Show that if $a$ and $h$ are positive numbers with $h&lt;a^2$ then   $$\sqrt{a^2+h}-a&lt;\frac{h}{2a}&lt;a-\sqrt{a^2-h}$$
Show that, if $a$ and $h$ are positive numbers, $h &lt; a^2$, then $$\sqrt{a^2 + h}-a &lt; \frac{h}{2a} &lt; a - \sqrt{a^2 - h}$$  I've been working on this problem for about 2 hours now, but I've made no progress. I'm not looking for an answer, but I just need some help to get me started since we didn't practice inequalities this complex in highschool. Thanks.   All I can tell is that we're supposed to take the square root of an expression at some point since one inequality ( h &lt; a^2 ) becomes two.   Edit : Thank you guys for the replies, but I'd appreciate only hints in the future (like Robert Israel) so that I can learn. Regardless, I found a different way to do it, so it's cool :)
Ok, so at my school, everybody was supplied with a Macbook Pro. I set up the Messages app and it was all great until the county decided to block it. Now, instead of receiving text messages on my phone they go to my Mac. I have no idea how to fix this because I can't open the Messages app. Please help?! Resampling GeoTIFF images to same resolution using QGIS I have two geotiff images that I would like to resample the same resolution. If I use ArcGIS Desktop this is easy to do but how do I do this using QGIS? How can I show that if the second fundamental form of a surface is identically equal to zero, then the surface is a plane? This is my question:  Let P be a plane considered as a surface in 3-space. Show that its second fundamental form is zero. Conversely, show that if the second fundamental form of a surface is identically zero then the surface is a plane.  So far I have that the general equation of the plane is $$ax+by+cz+d=0.$$ Then $$M(u,v)=(u,v,\alpha u + \beta v + \gamma),$$ where $a,b,c,d,\alpha,\beta, \gamma$ are all constants and $c\not=0$. Then $M_u=(1,0,\alpha)$ and $M_v=(0,1,\beta)$. I know that $\vec{n} =\frac{M_u \times M_v}{|M_u \times M_v|}$ which gives $$ \vec{n}=\frac{(-\alpha,-\beta, 1)}{\sqrt{\alpha^2+\beta^2+1}}.$$ Hence, the unit norm is a constant. $$ M_{uu}=M_{uv}=M_{vv}=0$$ Hence $l=m=n=0$, so the second fundamental form is $$ ldu^2+mdudv+ndv^2=0.$$  Now I need to show the converse, but I don't know how. Please help :) Logical explanation why exponentiation operation is not commutative and associative Considering Peano axioms we'll define addition, multiplication and exponentiation operations.  We can then prove that addition and multiplication operations are commutative and associative. The proof for addition operation is pretty straightforward but for for multiplication it a bit more difficult. For example, to prove associativity, we have to prove distributivity first.  Moreover, for the definition itself it is not obvious that commutativity should take place, because the definition is not symmetric.  $a \cdot 0 := 0$  $a \cdot S(b) := a \cdot b + a$  Here $S(b)$ is the "next" function from Peano axioms  This definition is not symmetric with respect to $a$ and $b$, so commutivity is not obvious.  For exponentiation operation neither commutivity nor associativity take place. And I am very curious about the logical explanation why that happens  NOTE: The explanation I am looking for is somehow similar to a Abel–Ruffini theorem which explains WHY there is no an algebraic solution for a general polynomial equation I forgot my username, how do I get it back? I extracted simple example:  require 'pp' x = 1..3 pp x.map do |i| {:value =&gt; i, :double =&gt; (i*2)} end pp x.map { |i| {:value =&gt; i, :double =&gt; (i*2)} }  pp(x.map do |i| {:value =&gt; i, :double =&gt; (i*2)} end) pp(x.map { |i| {:value =&gt; i, :double =&gt; (i*2)} })   I am wondering why first pp produces:  [1, 2, 3]   While all the oders are giving:  [{:value=&gt;1, :double=&gt;2}, {:value=&gt;2, :double=&gt;4}, {:value=&gt;3, :double=&gt;6}]   I assume it has something to do with operator precedence. Where can I find good explanation? Prevent Windows 7 from sleep + power button hook I'm used to not shutdown Windows but to sleeping it. I set up power button (PB) action to do it. When I want to shutdown I must go to start -> Shutdown. That's OK.  Now I think of a mechanism that will be hooked up to a PB. Should be started after I press PB but before OS will go to sleep. Something that will check few conditions (i.e. running backup, running AV scan, ...) and if true (something is running) prevent go to sleep.  