In a clause like this one:  the success of the book resides in the clear explanations of complex subjects  is "in" the right preposition? Or is it better to use "on", like in:  the success of the book resides on the clear explanations of complex subjects  Thanks!
As it is common with people from my country, I have an immense difficulty with prepositions in English, especially with the use of in and on.   When the preposition indicates the position of the object it is a bit easier to decide which one to use, but as it requires thinking from me, I still say things like "I put the bill in the fridge", "my pencil is on my bag", etc, even though if I think about it, I would know how to say it correctly.  Of course, even in those cases where the preposition indicates position the decision is not always clear, as is the case for example with "on the bus" (you are not on top of the bus, you are inside the bus) or "he is in bed" (you are not inside the bed, you are on top of your bed).  Other cases are even less clear, and I believe there is no real rule to decide what should be said; the person should just know it. Two examples are "in a bad mood", "you are on my mind".  So my first question is whether there is really no rule to help me. The second question refers to two of the cases where I still don't know whether to use in or on:   good luck on/in your exam in/on the list   The last one is especially important to me, as I have asked a number of native speakers, and each time I get a different answer. The only consensus seems to be the case where you say "on top of the list". Apart from that, some people will say that I should say "his name is in the list" whereas others think I should say "his name is on the list". And yeah, a couple of people did tell me that I can use either.  Can someone help me?
I am trying to pull logs and doing few others things directly. written a small line BASH but not working. I am sure thats with IF Condition.    sed -rne '/21:25:07/,/21:50:07/ p'  server.log.2015-04-21 &gt; /tmp/filename.log ; du -sh  /tmp/filename.log ;  if [`du -sh  /tmp/filename.log` -gt 0] then gzip /tmp/filename.log ; Let $x$ and $y$ be positive integers such that $xy \mid  x^2+y^2+1$. Show that $$ \frac{x^2+y^2+1}{xy}= 3 \;.$$ I have been solving this for a week and I do not know how to prove the statement. I saw this in a book and I am greatly challenged. Can anyone give me a hint on how to attack the problem? thanks What five odd integers have a sum of $30$? Show $R \setminus S$ is a union of prime ideals I'm stuck on the following question:     Let $R$ be a commutative ring with $1$, and $S \subseteq R$ a saturated multiplicative set (that is, $1 \in S$ and $x, y \in S$ if and only if $xy \in S$).  Show that $R \setminus S$ is a union of prime ideals.     Here's my attempt so far.  If $a \in R \setminus S$, then the ideal $Ra$ does not intersect $S$, because if $ra$ were in $S$ for some $r \in R$, then $a$ would have to be in $S$.  Thus, it follows by Zorn's lemma that there exist ideals $I$ of $R$ which are maximal with respect to the property that $I \cap S = \emptyset$.  Let $I$ be any such ideal.  If I can show that $I$ is prime, the conclusion will follow.  Let $xy \in I$.  I want to show that $x$ or $y$ is in $I$.  Suppose $x \not\in I$.  Then $I + Rx$ intersects $S$, so there exists $s \in S, a \in I, r \in R$ such that $s = a + rx$.  Then $sy = ay + rxy \in I$, so $sy$, and hence $y$, cannot be in $S$.  Similarly if $y$ is not in $I$, then $x \not\in S$.  So we are reduced to proving the following is impossible: $xy \in I$, $x$ and $y$ are not in $S$, but they aren't in $I$ either. I was wondering if there is an easy way to trigger an e-mail alert on Windows Server 2008 when any logical disk partitions become low on space.  I have 2 SQL servers that have come close to running out of disk space because of the DB log files.  Thanks, Ryan Execute command in remote active terminal How to install Hp laser jet 1020 plus printer? Why we usually do not exceed ARMA(5,5) models in practice? Is there any mathematical justification for this? How to suppress scientific notation when printing float values? Here's my code:  x = 1.0 y = 100000.0     print x/y   My quotient displays as 1.00000e-05.  