Prove that for all $a,b \in\mathbb{R}$, $a·b=0$ if and only if $a=0$ or $b=0$.
Showing that every field is an integral domain.
Let $S = \{ax + by: x,y \in \mathbb{Z}\}$  and let $e &gt; 0$ be the smallest element in $S$. Prove that $e \mid a$, and hence prove that $e = \gcd(a,b)$  I'm afraid I can't provide much of my initial working here because I'm at a loss of how to start.  I suppose that to show $e\mid a$ I would have to show that $eg = a$ for some $g \in \mathbb{Z}$. But I have no idea how I would do this. Why is a 1" sensor actually 13.2 × 8.8mm? I'm wondering how sensor sizes are calculated. The Sony RX100 reportedly has a sensor size of 1" which is 13.2 x 8.8 mm according to the press release and various sites.  How does that work? That's a pretty small inch.  The Wikipedia page on  has this chart:   which doesn't really help. (The 1" format is labelled Nikon CX. Chart by Moxfyre under the CC-BY-SA license.)   If full frame/35 mm refers to the horizontal size, what does 1" refer to? New Graphics Card Preventing Boot Up Partitioning sum of squares A limit question (JEE $2014$) The following is a JEE (A national level entrance test) question:  Find the largest value of the non-negative integer ( a ) for which:  $$ \displaystyle \lim_{x \to 1} \left( \dfrac{-ax + \sin(x-1) + a} { x + \sin(x-1) -1 } \right)^{\dfrac{1-x}{1-\sqrt x} } = \dfrac 1 4 $$  On solving this, we get $ a = 0 $ or $ a = 2 $. So we take the answer as $ 2 $. But the official answer key says that the answer is $ 0 $. The reason given (not in the official key - they simple contain the answer. I saw the reason in an "unofficial" solution to the question paper) is that if $ a = 2 $, then the term $\displaystyle \dfrac{-ax + \sin(x-1) + a} { x + \sin(x-1) -1 } $ tends to a negative value ( $-0.5$ ). So $ a = 0$.  BUT:  By putting $a = 2$ in the limit, and inputting that to wolframalpha (), the answer turns out to be $0.25$. So, by putting $ a = 2$ also, we get the same value of the limit - $0.25$.  How to solve this dispute? Also, could you please check with Mathematica? Unfortunately, I don't have access to it!  (This is very important as the change in the answer key will affect the rankings in a dramatic way) I have searched in %appdata%, pulling up the list of files there, but there is no file labelled .minecraft. Searching %appdata%\.minecraft returns "no results found".   I also opened Minecraft to click on the resources button and could not locate that either. How to prove that $\lim_{n\rightarrow \infty}\frac {n^{2}}{2^{n}}=0$ & $\lim_{n\rightarrow \infty}\frac {n^{2}}{n!}=0$? Prove that:  $$\lim_{n\rightarrow \infty}\dfrac {n^{2}}{2^{n}}=0$$  I think I need to show $2^n \geq n^3$ $\forall n \geq10$. Is it true?  What about: $$\lim_{n\rightarrow \infty}\dfrac {n^{2}}{n!}=0$$  What to do with $n!$? I know that IList is the interface and List is the concrete type but I still don't know when to use each one. What I'm doing now is if I don't need the Sort or FindAll methods I use the interface. Am I right? Is there a better way to decide when to use the interface or the concrete type? Every vector space has a basis Prove that every vector space has a basis. I am going to use Zorn's lemma for this also here is a necessary definition regarding totally ordered subsets: one element will be contained in the other.   proof: A basis for $V$ is a maximally linearly independent set. Let $X$ be equal to all linearly independent subsets of $V$. On, $X$ define $A\leq B$ means $A\subset B$. Let $S$ be a totally ordered subset of $X$, let $T = \bigcup_{A_{s}\in S} A_s$ then $A_s\subset T$ for all $s\in S$. Does $T\in S$? i.e. is $T$ a linearly independent set?. Let $$O = \alpha_1 v_1 + \alpha_2 v_2 + \ldots + \alpha_n v_n \ \ \text{where} \ \ v_i\in T = \bigcup_{A_{s}\in S} A_s$$ Then there exists $S_i$ such that for all $i\in A_{s_{i}}$ all $i$ $A_{s_{1}},A_{s_{2}},\ldots, A_{s_{n}}$ (these will have one contained in the other then we would take the larger). So, there exists $A_{s_{k}}$ such that $A_{s_{k}}\supset A_{s_{i}}$ all $i$ just compare two at a time. Therefore, $V_i \in A_{s_{k}}$ all $i$, but $A_{s_{k}}$ is a linearly independent set. So, $\alpha_i = 0$ all $i$. Hence, $T\in X$ so $T$ is an upper bound for $S$. Hence every totally ordered subset of $X$ has an upper bound, hence by Zorn's lemma, there exists a maximal element and this would be the basis.  