I want to show that for $\gcd(a,b)=1$, $a,b \in \mathbf{Z}$, $\gcd(b^2+ab,a^2+2ab)=1$. I think the first step is something like this $$\gcd(b^2+ab,a^2+2ab)=\gcd(b(a+b),a(a+b)+ab). $$ I have two properties that can be used. There are 1/ $\gcd(a+b,c)=\gcd(a,c)$ if $b \vdots a$. 2/ $\gcd(ab,c)=\gcd(a,c)$ if $\gcd(b,c)=1$. Please help me. Thank you.
I'm reading the proof of the multiplicative property of $$s_k(n)=\sum_{d|(n,k)}f(d)g\bigg( \frac kd\bigg)$$ The book wrote that in order to understand the proof, we need to know if $a,b,k,m$ are integers such that $(a,k)=(b,m)=1$, then $$(ab,km)=(a,m)(b,k).$$  I have written proof, but I don't know is it consistent, and this seems used more tools than it necessarily has to.   My proof: Let $$S_1=\{d: d|(a,m), d&gt;0\}\\S_2=\{d: d|(b,k), d&gt;0\}\\S=\{d:d|(ab,km), d&gt;0\}.$$ I will prove the statement by showing that the function $\phi:S_1\times S_2\to S$ defined as $$\phi(d_1,d_2)=d_1d_2$$ is bijective, and complete the proof by connecting the maximal elements from these partial ordering sets.  Obviously, for $d_1\in S_1, d_2\in S_2$, this tells us $d_1 $ is both divisor of $a$ and $m$, and $d_2$ is both divisor of $b$ and $k$, hence $d_1d_2$ is both divisor of $ab$ and $km$, so we can tell $\phi(S_1\times S_2)\subseteq S$.  To show one-to-one, if we have $c_1,d_1\in S_1$ and $c_2,d_2\in S_2$ such that $$\phi(c_1,c_2)=\phi(d_1,d_2)$$ then $$c_1c_2=d_1d_2$$  In particular, $c_1|d_1d_2$, we need to show $c_1|d_1$. Indeed, if $(c_1,d_2)&gt;1$, this implies $a,k$ have common factor larger than $1$, contradicts the assumption. Therefore $c_1|d_1$. A similar argument implies $d_1|c_1$, therefore $c_1=d_1$. Again a similar argument implies $c_2=d_2$, which shows one-to-one.   To show onto, let $d\in S$, since $d|ab$, we can split $d$ into $d_1d_2$ such that $d_1|a$ and $d_2|b$. Since $(a,k)=1$, we see $(d_1,k)=1$, by property of $S$ we see also $d_1|m$, therefore $d_1\in S_1$, a similar argument implies $d_2\in S_2$. So every number in $S$ has a preimage in $S_1\times S_2$  This completes the proof. Is there any mistake in the proof?     Edit: A proof about the statement: If $d|ab$ then there are $d_1,d_2$ such that $d=d_1d_2$ and $d_1|a, d_2|b$.  Let $ab=p_1^{e_1}\cdots p_r^{e_r}$ for distinct prime factors $p_1,\cdots,p_r$, then we can write $$a=p_1^{x_1}\cdots p_r^{x_r}\\b=p_1^{y_1}\cdots p_r^{y_r}$$ where some of $x_i, y_i$ can be zero, and $x_i+y_i=e_i$ for all $i=1,\cdots,r$.  Since $d|ab$, we can write $$d=p_1^{f_1}\cdots p_r^{f_r}$$ with $f_i\leq e_i$ for all $i=1,\cdots,r$. Then since $$f_i\leq x_i+y_i$$ Split into two cases:   Case 1: $f_i&gt; x_i$, then we can choose $g_i=x_i$ and $h_i=f_i-x_i$.  Case 2: $f_i\leq x_i$, then we can choose $g_i=f_i$ and $h_i=0$.  Then the choice $$d_1=\prod_{i=1}^r p_i^{g_i}, \quad d_2=\prod_{i=1}^r p_i^{h_i}$$ will prove the statement.
