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For some reasons, homomorphism is a very hard area for me to make improvements. I've been hitting the brick wall for almost 2 hours.   Prove that no homomorphism exists from $Z_{16}\oplus Z_2$ onto $Z_4\oplus Z_4$.  Assume a homomorphism $\phi$ exists from $Z_{16}\oplus Z_2$ onto $Z_4\oplus Z_4$. Then $\phi$ is an isomorphism.  Note that the $\left | \ker \phi \right |=2$ The $\ker \phi$ is also a normal subgroup for $Z_{16}\oplus Z_{2}$. We want possible normal subgroup of order 2. I.e., 2 elements in each normal subgroups.  By Lagrange's theorem, the order of each element in a group divides the order of a group so the possible order of the elements are 1 or 2. If the elements has order 1, then the Ker \phi cannot have 2 elements.  Hence, $\ker\phi$ = $\left \{ (0,0),(8,0) \right \}$ ,$\left \{ (0,0),(0,8) \right \}$,$\left \{ (0,0),(8,1) \right \}$, possibly.  By the First isomorphism theorem:  $\phi: Z_{16}\oplus Z_2 \rightarrow Z_2\oplus Z_2$  $\Psi: G/\ker \phi \rightarrow \phi\left ( Z_16\oplus Z_2 \right )=Z_2\oplus Z_2$  $g\ker \phi  \mapsto \phi(g)=\Psi(g\ker \phi)$  Any help is appreciated.  Edit: I know that $(2,0)$ has order $8$ in $Z_{16}\oplus Z_{2}$ but any elements in $Z_{4}\oplus Z_{4}$ does not have order $8$ which would have solved the question at the outset. But I would like a different route. I have created a fishnet grid (3km x 3km) over a vector polygon.   I now want to clip the fishnet over the vector polygon retaining all grid squares within the boundary of the polygon.   Therefore I want all squares that would typically be clipped into partial squares to remain as whole squares.   Does anybody know how to do this? What is a buffer overflow and how do I cause one? I have heard about a buffer overflow and I would like to know how to cause one.  Can someone show me a small buffer overflow example? New(And what they are used for?) Where to put my code: plugin or functions.php? Seeking shapefile of city boundaries in US? Should I include a link to my Stack Overflow profile on my résumé? Form inside a table Why doesn't "cd" work in a shell script? I just want to write a script which changes my directory.  I put the below commands in the file /home/alex/pathABC  #!/bin/sh cd /home/alex/Documents/A/B/C echo HelloWorld   I did   chmod +x pathABC   In the Terminal, while in /home/alex, I ran ./pathABC, but the output is just HelloWorld and the current directory is not changed.  So what is wrong? Maximum Likelihood Estimator for $\theta$ when $X_1,\dots, X_n \sim U(-\theta,\theta)$ Exercise :     Calculate a Maximum Likelihood Estimator for the model $X_1,\dots, X_n \; \sim U(-\theta,\theta)$.   Solution :  The distribution function $f(x)$ for the given Uniform model is :  $$f(x) = \begin{cases} 1/2\theta, \; \; -\theta \leq x \leq \theta \\ 0 \quad \; \; , \quad\text{elsewhere} \end{cases}$$  Thus, we can calculate the likelihood function as :  $$L(\theta)=\bigg(\frac{1}{2\theta}\bigg)^n\prod_{i=1}^n\mathbb I_{[-\theta,\theta]}(x_i)= \bigg(\frac{1}{2\theta}\bigg)^n\prod_{i=1}^n \mathbb I_{[0,\theta]}(|x_i|) $$  $$=$$  $$\bigg(\frac{1}{2\theta}\bigg)^n\prod_{i=1}^n \mathbb I_{[-\infty,\theta]}(|x_i|)\prod_{i=1}^n \mathbb I_{[0, +\infty]}(|x_i|)$$  $$=$$  $$\boxed{\bigg(\frac{1}{2\theta}\bigg)^n\prod_{i=1}^n \mathbb I_{[-\infty,\theta]}(\max|x_i|)}$$  Question : How does one derive the final expression in the box from the previous one ? I can't seem to comprehend how this is equal to the step before.  Other than that, to find the maximum likelihood estimator you need a $\theta$ sufficiently small but also $\max |x_i| \leq \theta$ which means that the MLE is : $\hat{\theta} = \max |x_i|$. Show that $a^2 \bmod b = (a \bmod b)^2 \bmod b$ Show that  $$a^2 \mathrm{\ mod \ } b = (a \mathrm{\ mod \ } b)^2 \mathrm{\ mod\ } b$$ for $ a, b \in \mathbb{Z}^+ $.  this was derived from an Informatics olympiad question. I have a problem. I have several large equations within the text. And when LaTeX does not manage to put the equation on the same page and moves it to the next one, he spreads the paragraphs of the text in the first page. How could I prevent this? I know about \pagebreak and \newpage, but I do not want to force the \newpage, as the text may move (if I add something later or so on) and the pagebreaks may not be valid. I would like to make LaTeX not to spread the paragraphs over a page automatically. Is it possible?  