I am not sure if my proof is right, I apologize if it is a bit messy. If it is right, any suggestions on cleaning it up would be greatly appreciated. How to launch the default (native) application for a given file from Java? problems installing mathtools Check out the screenshot below, I want to extend my C partition (where Windows is installed) into the unallocated free space shown. This space was cleared up by shrinking the D partition. I can extend the D partition if I want to, but not the C.  Why not? Debugging a custom-bib bst file that crashes on book items I created my own bibliography file with custom-bib (makebst), but when I run it with a reference to a book I get:  $ bibtex PhD [...normal output] Database file #1: PhD.bib You can't pop an empty literal stack for entry oecd2001brc while executing---line 1504 of file wdsphd.bst   The only thing on that line is a high-level ITERATE call.   The style was created with these options:  %% merlin.mbs  (with options: `head,exlang,annote,seq-no,nm-rvvc,dt-beg,yr-par,xmth,tit-qq,qx,bt-qq,vol-bf,volp-sp,num-xser,pub-date,isbn,issn,doi,english,pp,ord,url,url-nl,nfss,,{}') %% english.mbs  (with options: `exlang,annote,seq-no,nm-rvvc,dt-beg,yr-par,xmth,tit-qq,qx,bt-qq,vol-bf,volp-sp,num-xser,pub-date,isbn,issn,doi,english,pp,ord,url,url-nl,nfss,,{}') %% merlin.mbs  (with options: `tail,exlang,annote,seq-no,nm-rvvc,dt-beg,yr-par,xmth,tit-qq,qx,bt-qq,vol-bf,volp-sp,num-xser,pub-date,isbn,issn,doi,english,pp,ord,url,url-nl,nfss,,{}')   The code for book items looks like:  FUNCTION {book} { output.bibitem   author empty$     { format.editors "author and editor" output.check     }     { format.authors output.nonnull       crossref missing$         { "author and editor" editor either.or.check }         'skip$       if$     }   if$   format.btitle "title" output.check   crossref missing$     { format.bvolume output       new.block       format.number.series output       new.sentence       format.publisher.address output     }     {       new.block       format.book.crossref output.nonnull     }   if$   format.edition output   format.isbn output   format.doi output   new.block   format.note output   fin.entry   write.url }   And this is the entry:  @book{oecd2001brc,  author = {OECD},  doi = {10.1787/9789264193550-en},  isbn = {9789264186903},  publisher = {OECD Publishing},  title = {{Biological Resource Centres: Underpinning the future of life sciences and biotechnology}},  url = {http://www.oecd-ilibrary.org/science-and-technology/biological-resource-centres\_9789264193550-en},  year = {2001} }   Is there a way to get a better stacktrace from bibtex? What I have now doesn 't really tell me what is going on. If it can be fixed simply, help would also be appreciated. Using divisibility theorems, prove that if $\gcd(a,b)=1$ and $a|c$ and $b|c$, then $ab|c$.   This is pretty clear by UPF, but I'm having some trouble proving it using divisibility theorems. I was looking at , and if this could be proven, then the solution would follow immediately. Adding Facebook Online Account - "application you are trying to use doesn't exist or has been disabled" How to turn off chrome's annoying auto-fill? What I want to show is the following statement.  For every prime of the form $2^{4n}+1$, 7 is a primitive root.  What I get is that  $$7^{2^{k}}\equiv1\pmod{p}$$ $$7^{2^{k-1}}\equiv-1\equiv2^{4n}\pmod{p}$$ $$7^{2^{k-2}}\equiv(2^{n})^2\pmod{p}$$ Thus $(\frac{2^n}{p})=(\frac{7^{2^{k-2}}}{p})=1$.  I think that $7$ is important because $7$ is a primitive root but I don't know how to use $7$. Derangements with repetitive numbers I have a very simple problem.  Lets assume that I have a well shuffled deck of 52 cards. I start drawing the top card always and when the card matches its rank I lose. J=11 Q=12 K=13.  If there were only 13 cards I could easily use the $\ \frac{!n}{n!}$ for derangements in order to solve this. The problem is that there are 52 cards so when I pass 13 I start from 1 again so I don't know what is the probability to win. Example of the game     1st card: 4 - Continue   2nd Card: A - continue   3rd Card: K - Continue   4th Card: K - Continue   5th Card: 6 - Continue   6th Card: 9 - Continue   7th Card: 10 - Continue   8th Card: A - Continue   9th Card: J - Continue   10th Card: 3 - Continue   11th Card: 2 - Continue   12th Card: 8 - Continue   13th Card: A - Continue   1st card: 5 - Continue   2nd Card 2 LOSE   So actually I have to count from 1 to 13 4 times and if I draw all 52 cards then I win. What's the probability? We have the matrix: $$\left[\begin{array}{rrr} 0.8&amp;0&amp;0\\ 0&amp;1.2&amp;0\\ 0&amp;0&amp;1\\\end{array}\right]$$ Finding the eigenvalues and eigenvectors are easy. But finding ${\bf M}^n$ when $n \rightarrow \infty$ is the hard part. Any suggestions? theorem numbering: section counter but remove the section number