Let $r$ be a rational no. with $0&lt;r&lt;1$. Then $r$ can be expressed as a sum of reciprocals of finitely many distinct positive integers . Show that this is true even for $r&gt;1$.  If there is a particular theorem or an article on this please share..
Let $x\in\mathbb{Q}$ with $x&gt;0$.  Prove that we can find $n\in\mathbb{N}^*$ and distinct $a_1,...,a_n \in \mathbb{N}^*$ such that $$x=\sum_{k=1}^n{\frac{1}{a_k}}$$
Why is a geometric progression called so? Just curious about why geometric progression is called so. Is it related to geometry? Simple property of a valuation on a field Double limit of $\cos^{2n}(m! \pi x)$ at rationals and irrationals I stumbled upon this "relation" (is the name correct?):  $$ \lim_{m \to \infty} \lim_{n \to \infty} \cos^{2n}(m! \pi x) = \begin{cases} 1,&amp;x\text{ is rational}\\ 0,&amp;x\text{ is irrational}\end{cases} $$  How is it called and why is it so? I'm really not asking for a proof since I fear it would be too complicated for me to understand, but rather for an "intuition". The ideal generated by $2$ and $ x$ in $\mathbb{Z}[x]$ Prove that $\sqrt{n} \le \sum_{k=1}^n \frac{1}{\sqrt{k}} \le 2 \sqrt{n} - 1$ is true for $n \in \mathbb{N}^{\ge 1}$ I'm trying to solve  proposed by the department of mathematics of Oxford University. I don't know how to give a valid proof for the third one which says the following:     Prove that for $n \in \mathbb{N}^{\ge 1}$: $$ \sqrt{n} \le  \sum_{k=1}^n \frac{1}{\sqrt{k}} \le 2 \sqrt{n} - 1 $$   My first approach has been to convert the summation into a function in terms of $n$ so that I could find the maximum and minimum values in the interval. However, there doesn't seem to be an easy way to express the summation as a function. Thanks in advance! I'm not sure how to prove these basic theorems in propositional calculus. Instead of using the standard axioms, we're supposed to use:   Deduction Theorem (if $\Phi, \alpha \vdash \beta$ then $\Phi \vdash \alpha \to \beta$),  Reductio (if $\Phi, \alpha \vdash \, $, then $\Phi \vdash \lnot \alpha$), Cut Rule (if $\Phi \vdash \alpha$ and $\Psi, \alpha \vdash \beta$ then $\Phi \cup \Psi \vdash \beta$),  Inconsistency Effect (if $\Phi \vdash \, $, then $\Phi \vdash \beta$ for every formula $\beta$), and  the Principle of Indirect Proof (if $\Phi, \lnot \alpha \vdash \, $, then $\Phi \vdash \alpha$),    as all the axioms can be deduced using these theorems.  I don't really know how to start the proofs without using the axioms:  i) prove that $\lnot(\alpha \to \beta) ⊢ \alpha$  ii) prove that $\lnot\alpha \vdash \alpha \to \beta$  Any suggestions on how to start these proofs or any insight at all would be greatly appreciated!  Thanks! Will spoiled food always make you sick? Preventing robots from crawling specific part of a page The Data Interoperability Tools .tbx appears in my system toolboxes, however when I double click quick export or import, a red X appears.  I have gone into the administrator settings and the tool doesn't appear to be installed. How do I install it?  I am using ArcGIS 10.2 for desktop trial.  After I attempt to evaluate extensions, I receive a message which states all requested features have already been authorized on this machine. markdown doesn't work properly Verbatim Environment in Figure Caption I was reading through the Wikipedia article on  and came across the fact that the most distant Quasar is 29 Billion Light years. This is what the  exactly says      The highest redshift quasar known (as of June 2011[update]) is ULAS-J1120+0641, with a redshift of 7.085, which corresponds to a proper distance of approximately 29 billion light-years from Earth.   Now I come to understand that the  singularity is believed to be around 13.8 Billion years ago.  So how is this possible? Does the presence of such a quasar negate the Big Bang Theory?   I'm not a student of Physics and was reading this out of (whimsical) curiosity. Is there something I'm missing here or the "proper distance" mentioned in the fact is a concept that explains this?  Edit: My Bad! Here's how..  