Why are 32 bit apps loaded on 64 bit OS instead of 64 bit versions
Why are processes in Windows 8 32 bit?
If you know it, also try to include the precise reason why the proof is fallacious. To start this off, let me post the one that most people know already:    Let $a = b$.  Then $a^2 = ab$  $a^2 - b^2 = ab - b^2$  Factor to $(a-b)(a+b) = b(a-b)$  Then divide out $(a-b)$ to get $a+b = b$  Since $a = b$, then $b+b = b$  Therefore $2b = b$  Reduce to $2 = 1$   As @jan-gorzny pointed out, in this case, line 5 is wrong since $a = b$ implies $a-b = 0$, and so you can't divide out $(a-b)$. I just installed ubuntu 11.10 from a usb stick.   After installation, none of the software which I had collected and run on 11.10 or 11.04 runs on it. Even my newly downloaded opera or chrome installations does not work.  I do not have a internet connection.  Whenever i try to install any .deb file by double clicking it,software center opens up but the install button is grayed out. I have just installed 12.04 LTS on my laptop and I'm kinda new in this. I am required to write some program in Fortran 77 and I need a Fortran 77 compiler. I have tried installing g77 as described in the Ubuntu website but was unable to do so. I have typed in the following in terminal:  sudo apt-get install g77   and get the following:  Reading package lists... Done Building dependency tree        Reading state information... Done Package g77 is not available, but is referred to by another package. This may mean that the package is missing, has been obsoleted, or is only available from another source  E: Package 'g77' has no installation candidate   May I know what is the problem? I have tried searching g77 in the software center but apparently it is not there. Does this mean it is not available for 12.04?  I'm currently using an Asus K43S laptop with Intel Core i5-2450M processor and my Ubuntu is installed alongside Windows. Why does SQL Server use a better execution plan when I inline the variable? Let $ f $ be a continuous function defined on $ [0,\pi] $. Suppose that      $$ \int_{0}^{\pi}f(x)\sin {x} dx=0,   \int_{0}^{\pi}f(x)\cos {x} dx=0 $$   Prove that $ f(x) $ has at least two real roots in $ (0,\pi) $ An example of a norm which can't be generated by an inner product I realize that every inner product defined on a vector space can give rise to a norm on that space.  The converse, apparently is not true.  I'd like an example of a norm which no inner product can generate. What does the notation $\Gamma \vDash \phi$ mean (in Mathematical Logic)? What does the notation $\Gamma \vDash \phi$ mean (in Mathematical Logic)?  I understand that $\Gamma$ represents a system of formulas, and that $\phi$ represents an individual formula.  I also know that $\Gamma \vDash \phi$ means " $\phi$ is a semantic consequence of $\Gamma$" - but I do not understand what this actually means. Can someone explain this idea to me in layman's terms?  Furthermore, suppose I had the following question:     Let $\Gamma \cup \{ \phi \}$. Does $\Gamma \vDash \phi$ hold for the set of formulae   $$ \Gamma = \{ p \rightarrow q, q \rightarrow r, r \rightarrow s \} \hspace{10 mm} \text{ where } \phi \text{ is } p \rightarrow s $$   Would I attempt this question by trying to prove that, for every formula $\psi \in \Gamma$, the valuation $v(\psi) = T$, based on the assumption that $v(\phi) = T$? A question occurred to me when browsing SE this evening, just curious.  What specifies the inside of a 3D construct? If I have a hollow sphere, what's to say that the world isn't the inside, and the center of the sphere isn't the outside. Picture the Asylum in the Hitchhiker's Guide to the Galaxy series, if you have read that.  Additionally, if the inside of a 3D object is defined as what is smaller than the outside, what happens when there is a a sphere with a volume equivalent to 51% of the universe? Showing that $\mathbb{R}$ and $\mathbb{R}\backslash\mathbb{Q}$ are equinumerous using Cantor-Bernstein I need to prove that $\mathbb{R}\backslash\mathbb{Q} \sim \mathbb{R} $  Using Cantor-Bernstein, need to show an injection from $\mathbb{R}\backslash\mathbb{Q}$ to $\mathbb{R}$ and from $\mathbb{R}$ to $\mathbb{R}\backslash\mathbb{Q}$.    $\mathbb{R}\backslash\mathbb{Q}$ is a subset of $\mathbb{R}$ so only need to show injection from $\mathbb{R}$ to $\mathbb{R}\backslash\mathbb{Q}$ to complete the proof.    Possible injection: $f:\mathbb{R}\to \mathbb{R}\backslash\mathbb{Q}$ defined as $f(x) = \pi x$ if $x$ is not a multiple of $\pi$; otherwise $f(x) = \sqrt{2} x$.    