$F$ closed and connected then... Let $F\subset X$ be a closed and connected subset of the metric space $(X,d)$ then for every pair of points $a, b\in F$ and each $\epsilon&gt;0$ there are points $a=z_0,z_1,\ldots z_n=b$ in $F$ such that $d(z_{k-1}-z_k)&lt;\epsilon$ for $1\leq k\leq n$.  Is the hypothesis that $F$ is closed needed? If two coins are flipped and one gets head, what is the probability that both get head? Efficiently select beginning and end of multiple contiguous ranges in Postgresql query I've got about a billion rows of data in a table with a name and an integer in the range 1-288.  For a given name, every int is unique, and not every possible integer in the range is present--so there are gaps.  This query generates an example case:  --what I have: SELECT * FROM ( VALUES ('foo', 2),               ('foo', 3),               ('foo', 4),               ('foo', 10),               ('foo', 11),               ('foo', 13),               ('bar', 1),               ('bar', 2),               ('bar', 3)      ) AS baz ("name", "int")   I'd like to generate a lookup table with a row for each name and sequence of contiguous integers.  Each such row would contain:  name   --  the value of the name column start  --  the first integer in the contiguous sequence end    --  the final value in the contiguous sequence span   --  end - start + 1  This query generates example output for the above example:  --what I need: SELECT *  FROM ( VALUES ('foo', 2, 4, 3),               ('foo', 10, 11, 2),               ('foo', 13, 13, 1),               ('bar', 1, 3, 3)      ) AS contiguous_ranges ("name", "start", "end", span)   Because I have so many rows, more efficient is better.  That said, I only have to run this query once, so it isn't an absolute requirement.  Thanks in advance!  Edit:  I should add that PL/pgSQL solutions are welcome (please explain any Fancy Tricks--I'm still new to PL/pgSQL). Why is the multiplicative group of a finite field cyclic? FFmpeg - Apply blur over face How can I show that $y'=\sqrt{|y|}$ has infinitely many solutions? Show that the first order differential equation $y'(x)=\sqrt{|y(x)|}$ with intial value $y(1/2)= 1/16$ has infinitely many solutions on the interval [−1, 1].  My thought were to show that this equation has two solutions (just by ansatz, or just looking at the interval $]0, 1]$ to get $y'(x) = \sqrt{|y(x)|}$) and deduce from that that there must be infinitely many solutions since the solution set forms a vector space. I've got a nagging feeling this isn't the way to go though. Can anybody give me a hint where to look? Why does Spring MVC respond with a 404 and report "No mapping found for HTTP request with URI [...] in DispatcherServlet"? Add a link to the timeline of a post For the past year I have been quite active on SO and a bit on meta, but I had never heard of the timeline of a post until today in a comment on meta.  There was an  two years ago of this feature as "experimental". ( for that question).  This feature was , answer: this exists, but the link has to be built manually.  I think this could be quite interesting sometimes and I would like to use this, but I will not do so if I have to build the link each time by myself.  This feature has existed for two years and is still "experimental"? Who is using it? Who knows about it?  My feature request: Implement a link in the posts to their timeline. I am sure this nice feature will be used. Admin privileges requiring for installing a package build though Packages Using osx 10.9 and packages for making package installer.  While installing a package made using Packages in /Applications location on a osx 10.9 machine, installer is asking for Admin password. Even though I had unchecked 'Require admin password for installation' while package making.  So, how to avoid this i.e a non-admin user should install this package. Package should install in /Applications location, so that non-admin user can use the app.  I check  and Who is bound to the Prime Directive? diff a ruby string or array How do I do a diff of two strings or arrays in Ruby? Why is there a discrepancy in disk usage reported by df and du? I have a Linux CentOS server, the OS+packages used around 5GB. Then, I transferred 97GB data from a Windows server to two folders on this Linux server, after calculated the disk usage, I see the total size of the two folders is larger than the disk used size.  Run du -sh on each folder, one use 50GB, the other one use 47GB  But run df -h, the used space is 96GB. (50GB + 47GB + 5GB) > 96GB  Is there any problem? Those two folders contain lots of files (1 million+). Thanks. Prove $p_n(x) \rightarrow \sqrt{x}$ uniformly as $n \rightarrow \infty$ Define $(p_n)_n$ recursively by $p_0(x) = 0$ and $p_{n+1}(x) = p_n(x) + \frac{1}{2}(x - p_n(x)^2)$.  Prove that $p_n(x) \rightarrow \sqrt{x}$ uniformly for $0 \leq x \leq 1$ as $n \rightarrow \infty$.  I'm not sure where to go with this.  I can show inductively that $\forall n,\  0 \leq p_n(x) \leq 1$ for $0 \leq x \leq 1$, and I think that $p_n(x)$ is increasing with $n$, but I have no idea whether this helps me.  I know that I need to show that $\forall \epsilon&gt;0: \exists N \in \mathbb{N}: \forall n \geq N: \forall x\in [0,1]: \big|p_n(x) - \sqrt{x}\big| &lt; \epsilon$ Edit:  I just had a thought.  