Let A and B be subsets of a universal set U.  If $A \subseteq B,$ then $B' \subseteq A'$ For the above question I'm trying to solve it by contrapositive proof. I was suggested by some friends that it can also be solved using proof by contradiction. Can someone tell me how this could be done by proof by contradiction?
What is the difference between a "proof by contradiction" and "proving the contrapositive"? Intuitive, it feels like doing the exact same thing. And when I compare an exercise, one person proves by contradiction, and the other proves the contrapositive, the proofs look almost exactly the same.  For example, say I want to prove: $P \implies Q$ When I want to prove by contradiction, I would say assume this is not true. Assume $Q$ is not true, and $P$ is true. Blabla, but this implies $P$ is not true, which is a contradiction.  When I want to prove the contrapositive, I say. Assume $Q$ is not true. Blabla, this implies $P$ is not true.  The only difference in the proof is that I assume $P$ is true in the beginning, when I want to prove by contradiction. But this feels almost redundant, as in the end I always get that this is not true. The only other way that I could get a contradiction is by proving that $Q$ is true. But this would be the exact same things as a direct proof.   Can somebody enlighten me a little bit here ? For example: Are there proofs that can be proven by contradiction but not proven by proving the contrapositve?
Database Encryption scan for database  was aborted. Reissue ALTER DB to resume the scan Inertial frames of reference I'm struggling with the notion of an inertial frame of reference. I suspect my difficulty lies with the difference between Newtonian and relativistic inertial frames, but I can't see it.   I've read that Newton's laws apply in any non accelerating frame of reference, which are called inertial frames. So, if I play pool on a train moving with uniform velocity, the balls behave in the same way as if I were playing pool in a pool hall. So the train is (to a good approximation, ignoring forces caused by the Earth's rotation and movement around the Sun etc) an inertial frame.   I've also read that in an inertial frame, Newton's first law is satisfied. So, if I slide a rock on a sheet of ice and I could somehow eliminate the frictional forces between the rock and the ice, the rock would carry on sliding for ever, as predicted by Newton's first law.   Question 1 - is the  sheet of ice also therefore a good approximation to an inertial frame?  Question 2 - are these two definitions of inertial frames saying the same thing in different ways?  Both these examples occur in gravitation fields, which doesn't seem to matter as both the train and the ice sheet are presumably good approximations to inertial frames.  Question 3 - is gravity irrelevant when defining these two inertial frames?  In special relativity, I've read (Foster and Nightingale) that an inertial frame is also one where Newton's first law holds. But as it's special relativity there cannot be an inertial frame if there's a gravitational field, so I'm assuming the above two examples are not inertial frames in special relativity.  Question 4 - how come you can use the same definition of an inertial frame (satisfied Newton's first law) but in the case of special relativity, the train and the ice sheet aren't inertial frames. Is this to do with not being able to synchronize clocks in a gravitational field?  In general relativity, I've read (Schutz) that a freely falling frame is (on Earth) the only possible (and local) inertial frame. So again the first two examples would not be inertial frames in general relativity.  Question 5 - am I right to think that my two examples aren't even approximate approximations to an inertial frame in relativity?  Question 6 - have I missed anything else that might be useful?  Many thanks  Edit. On reflection, am I right in thinking that because Newtonian mechanics assumes universal, absolute time we don't need to worry about synchronizing clocks in a Newtonian inertial frame. Therefore we don't need to worry about gravity in a Newtonian inertial frame, because in such a frame gravity does not affect time.  This is not the case in spacetime, as here gravity does affect time and the only way to have synchronized clocks in an inertial frame in spacetime is: 1. have no gravitational field, or 2. use a local, freely falling frame.  