Not sure if $f$ actually is an injection... Suppose $f(x)$ is an irreducible polynomial over $\mathbb Z$ of degree $n$. Is it always the case that there exist distinct $x_1,\ldots,x_{2n+1}\in \mathbb Z$ such that $f(x_1),\ldots,f(x_{2n+1})$ are all prime?  So far I've tested a few irreducible polynomials using mathematica, and this has held. The motivation is to use this as a test for irreducibility, as if $f(x)$ is not irreducible then it takes on at most $2n$ prime values, as either factor takes on $1$ or $-1$ at most $2n$ times. (I'd also be interested in learning if this bound can be improved).  This could potentially be expended to other rings of integers, i.e. $\mathbb Z[i]$, but in that case the minimal number of primes needed to certify that $f(x)$ is irreducible would increase as more units appear. The question is based on the following exercise in real analysis:     Assume that $A,B\subset{\Bbb R}$ are both bounded and $x&gt;0$ for all $x\in A\cup B$. Show that $$ \sup(AB)=\sup A\sup B $$   where    $$ AB:=\{ab\in{\Bbb R}:a\in A, b\in B\}. $$   Since $0&lt;a\leq\sup A$ and $0&lt;b\leq\sup B$ for all $a\in A$ and $b\in B$, we have $$ ab\leq\sup A\sup B $$ for all $ab\in AB$ which implies that $\sup AB\leq\sup A\sup B$. I have trouble with another direction: $$ \sup AB\geq\sup A\sup B $$ I was trying to show that for every $\epsilon &gt;0$, $\sup AB-\epsilon \geq \sup A\sup B$. If one uses the definition of supremum, one has the estimates that for every $\epsilon&gt;0$, $$ \sup A-\epsilon\leq a, \quad \sup B-\epsilon\leq b $$ for some $a\in A,\ b\in B$. It follows that $$ \sup A\sup B\leq (a+\epsilon)(b+\epsilon)=ab+\epsilon(a+b)+\epsilon^2\leq \sup AB+\epsilon (a+b)+\epsilon^2 $$ which seems quite close to what I want. How can I go on? How it this possible? Even if the gif is fake, the Mythbusters did it and with a large sail it really moves forward. What is the explanation? Marketing Cloud IP Warming - is it critical that we send up to the weekly limits by domain? Updating to latest Blender version I currently have version 2.74. How do I update to version 2.76? I do not see any links in the UI to update my version of Blender Example of a relation that is symmetric and transitive, but not reflexive Can you give an example of a relation that is symmetric and transitive, but not reflexive?   By definition,   $R$, a relation in a set $X$, is reflexive if and only if $\forall x\in X$, $x\,R\,x$.  $R$ is symmetric if and only if $\forall x, y\in X$,  $x\,R\,y\implies y\,R\,x$. $R$ is transitive if and only if $\forall x, y, z\in X$, $x\,R\,y\land y\,R\,z\implies x\,R\,z$.    I can give a relation $\leqslant$, in a set of real numbers, as an example of reflexive and transitive, but not symmetric. But I can't think of a relation that is symmetric and transitive, but not reflexive. Why does shrinkage work? In order to solve problems of model selection, a number of methods (LASSO, ridge regression, etc.) will shrink the coefficients of predictor variables towards zero. I am looking for an intuitive explanation of why this improves predictive ability. If the true effect of the variable was actually very large, why doesn't shrinking the parameter result in a worse prediction? Who was/were the first Jedi? How can I get rid of this black hole in my minecraft world? I have interesting bug that I have experienced before but has gone away on its own. This one is not going away though.  I have a giant stripe or chunks missing from my desert biome, I would say probably 7 chunks long.  I'm guessing the chunks are corrupted or something, what was hoping to figure is whether it can repaired or reverted.  Here is a picture of the problem. Show that a norm on an inner product space satisfies parallelogram law. Hence use the parallelogram law to show that the space of continuous real functions defined on the interval $[a,b]$ is not a Hilbert space. Here I did the first part. Let $X$ be an inner product space and $x,y\in X$ consider $\Vert x+y\Vert ^2$ and $\Vert x-y\Vert ^2$ Then on adding I get $\Vert x+y\Vert ^2+\Vert x-y\Vert ^2=2(\Vert x\Vert ^2 +\Vert y\Vert ^2)$ For the second part I don’t know how can a parallelogram law prove $C[a,b]$ is not a Hilbert space, help me please. What is the best way to vertically center the content of a div when the height of the content is variable.  In my particular case, the height of the container div is fixed, but it would be great if there were a solution that would work in cases where the container has a variable height as well. Also, I would love a solution with no, or very little use of CSS hacks and/or non-semantic markup.