What would the full proof look like for $𝐵\subset 𝐴$? If   $A = \left\{6a + 4b\in\mathbb{Z} : a, b \in\mathbb{Z}\right\}$ and $B = \left\{2a \in\mathbb{Z} : a \in\mathbb{Z}\right\}$. Show that $A = B$  So far I have:  $A\subseteq 𝐵$  Suppose $x\in A$. $x = 6a + 4b$ for some $a, b \in\mathbb{Z}$. $x = 2(3a + 2b)$. Let $k1 = (3a + 2b)$ where $k1\in\mathbb{Z}$. x = 2k1. 2x, x ∈ ℤ. Then $x\in B$.  $B\subseteq 𝐴$ Suppose $y\in B$. $y = 2a$ for some $a\in\mathbb{Z}$. $y = 6a + 4(−a)$. Then $y\in A$.  I need help with showing steps for $B\subseteq A$.  Thank you all for the help!
1) How do I prove the following:   Let A = {6a + 4b ∈ Z : a, b ∈ Z} and B = {2a ∈ Z : a ∈ Z}. Show that A = B.  Thank you all for the help!
How to prove the property of the Lemoine point of a triangle? From Wolfram , I know there is a Lemoine point of triangle, also called symmedian point, the sum of squared distances of this point to all the three sides is algebraically minimum.    How to prove such a property? Can anyone explain why following code won't compile? At least on g++ 4.2.4.  And more interesting, why it will compile when I cast MEMBER to int?  #include &lt;vector&gt;  class Foo {   public:       static const int MEMBER = 1;   };  int main(){       vector&lt;int&gt; v;       v.push_back( Foo::MEMBER );       // undefined reference to `Foo::MEMBER'     v.push_back( (int) Foo::MEMBER ); // OK       return 0; } Is it possible to play D&D 5e solo? Is it at all possible to run a solo game where you act as the DM/GM for yourself? I know there was a 4th edition adventure published in an old issue of one of the D&amp;D magazines but recently I read somewhere (Reddit I think) that the games Outbreak: Undead and Eclipse phase were good games to play solo.  Can anyone with experience with running a game of D&amp;D 5e for themselves explain how to do so? Alternative ways to say "if and only if"? There are some scenarios about which I would like to get some confirmation:   when defining a concept A,      We call A, if ... [definition of   concept A]   Does "if" here mean equivalence instead of just sufficiency? Is it incorrect to replace "if" with "if and only if"?    For precisely what condition is    B satisfied?   Does "precisely" mean asking for necessary and sufficient condition for B? Some other ways to say "if and only if" / "necessary and sufficient"?   Some references that summarize some standard terminologies in Mathematics  such as this one?  Thanks and regards! Equation of the locus of the foot of perpendicular from any focus upon any tangent to the ellipse ${x^2\over a^2}+{y^2\over b^2}=1$ Find the equation of the locus of the foot of perpendicular from any focus upon any tangent to the ellipse ${x^2\over a^2}+{y^2\over b^2}=1$. will it also be an ellipse? Once a lead is converted it automatically creates an account, contact and one opportunity. But in my case i want to display the result without creating an Account. Please provide me with a sample code.  Any help will be appreciated... Prove $[A,B^n] = nB^{n-1}[A,B]$ I am trying to show that $[A,B^n] = nB^{n-1}[A,B]$ where A and B are two Hermitian operators that commute with their commutator. However, I am running into a little problem and would like a hint of how to proceed.  If A and B commute then $[A,B] = ABA^{-1}B^{-1} = e$ where e is the identity element of the group.  $$\therefore AB=BA$$  $$n=1; [A,B^1] = (1)B^0[A,B] = e$$ This statement is certainly true. however moving on to $n = 2$ I find...  $$[A,B^2] = AB^2A^{-1}B^{-2} = ABBA^{-1}B^{-1}B^{-1} = BBAA^{-1}B^{-1}B^{-1}$$  Where in the last step I have used the fact that A and B commute to rearange the terms. However, it is plain to see that this last term simply reduces to the identity as well and for the n = 2 case we have:  $$[A,B^2] = e \ne 2B[A,B] = 2Be = 2B$$  Clearly I have assumed something I shouldn't have. The fact that there is a multiplicative factor of n implies I should be adding things, but I thought if I kept it as general as possible, the answer should just fall out naturally. I don't want an answer please, only guidance. When I place a multipage pdf file into InDesign CC, the pages go outside of my document settings, making this function useless.  What do I have to do to make the placed  pdf pages go within my document settings, which in this case is 11 in. Length X 8.5 in. width, .5 in. margins on all sides, with two columns per page. I would greatly appreciate an answer to this, Thank you. The Canon L lenses are much more expensive and presumably much better quality than non-L lenses. What makes an "L" lens? How much better are they over a similar length/speed "ordinary" canon lenses? If $n = p_1^{a_1} ... p_r^{a_r}$ the set of nilpotents of $\mathbb{Z}_n$ is $(p_1 p_2 ... p_{r-1} p_r)\Bbb Z_n$ What is $A^-$ in topology? Hartshorne Algebraic Geometry Proposition 1.5 (page5) In a notherian topological space $X$, every nomempty closed subset $Y$ can be expressed as a finite union $Y=Y_1 \cup \cdots \cup Y_r$ of irreducible closed subsets $Y_i$. If we require that $Y_i \nsupseteq Y_j$ for $i \neq j$, then the $Y_i$ are uniquely determined. They are called the irreducible components of $Y$.    