Is there any way to suppress scientific notation and make it display as 0.00001? I'm going to use the result as a string. How to make a window always stay on top in .Net? What to do when you discover a mistake in the archived copy of your PhD thesis? I've just defended my PhD in mathematics and started a postdoc. While working  on an improvement of one of the results of my thesis, I realized that there are several minor mistakes and a big bug in a proof that invalids a minor result in the thesis (about 3-4 pages out of 110). Unfortunately neither I nor my advisor or referees figured it out the mistake. Though the result is minor, it is announced in the introduction and the manuscript is on-line on an ArXiv-like server, so that I could publish a new version but not cancel the one on-line.   What is the best thing to do? Upload an errata? Upload a "revised" version of the thesis?  Publish a "revised version" of the thesis on my web-page?  Can this damage my future career, making me look not "reliable" as a researcher? "ImportError: No module named _struct" configuring python-minimal When I run sudo apt-get upgrade, the dpkg error appears and interupts the process.  Error details:  Setting up python-minimal (2.7.3-0ubuntu7.1) ... Traceback (most recent call last):   File "/usr/local/lib/python2.7/runpy.py", line 162, in _run_module_as_main     "__main__", fname, loader, pkg_name)   File "/usr/local/lib/python2.7/runpy.py", line 72, in _run_code     exec code in run_globals   File "/usr/local/lib/python2.7/compileall.py", line 16, in &lt;module&gt;     import struct   File "/usr/local/lib/python2.7/struct.py", line 1, in &lt;module&gt;     from _struct import * ImportError: No module named _struct dpkg: error processing python-minimal (--configure):  subprocess installed post-installation script returned error exit status 255 dpkg: dependency problems prevent configuration of python:  python depends on python-minimal (= 2.7.3-0ubuntu7.1); however:   Package python-minimal is not configured yet.   And this problem leads to a series of dependency problems when processing the following packages:    python  gwibber-service  libgwibber3  libgwibber-gtk3  gwibber  gwibber-service-facebook  gwibber-service-identica  gwibber-service-twitter  python-all  python-dev  python-all-dev  python-apt  python-problem-report  python-apport  python-libxml2  unity-lens-gwibber  unity-scope-video-remote   Finally, nothing I can do for this problem. And several days later, the root directory is full and I even could not uninstall any software because of this problem. Can the default font be changed? I have been playing with the settings for the text object recently, and every time I add a new text object, I need to change the font to my favorite one. (Eurostile LT ExtendedTwo) Is there any way to change the default font so I do not need to change it every time I add a new text object?   Example: How long can I store a food in the pantry, refrigerator, or freezer? So, I have proven already that every open set in $\mathbb{R}$ can be written as a countable union of disjoint open intervals; i.e., that $\mathcal{O} = \bigcup_{i=1}^\infty (a_{x_i}, b_{x_i})$.  Now, consider the closed interval $\left [ a + \frac{1}{n}, b - \frac{1}{n}\right ] \subset (a, b)$.  Since the midpoint of $(a,b)$ is $\frac{b-a}{2}$, we need $\displaystyle n &gt; \frac{2}{b-a}$.  Otherwise, $\displaystyle a + \frac{b-a}{2} = a + \frac{1}{ \frac{2}{b-a}} = b - \frac{1}{ \frac{2}{b-a}} = b - \frac{b-a}{2}$, and our interval will be degenerate.  Note that $\forall (a,b) \in \mathcal{O}$,   $\displaystyle \bigcup_{n = \frac{b-a}{2}+1}^{\infty} \left[ a + \frac{1}{n}, b - \frac{1}{n} \right] = (a,b)$.  So, we have $\displaystyle \mathcal{O} = \bigcup_{i=1}^\infty \left ( \bigcup_{n = \frac{b-a}{2} + 1}^\infty \left[ a_{x_i}+\frac{1}{n}, b_{x_i}-\frac{1}{n} \right] \right)$.  But, I don't like this. It looks like I'm still just taking a union of open sets.  Please help me get this to look like the union of closed sets.  (For unbounded intervals, I know what to do - it should follow from this relatively easily). If a person holds prejudice against people because of their nationality, would that be considered racist? Would it be considered racist if a store owner believes all Canadians are thieves and does not let any Canadians into his store? Racism pertains to discriminating based on race, and (correct me if I'm wrong) Canadians are not a race. What would be an appropriate word to describe an individual such as the  store owner?  The discrimination is specific to Canadians, not foreigners in general. I'm looking for a word that could be applied to someone who lives in the country they discriminate against. For example if a Canadian announces "all Canadian culture is stupid!" what would that be called? My display driver's maximum screen width is 1280 pixels.   I need to test a website with resolution 1920&times;1080 pixels. Or 3000&times;5000.   Different browsers is optional. How to perfectly align UV coordinates automatically? I have this setup:   The mesh is only the visible top side, with a mirror modifier for the bottom part. Now I want to unwrap it like in the picture. I want the inner circle (red) to go to the top of the UV map...  The problem is that no matter what automatic unwrapping I use, it never works as intended.  The closest I've gotten was by using "follow quads", but that skewed the UV map a lot:   I used the result from the second image and manually adjusted the vertices with a lot of scaling and "align auto", but its very far from perfect as you can see in the first image.  Here's what I want exactly:   The uv vertices should be distributed evenly (horizontally) The middle line (yellow) should be on a "natural" y height (so there's no stretching)   I hope it's clear what I want to achieve. I'm sure there must be some automatic way to perfectly align and distribute the UVs in this way.  By the way: If you look closely you can see that I already placed a uv seam (aligned with the green y axis in the screenshot) but that didn't really help. When I analyze my variables in two separate (univariate) logistic regression models, I get the following:  Predictor 1:    B= 1.049,    SE=.352,    Exp(B)=2.85,    95% CI=(1.43, 5.69),    p=.003    Constant:    B=-0.434,    SE=.217,    Exp(B)=0.65,                            p=.046  Predictor 2:    B= 1.379,    SE=.386,    Exp(B)=3.97,    95% CI=(1.86, 8.47),    p&lt;.001    Constant:    B=-0.447,    SE=.205,    Exp(B)=0.64,                            p=.029   but when I enter them into a single multiple logistic regression model, I get:  Predictor 1:    B= 0.556,    SE=.406,    Exp(B)=1.74,    95% CI=(0.79, 3.86),    p=.171 Predictor 2:    B= 1.094,    SE=.436,    Exp(B)=2.99,    95% CI=(1.27, 7.02),    p=.012    Constant:    B=-0.574,    SE=.227,    Exp(B)=0.56,                            p=.012   Both predictors are dichotomous (categorical). I have checked for multicollinearity.  I am not sure if I have given enough info, but I cannot understand why predictor 1 has gone from being significant to non-significant and why the odds ratios are so different in the multiple regression model. Can anyone provide a basic explanation of what is going on? Baby Rudin claim: $1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}...$ converges This sequence is a rearrangement of the series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}...$. Note that at this point in the text we do not have any theorem about the convergence of rearrangements.  Let $\{s_n\}$ be the sequence of partials sums of the series then for $n \ge 0$  $s_{3(n+1)} = \sum ^n _ {k=0} \frac{1}{4k+1} + \frac{1}{4k+3} - \frac{2}{4k+4}$  We can view it as the sequence(on $n$) of partials sums of  $\sum_0 a_n = \sum_0 \frac{1}{4n+1} + \frac{1}{4n+3} - \frac{2}{4n+4}$  Where $|a_n| = a_n = \frac{1}{4n+4}\{\frac{3}{4n+1}+\frac{1}{4n+3}\} \le \frac{1}{4n^2}$.  By the comparison test $s_{3(n+1)}$ converges to some real $\alpha$.  But $s_{3(n+1)+1} = s_{3(n+1)}+ \frac{1}{4n+5}$ and $s_{3(n+1)+2} = s_{3(n+1)}+ \frac{1}{4n+5}+\frac{1}{4n+7} $hence we have a partition of $\{s_n\}$ into subsequences which tend to $\alpha$ and this implies $s_n \rightarrow \alpha$.  Is my proof correct? Any alternative solutions are appreciated.