Am I on the right track here? If I squeeze something really hard, will it ever become two-dimensional? I am a beginner using QGIS (1.8.0 Lisboa) on a Mac (OS X 10.6.8 Snow Leopard). I work on a file with two layers in the following sequence (top to bottom):    a vector polygon layer with a few outlines of various buildings in France and Italy (CRS: EPSG:4326 - WGS 84 - "on the fly" re projection is activated ) a Google satellite layer uploaded via open layer plugin.   Sometimes when changing the scale (scrolling in and out) the Google layer updates a little wrong in size or position compared to the polygon layer. But as soon as I "pan" the map everything reconstructs right. Now the problem I have concerns the print composer.  The result of any kind of export - pdf, or image or svg - of my map (vector layer + Google layer in the background) differs from what is shown in the preview of the print composer: the vector layer and the Google layer positions are mismatching and in a different scale. I can send some pictures showing screenshots of the print composer preview and the final results if required.  I tried almost everything:    new files with different CRS settings "cache" update preview "render" update preview different paper sizes,  different scales,   The problem appears only using an open layer map (Google, OpenStreetMap...). If I try the same using two vector layer (shapefile) - one with the buildings, another one with the city in background - everything works out great.  ...exasperating! Since it is a real important research project I would be very grateful for some help and tips. spectral radius in normed space is subadditive. I am reading the book "characterizations of c*-algebras" by Doran and Belfi. On the page 299, the book reads "It can also be shown that if $xy=yx$, then $\nu(x+y)\le\nu(x)+\nu(y)$." The definition of $\nu$ is as follows: in normed algebra,  $$\nu(x)=\limsup_{n\to\infty}\|x^n\|^{\frac1n}$$  Note that the algebra is not assumed to be complete and is also not assumed to be complex scalar(we can have real scalars)  This is my trial: I just tried to mimic the proof of the Minkowski inequality, but failed.  Thank you I'm starting to do designs meant for printing, but I'm fairly new to this medium so I don't really understand how bleed works and how much bleed should I specify for:   Business cards Large format prints (42 inches and bigger) Magazines   Should I extend the design into the bleed areas? How to trigger workflow from button in O365 I created a script editor webpart. I added a button which start a js function in the onClick event. I would like to start a workflow which is implemented on some list. Is this possible? In which Comic Book series is Batman actually Thomas Wayne? I've heard people say that there is a series of comics where Batman is actually Thomas Wayne instead of his son, Bruce.   Where can I find this? What is the series called? Proving the measure of an increasing sequence of measurable sets is the limit of the measures Show that if $A_1\subseteq A_2\subseteq A_3\subseteq\cdots$ is an increasing sequence of measurable sets (so $A_j\subseteq A_{j+1}$ for every positive integer $j$), then we have $$m\left(\bigcup_{j=1}^\infty A_j\right)=\lim_{j\to\infty}m(A_j)$$  Here is my proof: According to the $\sigma$-algebra property, $\bigcup_{j=1}^{\infty}A_j$ is a measurable set, so it makes sense to talk about $m(\bigcup_{j=1}^{\infty}A_j)$. Firstly, I prove that $\lim_{j\to\infty}m(A_j)\leq m(\bigcup_{j=1}^{\infty}A_j)$. This is because for any given positive integer $N$, $A_N\subseteq \bigcup_{j=1}^{\infty}A_i$, according to monotonicity, we have $m(A_N)\leq m(\bigcup_{j=1}^{\infty}A_i)$. Take the limit,we will have $\lim_{j\to\infty}m(A_j)\leq m(\bigcup_{j=1}^{\infty}A_j)$. Secondly, I prove that $m(\bigcup_{j=1}^{\infty}A_j)\leq \lim_{j\to\infty}m(A_j)$. For any given positive integer $N$, $\bigcup_{j=1}^N A_j = A_N$. According to monotonicity,we have $m\left(\bigcup_{j=1}^N A_j\right)=m(A_N)\leq \lim_{j\to\infty}m(A_j)$. Take the limit, we will have $m\left(\bigcup_{j=1}^\infty A_j\right) \leq \lim_{j\to\infty} m(A_j)$. Combine the above two arguments, we will see that $$m\left(\bigcup_{j=1}^\infty A_j\right)=\lim_{j\to\infty} m(A_j)$$$\Box$  The above is my proof, unlike many books, my proof does not use the property of countable additivity. So I doubt my proof is correct. Who can point out where are my mistakes? How do I unlock Rosalina in Mario Kart Wii? How do I unlock Rosalina from Super Mario Galaxy as a playable racer in Mario Kart Wii? Compare classifiers based on AUROC or accuracy? I received a journal paper for a review. The paper suggests a solution to an important problem, but the solution is very complicated and has mediocre performance. While reading the paper, I thought of a different solution, which is much simpler and has better performance (to give some scale: their solution is 1 page to write and 5 pages to prove; my solution is 5 lines to write and half a page to prove).  