I understood the existence of such a representation. But I can't understand a sentence in  the uniqueness part, which I highlighted: Let $Y=Y_1 \cup \cdots \cup Y_r$ and $Y=Y&#39;_1 \cup \cdots \cup Y&#39;_s$ be two such representation. Then we can deduce that $Y_1=Y&#39;_1$ (the details are in the textbook). Now let $Z=(Y-Y_1)^-$, then $Z=Y_2 \cup \cdots \cup Y_r$ and also $Z=Y&#39;_2 \cup \cdots \cup Y&#39;_s$.  What is the operator $^-$? I understood it as a closure, is it right? But then how can I deduce that $Z$ is represented as $Z=Y_2 \cup \cdots \cup Y_r=Y&#39;_2 \cup \cdots \cup Y&#39;_s$? The definition of "Lie group" typically restricts to a smooth manifold.  If we instead define a "Lie group" to be a topological manifold such that multiplication and inversion are continuous, is the manifold necessarily smooth?  Is the smooth structure unique if we want a smooth Lie group?  I believe the answer is yes, since a connected Lie group structure is determined by the Lie algebra, but my search attempts failed. Say, teamA score is 368 (369 to win), and teamB is batting at 368 (or 366). One run (or 3 runs) will make them win. The batsman hits the ball for a six, so score is 374 (or 372) which is more than the required 369. Even then the runs are counted for the team and for the batsman.  Now, on the last ball, suppose the batsman did not hit a six, but ran for a single : teamB will win. Can the batsmen continue to run until the ball is dead ?  Why will they want this ? (A) If batsman is at 98 and wants to reach a century. (B) If batsman has a career total of 9998 and wants to reach 1000 runs. (C) If the batsman has a very high record strike rate, and wants to remain at that record strike rate. (D) In the championship, this is the last quarterfinal, &amp; even winning will put teamB equal to teamC, and the better runrate at the end of the quarterfinals will enable them to move to the semifinals. (E) Many similar situations can happen.  So coming back to the question : Do the laws of cricket allow running on the last ball, even after winning ?  In most of the matches I have seen, after winning, the batsmen do not even look where the ball is and start celebrating the win. If the rules did not allow them to run more, then fine. If the rules did allow it, then has there been any game where they did run more ? What causes javac to issue the "uses unchecked or unsafe operations" warning Why do we use ./filename to execute a file in linux?  Why not just enter it like other commands gcc, ls etc... This  asks how to install FF32 bit while keeping the 64 bit.  I want the much more simple thing: is there a canonical way to install firefox 52 ESR as "default" firefox in my 16.04?  I found this  ... but there are no packages; probably no wonder given the fact that the linked blog article dates back to 2012.  So, 2017, Ubuntu 16.04; how to install FF 52 ESR (as I simply need the java plugin to work for me).  ( my current solution: I downloaded the TAR file from mozilla, extracted that to /opt; and then changed /usr/bin/firefox to be a link to that ESR version of firefox. works, but ugly ) How is floating point stored? When does it matter? In follow up to , it appears that some numbers cannot be represented by floating point at all, and instead are approximated.  How are floating point numbers stored?  Is there a common standard for the different sizes?  What kind of gotchas do I need to watch out for if I use floating point?  Are they cross-language compatible (ie, what conversions do I need to deal with to send a floating point number from a python program to a C program over TCP/IP)? I use \includegraphics command to insert image to the document. But how can I add a border around this image w/o any margin between border lines and image? It seems to be very easy, but I can't find it in the documentation. Distribution of the digits of Pi Can anything be stated about the distribution of the digits of Pi, i.e., if I were to sample n digits of Pi, can anything be said about the probability to observe certain digits, or is there any reason to assume that they would not be evenly distributed? This is purely a curiosity question. Prediction interval for binomial random variable What is the formula (approximate or exact) for a prediction interval for a Binomial random variable?  Assume $Y \sim \mathsf{Binom}(n, p)$, and we observe $y$ (drawn from $Y$). The $n$ is known.  Our goal is to obtain a 95% prediction interval for a new draw from $Y$.   The point estimate is $n\hat{p}$, where $\hat{p}=\frac{y}{n}$. A confidence interval for $\hat{p}$ is straightforward, but I cannot find a formula for a prediction interval for $Y$. If we knew $p$ (rather than $\hat{p}$), then a 95% prediction interval just involves finding the quantiles of a binomial. Is there something obvious I'm overlooking?