Optionally: It would be nice to start the same "mechanism" before restart and shutdown (invoked not only from Start menu, but also by other software (i.e. installers) but with option to allow some sources (i.e. UPS software)  Generally: How to run something after PowerButton but before Sleeping. How to code (C#.NET, vbs, batch (preferred)) sleep prevention (cancellation). Badge Name is Blue? Now some random badges on the Profile page for the mobile web is now blue! This is seemingly weird as the default and now permanent color for Badge name text is white. Here's the problem here:     Ahhhh! Now it has invaded the desktop/computer version too! Now seemingly, every version has badges with blue text. Android app has the same issues too! Can the badge color be fixed from blue to white on the mobile version please? Irreducibility of $X^n - 2$ over $\mathbf Z[i]$. I was “playing” with Eisenstein's criterion and the polynomials $X^n-p$ ($p$ being a prime number).  Over $\mathbf Z$, the classical Eisenstein's criterion directly applies and shows that $X^n - p$ is an irreducible element of $\mathbf Z[X]$. The question was “what happens over $\mathbf Z[i]$?”  We know that in this new ring, a prime number $p$ has one of three different fates:   if $p \equiv -1\ (\mathrm{mod}\, 4)$, $p$ still is an irreducible element of $\mathbf Z[i]$, or equivalently $(p) \subset \mathbf Z[i]$ still is prime. (Unless I'm mistaken, number theorists say that $p$ is inert). Eisenstein's criterion still applies and the polynomial is irreducible over $\mathbf Z[i]$. if $p \equiv 1\ (\mathrm{mod}\, 4)$, $p$ can be written as a sum of two squares: $p = n^2 + m^2$ and is therefore reducible on $\mathbf Z[i]$: $p = (n+im)(n-im)$. These two factors aren't associated (the units in $\mathbf Z[i]$ being $\pm 1$ and $\pm i$) so $(p)$ splits: $(p) = (n+im)(n-im)$ is the product of two different prime ideals. One can still use Eisenstein's criterion with either of these prime ideals, and $X^n - p$ is irreducible over $\mathbf Z[i]$. if $p = 2$, the oddest of all primes, one still has a factorisation $p = (1+i)(1-i)$ but this time, $1+i$ and $1-i$ are associated. In NT slang, $(2)$ is ramified: $(2) = (1+i)(1-i) = (1+i)^2$ and one cannot directly use Eisenstein's criterion.   So here's my question: how do we prove (or disprove) the irreducibility of $X^n - 2$ over $\mathbf Z[i]$? Roots in a polynomial ring The 3 different equals How to write a simple jQuery plugin Catenatives followed by infinitives and gerunds What is the difference in meaning when the catenative verb “like” is followed by an infinitive, or by a gerund? For example:      Do you like ski jumping?    vs.      Do you like to ski jump?   Also, what is the difference between:      My brother taught me to read and write.    vs.       My brother taught me reading and writing. Execute command from file in current shell I want to executed command from file (which is script) in current shell in terminal.  Example: I have file ch_dir contains following:  #!/bin/bash cd /usr   Now if I execute file usually as following then it executes commands in different shell:  pandya@pandya-desktop:~$ ./ch_dir pandya@pandya-desktop:~$   Here cd /user is executed in different shell.  But I want to execute in Current shell. How do I do that?  Note: Here cd is only used to explain. Question's aim is to execute any command in current shell Find duplicate files Is it possible to find duplicate files on my disk which are bit to bit identical but have different file-names? I understand that division by zero isn't allowed, but we merely just multiplied $f(x) = 1$ by $\frac{x-1}{x-1}$ to get $f(x) = \frac{x-1}{x-1}$ and $a\cdot 1 = 1\cdot a = a$ so they're the same function but with different domain how is this possible?  Or in other words why don't we simplify $f(x) = \frac{x-1}{x-1}$ to $f(x) = 1$ before plotting the points. Is it just defined this way or is there a particular reason ?  Note: my book says the domain of $f(x) = 1$ is $\mathbb{R}$ and the domain of  $f(x) = \frac{x-1}{x-1}$ is $\mathbb{R}$ except $1$. All the info about TLP on the surface looks more like advertisement and marketing to me, I can't see any detailed comparison to default configurations. What are the precise differences compared to default (K/L)Ubuntu installs under the hood, are they really effective?  Update: And if they really are, why aren't they a part of a default config? How to get my own IP address and save it to a variable in a shell script? I get this error in the deployment step of a share-point farm based project. Most of the suggested solutions on the net are a simple Visual Studio re-start, however, a simple VS restart won't fix this.   Any ideas on what might be causing this? Question : $$\begin{align}a+b+c &amp;= 0\\ a^3 +b^3 +c^3 &amp;= 12\\ a^5 +b^5 +c^5 &amp;= 40\end{align}$$ Then , $a^4 + b^4 + c^4 = ?$  My try : As a common perspective I just went to find any trick related to it and my first step is as usual as a common man will think $3abc = 12$ and got $abc = 4$ After that I am unable to link it with any other equation