My question is: How come the three words "awesome," "awe," and "awe-inspiring" denote positivism with the root awe, they are virtuous too,but the 4th word "awful" denotes negativity. Even the spelling of "awe" gets pruned to "aw", as if the "e" of "awe" chewed up the virtues of "awe" like a rat and ran away with them. It leaves me clueless. Hope you help me locate the rat.
How did the English language come to this?       The play was awful.     Is the complete opposite of     The play was awesome.   But if you break it down to awe followed by ful or some, it doesn't make sense at all. Can someone shed a light on this?  Is there a rule to this, or is it one of those things you need to just memorize?
Show that there is a positive integer $n$ such that $G \cong \ker(\phi^{n}) \times \phi^{n}(G)$ Let $G$ be a finite abelian group and let $\phi: G \rightarrow G$ be a group homomorphism. I am trying to show that there is a positive integer $n$ such that $G \cong \ker(\phi^{n}) \times \phi^{n}(G)$.  I know that since $G$ is abelian we have that $\ker(\phi^{k})$ and $\phi^{k}(G)$ are normal subgroups of $G$ for any $k$  I also suspect we have the following towers which stabilize at $m$ and $m'$  $$\phi(G) \unrhd \phi^{2}(G) \unrhd \cdots \unrhd \phi^{m}(G)=\phi^{m+1}(G)=\cdots$$  $$\ker(\phi) \unlhd \ker^{2}(\phi) \unlhd \cdots \unlhd \ker^{m'}(\phi)=\ker^{m'+1}(\phi)=\cdots$$  I know that if given a group of the form $HK$ where $H$ and $K$ are normal subgroups of $HK$ and $H\cap K=1$ then $H \times K \cong HK$. I want to apply this to this situation but I am unable to show how write $G$ as a product $HK$  Resolution One of the comments directed me to the following article:  The last part of which answers the question. Below is what is says:  Choose $n=\max(m,m')$, then we have for $x \in \ker^{n}(\phi) \cap \phi^{n}(x)$, this means that $x=\phi^{n}(y)$ for some $y \in G$. This gives:  $0=\phi^{n}(x)=\phi^{2n}(y)$ which means that $y \in \ker^{2n}\phi=\ker^{n}{\phi}$. and then we have that $0=x=\phi^{n}(y)$.  The answer below kindly points out that every element $x$ is contained in one of the cosets of $G/\ker^{n}(\phi)$, this means that $x=k+g$ for some $k \in \ker^{n}(\phi)$ and $g \in G$. We also have that $g=\phi^{n}(h)$ for some $h \in G$. Writing $x=k+\phi^{n}(h)$ essentially shows that $G=\ker(\phi^{n}) + \phi^{n}(G)$. Using the fact above, the claim follows. i have created a schedulable class which has to trigger a batch job. The run method for the batch class will get list of data to process from a webservice callout. Now, i got the exception below. Any suggestions please.     System.CalloutException: Callout from scheduled Apex not supported.   global class scheduledDeltaBatchable implements Schedulable {    global void execute(SchedulableContext sc) {     ID batchprocessid = BatchUpsertSAPCustomerMasterBPSync.run(SAPServices.GetSAPCustomerMasterBPSync('Delta'));    } } I'm wondering if it's possible to hide a folder without renaming it with a dot on the Desktop.  Ubuntu 12.04  Thanks How to prove that if $\forall x \in (a,b)$ Lebesgue integral $\int_{(a,x)}fd\lambda=0$, then $f(x)=0$ $\lambda$-almost everywhere? Let $f:(a,b)\rightarrow\mathbb{R}$.  The statement to prove is that if $\forall x \in (a,b)$ Lebesgue integral $\int_{(a,x)}fd\lambda=0$, then $f(x)=0$ $\lambda$-almost everywhere.  So if it wouldn't be true then we would have $\forall x \in (a,b):\int_{(a,x)}f_+d\lambda=\int_{(a,x)}f_-d\lambda\neq0$, so every interval contains positive and negative values of the function. It is  such an $f$ so my guess would be that this function is not Lebesgue measurable and we can't take this integral.  Thanks in advance! If $n&gt;1$ is an integer, then $\sum \limits_{k=1}^n \frac1k$ is not an integer.  If you know , then you know there must be a prime $p$ between $n/2$ and $n$, so $\frac 1p$ appears in the sum, but $\frac{1}{2p}$ does not. Aside from $\frac 1p$, every other term $\frac 1k$ has $k$ divisible only by primes smaller than $p$. We can combine all those terms to get $\sum_{k=1}^n\frac 1k = \frac 1p + \frac ab$, where $b$ is not divisible by $p$. If this were an integer, then (multiplying by $b$) $\frac bp +a$ would also be an integer, which it isn't since $b$ isn't divisible by $p$.  Does anybody know an elementary proof of this which doesn't rely on Bertrand's Postulate? For a while, I was convinced I'd seen one, but now I'm starting to suspect whatever argument I saw was wrong. How can I iterate over each file in a directory using a for loop?  And how could I tell if a certain entry is a directory or if it's just a file? Brown sugar instead of white sugar Shimano Deore front derailleur occasionally sticking I have a question about my front derailleur sticking occasionally.  I've had this bike for a 1.7 years and have ridden it through 2 winters.  It has only been after this last winter that my front derailleur has been sticking on occasion where it's like grit has jammed in somewhere (which I'm sure it is the case).  When I'm on the fly I just knock it with my heel and it will downshift and I can push the shifter hard enough to go back up when I need to.  The first time I decided to overhaul both the front and back ones.  Took them apart, cleaned/degreased, regreased the two back cogs, and used WD-40 professional silicone on the rest of the parts as to not attract more dirt.  