URL of a link before following it. I'm pretty sure that I'm not the only one who dislikes blind-clicking -tapping links.   When I use SE in the mobile browser, I routinely long-press links to find out where they go before following them. I wish I could do the same thing in the app, with the same or a similar method.  (It's important that this won't conflict with text selection, since you can't long-press on links to select them now anyway.)
$A=\begin{pmatrix}0&amp;0&amp;1\\ 0&amp;1&amp;0\\ 1&amp;0&amp;0\end{pmatrix}$  How do I show that there are no solutions for $X^2=A$ in $M_3(R)$?  I tried $X^2=\begin{pmatrix}a_{11}&amp;a_{12}&amp;a_{13}\\ a_{21}&amp;a_{22}&amp;a_{23}\\ a_{31}&amp;a_{32}&amp;a_{33}\end{pmatrix}^2=A$, but I couldn't seem to find anything that shows there are no solutions. How should I approach this? Technically, NULL = NULL is False, by that logic no NULL is equal to any NULL and all NULLs are distinct. Shouldnt this imply that all NULLs are unique and a unique index should allow any number of NULLs? My "Meta" rep on Photo.SE differs from my main rep Hey guys I found one or two papers which use ridge regression (for basketball data). I was always told to standardize my variables if I ran a ridge regression, but I was simply told to do this because ridge was scale variant (ridge regression wasn't really part of our course, so our lecturer skimmed through it).  These papers I read didn't standardize their variables, which I found a bit surprising. They also ended up with large values of lambda (around the 2000-4000 level) via cross-validation, and I was told that this is due to not standardizing the variables.  How exactly does leaving the variable(s) unstandardised lead to high lambda values and also, what are the consequences of not standardizing the variables in general? Is it really such a big deal?  Any help is much appreciated. Custom ordering of duplicate targets is not necessarily honored on mobile apps Does Wasserstein distance require the source and target distributions to have the same mass? If we minimize the Wasserstein loss, $$W_1 (P_S, P_T) = \underset{\gamma \in \Pi}{\text{min }} \sum_{x,y} |x-y|\gamma(x,y)$$ which means we are looking for the coupling $\gamma(x,y)$ that minimizes the cost of moving mass from the source distribution $P_S$ to the target distribution $P_T$,  Do the masses of the source and target distributions have to be identical? If so or not, how are the masses of $P_T$ and $P_S$ accounted for mathematically in, or when minimizing, the shown formula, or what might be missing/overlooked by the Wasserstein approach overall due to the mass differences? How can I make an object follow a path exactly Least rank of a matrix Let $A$ be a $7\times 7$ matrix such that it has rank $3$ and $a$ be $7\times 1$ column vector.  Then least possible rank of $A+(a a^T)$ is? ($a^T$ is transpose of the vector)  Intuitively I think it's $2$ as $a a^T$ has rank $1$, and the most it could affect is one row of $A$. Oversized \underbrace's label causes unwanted spacing Consider this:  \left( x-1 \right) \underbrace{ \left( ? \right) }_{ \text{what is this?} }   The label is longer than the braced portion, which causes unwanted space between the parentheses.  Is there a way around that, so that the parentheses "touch" as they would normally?  Please note that I just started on LaTeX, and I'm using TexMaths inside LibreOffice Writer. why each left ideal of this is minimal? How to justify $\phi(n) \ge \sqrt{n}$ If $\phi(n)$ is the Euler-totient function, how can I show that $\phi(n) \ge \sqrt{n}$? C# Unsafe/Fixed Code Positive semidefinite cone is generated by all rank-$1$ matrices. How to make a GIS inventory? My office will be seeing a big change in its GIS section. This section has been operational since the 1980's and has a huge collection of GIS data (i.e., shapefiles, raster files, data, etc) but never been through any inventory. Now it will happen.   Is there any automated way to extract all the information about the GIS data (i.e., shapefile, arc-info coverage, layer file, *.mxd, gdb, raster file, and more) from a PC to an Excel file? The information may include date of creation, date last edited, folder or container name, etc. import Excel sheets by name Two finite abelian groups with the same number of elements of any order are isomorphic Suppose $G, H$ are finite abelian groups with the same number of elements of any given order. Show they are isomorphic.   Since finite abelian groups are isomorphic if and only if their Sylow subgroups are, we may restrict our attention to the case where $G, H$, are $p$-groups, but I can't quite make it past there... Looking at your network profile, there is no number anywhere in your entire profile that has a sum of all your reputation across the network except for the flair page, which isn't always accurate because it's cached, and isn't exact because it rounds. Yet, there's so much empty space on the profile page.  My simple idea is to list the user's total reputation and badge counts below their Gravatar image like on any normal profile page across the network (keeping the rule of only accounts with over 200 reputation). It would make it incredibly easy to find and I wouldn't have to go search for myself on Area 51 to figure it out. Roots of $\,ax^2+bx+c\,$ are irrational if $a,b,c$ are odd integers How would I show the following.  For all odd integers a,b, and c if z is a solution of $ax^2+bx+c=0$ the z is not rational.  How would I show this? This is what I did  Let a be an odd integer then for a integer k it has the form $a=2k+1$  then $b=2t+1$ for if c integer.  then $c=2r+1$ if s is an integer then I do  $(2k+1)(x)^2+2t+1(x)+(2r+1)=0$  But I find my self stuck on what to do any hint are appreciated. Could there be an eighth Horcrux? Simple way of finding the locus of $z$? I am tasked to sketch the locus of complex numbers $z$ that satisfies $$(z-2+i)^2 + (z^*-2-i)^2 + 2zz^* + 6 = 0$$ I was wondering if there’s a special trick to this question, or if this is only solvable by plugging in $z=x+iy$ ? I noticed that the second term is the conjugate of the first, and wanted to substitute $w = z-2+i$, however this gives me a complicated expression for $zz^*$ which makes it more confusing. Is there a trick I’m missing out? Thanks! Edit: I have already solved it through direct substitution (parabola intersecting real axis at $x=3$ and $x=-1$) but it took a good 5 minutes or so, expanding everything correctly, which seems rather long (I was attempting it under exam conditions). I was just wondering if it is possible to exploit the $w$ substitution I made in particular.