I am into finding a convincing formula, or at least a rule of thumb, about when to use "to" and when to use "for" after adjectives - such as easy, difficult, important, interesting, and useful - and before personal pronouns and nouns.  I have gone through numerous and different advice by English-speaking people on the "to vs. for" problem for non-English speakers.  There were some cases in which they seemed interchangeable while there were others where they did not. Many people said the issue could be more about a preference or tendency than a rule, thus leaving English learners no choice but to learn them case by case. I came across some advisers who came up with a few rules, some of which I found very helpful while others seemed to contradict one another.   In the following examples, "to" and "for" seem to be interchangeable, and yet slightly different in meaning.   I reason that a marked difference between "to me" and "for me" would be found in whether "me", the narrator, has experienced it, the subject of the sentence, or not.   (A) English is not easy "to me".   This implies that English is not easy "in my view (in the view of the narrator)".  (B) English is not easy "for me".   This implies that English is not easy "in my experience (in the experience of the narrator)".  How does my reasoning sound to you? Any comments will be greatly appreciated.
Which one should be used?      To me, it makes no difference, but I'm not really sure why.   vs     For me, it makes no difference, but I'm not really sure why.
Does there exist an automorphism of $\mathbb{C}$ that does not keep $\mathbb{R}$ fixed? Showing group with $p^2$ elements is Abelian I have a group $G$ with $p^2$ elements, where $p$ is a prime number. Some (potentially) useful preliminary information I have is that there are exactly $p+1$ subgroups with $p$ elements, and with that I was able to show $G$ has a normal subgroup $N$ with $p$ elements.   My problem is showing that $G$ is abelian, and I would be glad if someone could show me how.   I had two potential approaches in mind and I would prefer if one of these were used (especially the second one).   First: The center $Z(G)$ is a normal subgroup of $G$ so by Langrange's theorem, if $Z(G)$ has anything other than the identity, it's size is either $p$ or $p^2$. If $p^2$ then $Z(G)=G$ and we are done. If $Z(G)=p$ then the quotient group of $G$ factored out by $Z(G)$ has $p$ elements, so it is cylic and I can prove from there that this implies $G$ is abelian. So can we show theres something other than the identity in the center of $G$?  Second: I list out the elements of some other subgroup $H$ with $p$ elements such that the intersection of $H$ and $N$ is only the identity (if any more, due to prime order the intersected elements would generate the entire subgroups). Let $N$ be generated by $a$ and $H$ be generated by $b$. We can show $NK= G$, i.e every element in G can be wrriten like $a^k b^l $. So for this method, we just need to show $ab=ba$ (remember, these are not general elements in the set, but the generators of $N$ and $H$).   Do any of these methods seem viable? I understand one can give very strong theorems using Sylow theorems and related facts, but I am looking for an elementary solution (no Sylow theorems, facts about p-groups, centrailzers) but definitions of centres and normalizers is fine. I was querying data from  using a LINQ query in LINQPAD and found that the data I got back is quite obsolete. I searched for Shekhar-Pro (me) by using a where clause on ID 399722 and the data returned was for 8 Jan 2011. Why was the data so old? Isn't the data updated in real time, or at least frequently? If not, where can I get realtime data? Every six months a new Ubuntu version will be released. Does anyone know at which time this will be? Why don't all Jedi stick around as force ghosts? In many of the Star Wars movies, Jedi come back as Force ghosts.    Why don't they stick around forever, giving advice to new generations, as well as countless knowledge from thousands of years? Except when we use right to denote direction, what is the difference between these two terms?  Also, which one is the preferred construction between these two     Am I right?   or     Am I correct? Can airlines deny boarding if my port of entry is different than the country which issued the Schengen visa? Few days back I did a Europe tour from India. As stated by many (I haven't confirmed official documents), I can apply for a visa from the country which is either my first port of entry in this case Paris or the longest stay which in this case was Italy.  Now following is what happened with me.   My flight was New Delhi to Paris via Istanbul with Turkish Airlines I Traveled from New Delhi to Istanbul (no problem here). When I was boarding to flight Istanbul to Paris, zhe airlines officials (the ones who checks the boarding pass before boarding the flight) told me that I could not board the flight because my visa was issued by Italy and I need to go to Italy First. They were denying the longest stay option of the visa. Moreover NONE of the staff of Istanbul airport were able to understand English so I had to go to a lot of problems before I managed to get tickets to Venice. I somehow managed to fly to Venice from Istanbul after almost 24 hours of struggle at Istanbul airport (without food and water).   So, my question is do the airlines staff have a right to deny flight boarding in this case? Neatest proof that set of finite subsets is countable? I am looking for a beautiful way of showing the following basic result in elementary set theory:     If $A$ is a countable set then the set of finite subsets of $A$ is   countable.   