A simple google search led me to  which says the farthest quasar found is 12.9 billion LYs and not 29 billion. So in the end we have just proven that wikipedia needs more moderation. Spacial curvature and expanding space If we take the analogy that in an empty space the space is just a flat sheet then if there is a single planet or a star then the flat sheet will curve below the planet leaving a curvature shaped like a hemisphere below the object, my question is, Does this curvature change as space expands? How to derive $[x_i, F(\vec p)] = i \hbar \frac {\partial F(\vec p)}{\partial p_i}$ indicates that the following relation is &quot;easily shown&quot;: $[x_i, F(\vec p)] = i \hbar \frac {\partial F(\vec p)}{\partial p_i}$, however I'm having some trouble showing it.  I think I'm just messing up the multivariable Taylor expansion (of $F(\vec p)$).  Can one of you walk me through it or link me to site that will?  Thanks.  Edit: Here's what I get (without using $x=i\hbar \frac \partial {\partial p}$ which I haven't proven yet): $$F(\vec p) = F(\vec 0) + \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}p_j + \frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} p_j p_k + \dots$$ so $$(x_iF(\vec p) - F(\vec p)x_i)\psi$$ $$= x_i[F(\vec 0)\psi -i\hbar \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}(\nabla \psi)_j + \hbar^2\frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} (\nabla \psi)_j (\nabla \psi)_k + \dots]$$ $$- [F(\vec 0)x_i\psi -i\hbar \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}(\nabla x_i\psi)_j + \hbar^2\frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} (\nabla x_i\psi)_j (\nabla x_i\psi)_k + \dots]$$ where $$(\nabla x_i\psi)_j=\frac {\partial x_i}{\partial x_j}\psi + x_i\frac {\partial \psi}{\partial x_j} = \delta_{ij}\psi + x_i\frac {\partial \psi}{\partial x_j}$$ and $$(\nabla x_i\psi)_j(\nabla x_i\psi)_k=(\delta_{ij}\psi + x_i\frac {\partial \psi}{\partial x_j})(\delta_{ik}\psi + x_i\frac {\partial \psi}{\partial x_k}) = \delta_{jk}\psi^2 + x_j \psi \frac {\partial \psi}{\partial x_k} + x_k \frac {\partial \psi}{\partial x_j} \psi + x_i^2 \frac {\partial^2 \psi}{\partial x_j \partial x_k}$$ Thus: $$(x_iF(\vec p) - F(\vec p)x_i)\psi$$ $$=  x_i[F(\vec 0)\psi -i\hbar \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}\frac {\partial \psi}{\partial x_j} + \hbar^2\frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} \frac {\partial \psi}{\partial x_j} \frac {\partial \psi}{\partial x_k} + \dots]$$ $$- [F(\vec 0)x_i\psi -i\hbar \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}(\delta_{ij}\psi + x_i\frac {\partial \psi}{\partial x_j}) + \hbar^2\frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} (\delta_{jk}\psi^2 + x_j \psi \frac {\partial \psi}{\partial x_k} + x_k \frac {\partial \psi}{\partial x_j} \psi + x_i^2 \frac {\partial^2 \psi}{\partial x_j \partial x_k}) + \dots]$$ From here it doesn't look like those higher order terms are all going to cancel out. Every finite group is isomorphic to some Galois group for some finite normal extension of some field. Groups in derived series are normal in the whole group I am considering a finitely generated, solvable group $G$ which has subexponential growth. Let $G=G_0\vartriangleright G_1\vartriangleright....\vartriangleright G_d\vartriangleright G_{d+1}=1$ denote the derived series of G.  It should be the case that $G_d$ is normal in $G$. Does anyone know why this is true? I don't know whether all of the assumptions above are necessary for the result. Another way to evaluate $\int\frac{\cos5x+\cos4x}{1-2\cos3x}{dx}$? What I've done is this:$$\int\dfrac{\cos5x+\cos4x}{1-2\cos3x}{dx}$$ $$\int \dfrac{\sin 3x}{\sin 3x}\left[\dfrac{\cos5x+\cos4x}{1-2\cos3x}\right]{dx}$$ $$\dfrac {1}{2}\int\dfrac{\sin 8x -\sin 2x +\sin 7x -\sin x}{\sin 3x - \sin 6x}$$ $$-\dfrac {1}{2}\int\dfrac{ \sin \frac{7x}{2} +\sin \frac{5x}{2} } {\sin \frac{3x}{2} }$$ $$-\int\dfrac{ \sin {3x}\cos \frac{x}{2} } {\sin \frac{3x}{2} }$$ $$-\int\dfrac{2\sin \frac{3x}{2} \cos \frac{3x}{2}\cos \frac{x}{2} } {\sin \frac{3x}{2} }$$ $$-\int {2\cos \frac{3x}{2}\cos \frac{x}{2} }$$ $$ -\left(\frac{\sin 2x}{2} +\sin x \right) +c $$ Is there any other way to do so ? Is it possible to do it by substitution ? Why is $1$ not a prime number? Why is $1$ not considered a prime number?  Or, why is the definition of prime numbers given for integers greater than $1$? What is the concept of erasure in generics in Java? Stone-Weierstrass theorem with $p(1/x)$