I tried to make an example:  \documentclass[a4paper,12pt,twoside]{book} \usepackage{amsmath} \begin{document}  LaLaLa\\\  LaLaLa\\\  LaLaLa\\\  LaLaLa\\\  \begin{eqnarray} 123&amp;+&amp;123 \\\ 123&amp;+&amp;123 \nonumber\\\ 123&amp;+&amp;123 \nonumber\\\ 123&amp;+&amp;123 \nonumber\\\ 123&amp;+&amp;123 \nonumber\\\ \end{eqnarray}  LaLaLa\\\  LaLaLa\\\  \begin{eqnarray} 123&amp;+&amp;123 \\\ 123&amp;+&amp;123 \nonumber\\\ 123&amp;+&amp;123 \nonumber\\\ 123&amp;+&amp;123 \nonumber\\\ 123&amp;+&amp;123 \nonumber\\\ 123&amp;+&amp;123 \nonumber\\\ 123&amp;+&amp;123 \nonumber\\\ 123&amp;+&amp;123 \nonumber\\\ 123&amp;+&amp;123 \nonumber\\\ 123&amp;+&amp;123 \nonumber\\\ 123&amp;+&amp;123 \nonumber\\\ 123&amp;+&amp;123 \nonumber\\\ 123&amp;+&amp;123 \nonumber\\\ 123&amp;+&amp;123 \nonumber\\\ 123&amp;+&amp;123 \nonumber\\\ \end{eqnarray}  \end{document} 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. It's possible to do a array of Matrix in Matlab? find $\left( \frac{x}{x+y} \right)^{2007} + \left( \frac{y}{x+y} \right)^{2007}$ Proof that the set of irrational numbers is dense in reals Power Supply Capacity Formula Given that I am building a home computer from scratch and I have several components (MB, CPU, HDD, Video, Sound, etc) which have specific power requirements, is there a formula for me to calculate the size/capacity of the Power Supply I will need? I want to be able to expand the setup later on also... As small as it may be, does every 'thing' have a gravitational pull?  That is, something with mass or energy. No matter how obsolete or negligible it may be, is it there? If so, how is it calculated? What does 'it' affect? Is there a word for someone who has others do all their work for them? In a way like a slave master. More though like say someone who had robots do all their work and they were never themselves productive and had no drive. If you've read R.U.R. you'll understand the question better. I'm looking for a word with more definition than just lazy.  Update 1: It think I may actually go with drone if anything. I think that definition fits what I'm looking for the best. I also liked the definition of the British term layabout that one person mentioned, but I will probably use drone. Thanks for all the help and sorry if I was confusing or unclear.  Update 2: I needed this word for use in an essay I was writing which I have now handed in. Thanks for the help guys. No more suggestions needed. What would be the the mathematical equivalent of this excel formula? =PMT() I need to create a JavaScript form that does the same calculation as this =PMT() function.   mortgage = 220000 rate= 4.75% term = 30   The example formula I have is =PMT(4.75%/12,30*12,220000*-1)  What would be the equation for this function? I tried to look up this function but it doesn't explain it very well. This is a follow up question to my previous question  Let $k$ be a field and $V \subseteq \Bbb{A}^n$ and $W \subseteq \Bbb{A}^m$ be algebraic sets. Then it should be true that $I(V \times W ) = I(V) + I(W)$ where by $I(V)$ here we mean the extension of $I(V)k[x_1,\ldots,x_{m+n}]$. Now I believe I have proven this (see the proof at the bottom of my question) but when I look at Martin's answer , it is instead claimed that we  actually have  $$I(V \times W )= \sqrt{I(V) + I(W)},$$  and for $I(V) + I(W)$ to be a radical ideal we need $k$ to be algebraically closed.           My question is: What's going on here? I believe my claim is true even without the assumption that $k$ is algebraically closed.           Here's a proof user Sanchez told me of, which I have simplified:       First it is clear that we always have $I(V) + I(W) \subseteq I(V \times W)$. For the reverse inclusion consider a polynomial $f \in I(V \times W)$. Then we can always write $$f = \sum_{i=1}^n f_ig_i$$ where $f_i \in k[x_1,\ldots,x_n]$ and $g_i \in k[x_{n+1},\ldots,x_{m+n}]$. Now take any $b' \in W$. If for all $i$ we have $g_i(b') = 0$ then since $b'$ is arbitrary, $g_i \in I(W)$ for all $i$. Then $f \in I(W) \subseteq I(V ) + I(w)$ and we are done. Otherwise suppose there exists $b \in W$  and $i$ such that $g_i(b) \neq 0$. Then wlog we may suppose that  $g_1(b) \neq 0$.   Next, $\sum f_ig_i(b) = 0 $ on all of $V$ by assumption of $f \in I(V \times W)$. So $\sum f_i g_i(b) = p$ for some $p \in I(V)$. Now write $$f_1 = \frac{ p - g_2(b) f_2 + \ldots g_n(b)f_n}{g_1(b)}.$$  Substituting this for $f_1$ in $\sum f_ig_i$, followed by taking things mod $I(V)$ we get an  expression with only $n-1$ terms $\mod{I(V)}$. Continuing this process we will finally get an expression with $0$ terms $\mod{I(V)}$ so that $f \in I(V)$. This shows $I(V \times W) \subseteq I(V) + I(W)$ which completes the proof.