What should I do now?  A. Write my solution in the review, hoping the authors will remember to "thank the anonymous reviewer"?  B. Recommend that the paper published as is, then write a paper of my own with my improved solution?  C. Something else?  EDIT: Before I found the superior solution, I thought of recommending "resubmit with major revision", with the reasons that I mentioned above: your solution is too complicated and its performance is too weak.  However, now I am not sure it is the good course of action. A major revision may take a long time. The authors might even decide to delay or stop working on the paper without telling the editor (or me). Meanwhile, I will have to postpone the publishing of the superior solution for an indefinite amount of time. What do you think? How would you do a "not in" query with LINQ? I have two collections which have property Email in both collections. I need to get a list of the items in the first list where Email does not exist in the second list. With SQL I would just use "not in", but I do not know the equivalent in LINQ. How is that done?  So far I have a join, like...  var matches = from item1 in list1 join item2 in list2 on item1.Email equals item2.Email select new { Email = list1.Email };   But I cannot join since I need the difference and the join would fail. I need some way of using Contains or Exists I believe. I just have not found an example to do that yet. Stop iCloud automatic download to harddrive How do I stop iCloud from downloading files to my hard drive? I opened up a folder and it immediately started downloading all contents Ted Shifrin on this site posed an interesting problem to me: show that   $$ \lim_{(x,y) \to (0,0)} \frac {|x|^ \alpha |y|^ \beta} {|x|^ \gamma + |y|^ \delta} \text {exists} \iff \frac\alpha\gamma + \frac\beta\delta &gt; 1, \,\,\,\,\text{where } \alpha, \beta, \gamma, \delta &gt;0$$  I think I've got the $(\Leftarrow)$ direction as follows: assume WLOG that $\gamma \leq \delta$. Then switch to polar coordinates and get   $$\frac {r^ {\alpha + \beta} (*) } {r^\gamma (1(*) + r^{\delta - \gamma}(*))},$$  where $(*)$ represents some trig stuff that is bounded near zero.   Edit: I need to make sure that $(1(*) + r^{\delta - \gamma}(*))$ is bounded below here.  The other direction is giving me some trouble. It seems we need a lower bound for the fraction (something to force to zero), and I'm not sure where to find one. In particular, I'm not sure what to do with the denominator. Any ideas? It's essential that the documents (should)/(ought to) be destroyed immediately.   Which one is more suitable one as I know "ought to" and "should" are synonyms. How do i show that $w=z^2$ maps $|z-1|\leq 1$ onto the cardioid $R=2(1+\cos \theta)$? The number of labelled graphs with all vertices of even degree I have a graph exam tomorrow, and there is a problem that said number of graph with labeled vertices and all of them of even degree is $ 2^{n-1 \choose 2} $.  According to that topic, it means the number of cut spaces is $ 2^{n-1} $. Well, I don't understand what is cut space yet, and why its count is $ 2^{n-1} $ And don't  know why we should divide the total number of graphs by this! I have an ArrayList that I want to output completely as a String. Essentially I want to output it in order using the toString of each element separated by tabs. Is there any fast way to do this? You could loop through it (or remove each element) and concatenate it to a String but I think this will be very slow. $\sum_{1}^{n}\left ( a_{i}b_{i} \right )\geq \prod_{1}^{n}\left ( a_{i}^{b_{i}} \right )$ Let $a,b,c,x,y,z&gt;0$ and $x+ y+ z= 1$. Show that $$ax+ by+ cz\geq  a^{x}.b^{y}.c^{z}$$ I have found the generalization of this inequality: For 2 sets of numbers $\left \{ a_{i} \right \}, \left \{ b_{i} \right \}&gt; 0$ and $\sum_{1}^{n}b_{i}= 1$, we have: $$\sum_{1}^{n}\left ( a_{i}b_{i} \right )\geq  \prod_{1}^{n}\left ( a_{i}^{b_{i}} \right )$$ First solution Simply compare the logs: as the natural logarithm is a concave function, one has $$\ln\bigl(a^x b^y c^z)=x\ln a+y\ln b+z\ln c\le \ln(ax+by+cz)$$ for all non-negative $x,y,z$ such that $x+y+z=1$. How about another solution? I really hope to see that. Thanks!