Then after about a month it got stuck again, and I couldn't see anything that could be jamming it (while it was still stiff as heck).  Didn't take it apart this time just used a tooth brush and silicone again and it's fine at the moment.  I'm assuming it may be that since that the roads have been grittier/dirtier from the winter with some recent rain and slush that it may just be normal.  I haven't been on the trails yet this year so I know it's not that.  Is this normal, has some kind of bur that I can't see come up and is catching grit?  EDIT:  I should also note that this is a hardtail mtb that I use for trails and winter commuting (a lot of salt and grit on the roads). Deleting 'garbage' nodes after polygon dissolve I work with QGIS 2.10.1 on Windows 7 (x64). I have dissolved polygons from an attribute value and the polygons geometries being not perfectly joined together, I get some 'garbage' nodes I need to delete. I have tried to edit the layer and using nodes tool, selecting and deleting the nodes, but it seems impossible to do (nodes are not deleted at all!). Any idea on how to process this kind of clean-up?  EDIT : you can  for testing. maximum of two uniform distributions I have a question. Let's suppose that the two random variables $X1$ and $X2$ follow two Uniform distributions that are independent but have different parameters:  $X1 \sim Uniform(l1, u1)$  $X2 \sim Uniform(l2,u2)$  If we define X3 as the maximum of X1, and X2, i.e., $X3 = max(X1, X2)$, what kind of distribution would it be? It is certainly not a uniform and I can calculate the cumulative probability, i.e., $p(X3 &lt;= a)$, because $p(X3 &lt;= a) = p(X1 &lt;= a) p(X2 &lt;= a)$. But, I wonder if there exists any density function that can model X3. Alternate property name while deserializing How can you duplicate weapons in Borderlands 2? I want to get 2 identical Unkempt Harolds for my Gunzerker, and I do not want to Tourge token grind. I have one already, but I want to get another. How would I duplicate with the newest version of PC BL2? Integrating $\int^0_\pi \frac{x \sin x}{1+\cos^2 x}$ prove or disprove: For a prime number p, there is a positive integer n such that $p+1 $= $n^2$ if and only if $ p = 3.$ Prove or disprove: For a prime number p, there is a positive integer n such that $p+1 $= $n^2$ if and only if $ p = 3.$  first im trying to proof if p is prime number there is positive integer n such that $p+1=n^2$ implies $p=3$  suppose there is positive n integer such that $p+1=n^2$ and prove that $p=3$. $p=n^2-1=(n+1)(n-1)$ since $p$ is prime $n-1=1$,$ n = 2 $ $p+1=4$ $p=3$ hence it is proved that $p=3$  now if $p=3$ implies there is positive integer n such that $p+1=n^2 $  $p=3$ and p is prime  $p+1=n^2$, $4=n^2$, $n2$ or $n=-2$ hence it is proved that there is positive integer $n$ such that $p+1=n^2 $  is my proof correct? and also for prime number can i write $n+1=1$ instead $n-1=1$? How to create the matrix attached that has 2x2 blocks in its diagonal and a square outlining them? Alien invasion movie, giant robot aliens with long legs capturing people How can I visualize the file system usage on Windows? How can I visualize which folders and files are taking up all of the space on my hard drive?  I'm getting some conflicting reports on the size of hard drive contents. Namely what is and isn't there and what folders are actually using the space.   I need to know which of the files or folders the culprits behind all this hidden bloat. Also there should be a print option to get it on paper. A   B   C   D   E   F             I don't even know. It just seemed weird.  Yes, I understand I could just avoid doing something like that. But the problem already exists with a single level.   Consider having 2 images in a post. Both exceed the maximum width of a post. One is nested in a list. Key to this issue is that my original images were much larger. They were automatically resized.  Image    List      I think it looks weird. How to solve configure: error: readline library not found? I have an Ubuntu 10.04 LTS server with a minimal install that I want to compile postgres on. With the minimal install the server did not have gcc so I had to do an apt-get install gcc but now I keep getting this error from the postgres ./configure.  configure: error: readline library not found   I can't seem to find the name of the package that contains the readline library. So I have two questions.   What is the name of the package that contains readline? Is there some way given the error message above for me to find the associated package from its standard name? Mail vs. mail what is the difference What is this business about mail and Mail? [On a CentOS box] Why mail does not work but won't return an error either while Mail works fine with the exact same syntax?  Mail is basically a link to mail, isn't it?  Sorry, this was a mixup, apparently (for historical reasons?!) there is a mail and a Mail  Nowadays it looks like they both point to the same thing, on my machine:  sr-linux&gt; ls -l `which mail` -rwxr-xr-x    1 root     mail        66492 Jun 24  2001 /bin/mail sr-linux&gt; ls -l `which Mail` lrwxrwxrwx    1 root     root           14 Nov 19  2001 /usr/bin/Mail -&gt; ../../bin/mail