I'm given a version of Dini's theorem that reads  Let $X$ be a compact metric space, $(f_n)_n$ a sequence of real-valued continuous functions on $X$, $f$ a continuous function on $X$ such that: (i) $f_n \rightarrow f$ pointwise on $X$, and (ii) $f_n(x) \geq f_{n+1}(x) $ for all $x \in X, n \in \mathbb{N}$. Then: $f_n \rightarrow f$ uniformly on $X$.  Obviously, (ii) is not true for my sequence, but what if I took $f_n(x) = -p_n(x)$ and $f(x) = - \sqrt{x}$?  Then I could prove (ii) to be true, I know that $[0,1]$ is compact, and $f_n(x)$ is real-valued.  The only part I would need to prove is (i), which is where I get confused.  Correct me if I'm wrong, but after that I could say that since $(-p_n)_n \rightarrow -\sqrt{x}$, then $(p_n)_n \rightarrow \sqrt{x}$.  Any advice? Rotating clears selected tags Simple: every other race in Middle Earth has rings, why do Hobbits and Orcs not? Why doesn't the auxiliary will qualify as future tense? According to A Student's Introduction to English Grammar, Tense is defined as follows:     Tense. A system marked by verb inflection or auxiliaries whose basic use to locate the situation in time: I liked it (past tense, past time), I like it(present tense, present time).   However, the author also points out:     ..., English is not one of them: it has no future tense. It does have several ways of talking about future time, and the most basic one does involve the auxiliary will.   These two quotes made me puzzled because, as noted above, Tense is denoted by inflection or auxiliaries and the auxiliary will seems to clearly meet the criteria. In conclusion, my question is   Why doesn't the auxiliary will qualify as future tense? Why does just splitting an Ethernet cable not work? I thought the Ethernet is logically a one-line communication bus (for argument's sake, I am excluding hubs). All machines attached on the bus hears the same signals and the machines themselves try to avoid collisions by randomly backing off.    If so, why would splitting one Ethernet line from my home router into two and connecting two computers not work? Why do I have to add a switch to it?  What the Internet said would not work.                                                             +----------+                                                +----------&gt;|computer 1|                                                |           +----------+     +------------------+       +---------------+     |4 port home router+------&gt;|simple splitter|     +------------------+       +---------------+                                                |           +----------+                                                +----------&gt;|computer 2|                                                            +----------+   What the Internet said I should do                                                             +----------+                                                +----------&gt;|computer 1|                                                |           +----------+     +------------------+       +---------------+     |4 port home router+------&gt;|    switch     |     +------------------+       +---------------+                                                |           +----------+                                                +----------&gt;|computer 2|                                                            +----------+   Is this because of the signal degradation (reduced electric current)?    Thank you for all the answers! The reason why I did not just use the two ports of my home router is...  The 4-port gigabit router is in my room, and I had put a computer in another room (also my room, though). Since a wired network is far more reliable and secure, I had bought a long Ethernet cable and and connected the computer to the router. Now I was thinking about adding another computer to that room. I could buy another long Ethernet cable, but then there will be two cables between the rooms. The one line already is a minor annoyance, so I thought if I could share the one line between the two computers in that room. A switch would work, but it requires power and is a little bit pricey. That is why I wondered why it would not work to simply split the physical Ethernet cable.  Apparently I do not completely understand how Ethernet and a switch work. I just have some bit of knowledge I heard in my college class. Group of order 15 is abelian How do I prove group of order 15 is abelian?  Is there any general strategy to prove that a group of particular order(composite order) is abelian? Show/Hide hidden files without restarting finder? I came to understand that the following is the way to toggle the visibility of hidden files:  defaults write com.apple.finder AppleShowAllFiles YES # replace YES with NO to hide hidden files killall -HUP Finder /System/Library/CoreServices/Finder.app   Is there a way to show/hide hidden files without killing the Finder? Showing $\int_x^{x+1}f(t)\,dt \xrightarrow{x\to\infty}0$ for $f\in L^2 (\mathbb{R})$ Show that if $f\in L^2 (\mathbb{R})$ then    $$\lim_{x \rightarrow \infty} g(x)=0,$$   where    $$g(x)=\int_x^{x+1}f(t)\,dt$$   Since $f \in  L^2 (\mathbb{R})$ then $f \in L^2[x,x+1]$ for every $x$. Then  $$|g(x)| \leq \left(\int_x^{x+1} f(t)^2 dt\right)^{\frac{1}{2}} \int_x^{x+1} 1 dt$$ $$|g(x)|\leq \left(\int_x^{x+1} f(t)^2 dt\right)^{\frac{1}{2}}$$