I proved it as follows but my proof is somewhat fugly so I was wondering if there is a neat way of showing it:  Let $|A| \le \aleph_0$. If $A$ is finite then $P(A)$ is finite and hence countable. If $|A| = \aleph_0$ then there is a bijection $A \to \omega$ so that we may assume that we are talking about finite subsets of $\omega$ from now on. Define a map $\varphi: [A]^{&lt;\aleph_0} \to (0,1) \cap \mathbb Q$ as $B \mapsto \sum_{n \in \omega} \frac{\chi_B (n)}{2^n}$. Then $\varphi$ is injective hence the claim follows. (The proof of which is what is the core-fugly part of the proof and I omit it.). A method to count occurrences in a list Use parameter expansion to modify output of another expansion I'm trying to use the Bash parameter expansions to modify the output of a command substitution or another parameter expansion. The following nested expansions work quite well in Zsh; but result in a &quot;bad substitution&quot; error in Bash: ${${PWD##*/}//trunk/latest}  or ${$(basename $PWD)//trunk/latest}  the output should be the last folder of the $PWD, replaced by latest when my current directory is trunk so /home/user/trunk should become latest Is there a Bash equivalent allowing to chain expansions without relying on variables or pipes? Or do Bash expansions only allows the input to be a string or a plain variable? Can Recovery Mode install OS X onto a blank SSD? I am trying to disable the touchscreen on my Asus S200E laptop.  In Ubuntu 16.04, I was able to do so with this:  Section "InputClass"     Identifier "evdev touchscreen catchall"     MatchIsTouchscreen "on"     MatchDevicePath "/dev/input/event*"     Driver "libinput" EndSection   How can this be done on Ubuntu 18.04? What would be the right position of "substantially" in the following:     1). Before verb: These optimal values substantially contribute to the success of the methodology.      2). After verb: These optimal values contribute substantially to the success of the methodology.   Here is my issue:  Fact1: When I use Google Ngram, I get , 0 occurrence of option 1.  Fact2: I found an excellent Grammar book that discusses Adverbs, their types and prefered positions. The author defines some rules that I'll summarize as follows:   Connecting adverbs, eg: however, then,..., Position: Beginning of clause. Indefinite frequency, eg: always, usually,..., Position: Mid position (after auxilary verb, before other verbs) sometimes beginning of clause. Focusing adverbs, eg: mainly, only,..., Position: Mid position. Adverb of certainty, eg: clearly, obviously, Position: Mid position. Adverb of completeness, eg: almost, nearly,..., Position: Mid position. Adverb of manner, eg: fast, slowly, nicely,..., Position: End clause, mid and beginning. Adverb of place, eg: upstairs, around, Position: End clause, sometimes beginning. Adverb of time, eg: daily, weekly, ... Position: Mostly end clause, sometimes beginning and mid position.   I think that "substantially" is an adverb of completeness. Therefore, the right position is before the verb "contribute" (option 1). However, Google Ngam indicates that option 1 is not widely used.  What do you think? Any suggestions? Latest NVIDIA driver on Ubuntu 16.04 Height of an HTML select box (dropdown) Why are processes in Windows 8 32 bit? I am looking for an example of a sequence which shows that, $a_{n+1} - a_n \to 0$ as $n \to \infty$ does not imply that sequence $a_n$ converges. I have a feeling that the sequence should be an oscillatory one but I am unable to think of an example. TikZ externalization with absolute path and MiKTeX distribution This code:  \documentclass[11pt,openright]{book} \usepackage[T1]{fontenc} \usepackage[latin9]{inputenc} \usepackage[british]{babel}  \usepackage{tikz} \usetikzlibrary{external} \tikzexternalize[prefix={"C:/Users/MyName MySurname/Documents/LaTeX/Figures/"}]  %\tikzexternalize[prefix={Figures/}]   \begin{document} \begin{tikzpicture}     \draw (0,0) circle (1cm); \end{tikzpicture} \end{document}   gives me the error:     ! I can't write on file `"C:/Users/MyName   MySurname/Documents/LaTeX/Figures/MyMWE-figure0.md5"'.   I'm working with Windows 10, MiKTeX 2.9 distibution.  Following the advice Tarass gave me on a comment of this question:  and as indicated here:  I've registered a user-managed TEXMF directory but I still get the same error.  With \tikzexternalize[prefix={Figures/}] all works perfectly but, since I'm using LyX, the files are stored in a LyX temp dir, hence I loose them when I close the program (making externalization useless), so I need to use an absolute path. This is a follow-up question to the discussion about the finite axiom of choice .  Suppose we have a countable collection of non-empty sets $\{A_1, A_2, A_3,\cdots\}$ Reasoning as indicated in that discussion, we can prove the existence of a choice function $c_n$ such that $c_n(i)$ belongs to $A_i$ for $i = 1,\cdots,n$  Again reasoning as suggested and using induction, we can prove the existence of a function $C$ defined over $\{1, 2, \cdots\}$ such that $C(n)$ is a choice function defined for $i=1,\cdots,n$ with $C(n)(i)$ an element of A_i. We can require in addition that $C(n)(i)$ and $C(n-1)(i)$ are equal for $i=1,\cdots,n-1$.  Now it would it seem that we can prove the countable axiom of choice by taking the union of $C(1)$, $C(2)$, $\cdots$  The question is as follows: why is that proof wrong? I suspect the answer may have something to do with the axiom of union. Keyboard not working after suspend to ram (sleep)