This is in refernce to testing if a given number is prime by square root method. Qs:  I understand the logic behind taking square root ( based on factor pair) but why do we consider only the prime numbers below the sq. root to test the divisibility and not all the numbers? 2). Also as any number with a square  root will have that sq. root as a factor, so why cant that be considered as another factor apart from (1 and itself) ( I guess may b we consider only integers ?)
I was going through the fundamental theorem in Number Theory where any non zero integer n can be represented as a product of distinct primes. A related problem with this theorem is to prove that for every such number, there exists a prime $p$ such that $p&lt; \sqrt n$. I was wondering if there is any mathematical proof that no prime $p$ exists for the number $n$ such that $p&gt; \sqrt n$.
My question comes from what I learned in the movies, considering I have not yet been able to read the books. I would just like to know why Severus Snape called himself the Half-Blood Prince? Did many people call him that or did he just think that it was a cool nickname? This question might be answered in the books so feel free to quote from it. Why has my database connection failed? Let $E$ be a metric $\mathbb R$-vector space, $I$ be a nonempty set and $(\mu_t)_{t\in I},(\nu_t)_{t\in I}$ be families of nonnegative finite measures on $\mathcal B(E)$.  that if $(\mu_t)_{t\in I}$ and $(\nu_t)_{t\in I}$ are tight$^1$ and bounded in total variation norm, then the family of convolutions $(\mu_t\ast\nu_t)_{t\in I}$ is tight and bounded in total variation norm as well.  Now I would like to show the converse: If $(\mu_t)_{t\in I}$ and $(\mu_t\ast\nu_t)_{t\in I}$ are tight, then $(\nu_t)_{t\in I}$ is tight. (Most probably we need to impose a boundedness assumption on $(\mu_t)_{t\in I}$ and on $(\mu_t\ast\nu_t)_{t\in I}$ or even a priori on $(\nu_t)_{t\in I}$.)  For the moment, I'm only able to prove the claim assuming that the $\mu_t,\nu_t$ are probability measures. Under this assumption, the proof goes as follows: Let $\varepsilon&gt;0$. Since $(\mu_t)_{t\in I}$ and $(\mu_t\ast\nu_t)_{t\in I}$ are tight, there is a compact $K\subseteq E$ with $$\sup\left(\sup_{t\in I}\mu_t(K^c),\sup_{t\in I}(\mu_t\ast\nu_t)(K^c)\right)&lt;\varepsilon\tag1.$$ Now $K-K$ is compact and \begin{equation}\begin{split}1-\varepsilon&lt;(\mu_t\ast\nu_t)(K)&amp;=\underbrace{\int_K\mu_t({\rm d}x)\nu_t(\underbrace{K-x}_{\subseteq\:K-K})}_{\le\:\nu_t(K-K)}+\underbrace{\int_{K^c}\mu_t({\rm d}x)\underbrace{\nu_t(K-x)}_{\le\:1}}_{&lt;\:\varepsilon}\\&amp;&lt;\nu_t(K-K)+\varepsilon\end{split}\tag2\end{equation} for all $t\in I$, which immediately yields $$\nu_t(K-K)&gt;1-2\varepsilon\;\;\;\text{for all }t\in I\tag3.$$  Can we drop the probability measure assumption?  EDIT I'm not able to obtain the claim, even with the help of the answer below. If $\varepsilon&gt;0$, the goal is to show that $$\nu_t(E)-\nu_t(K-K)=\nu_t((K-K)^c)&lt;\varepsilon\tag4.$$ If I'm choosing all the constants as generic as possible, we obtain $$\mu_t(E)\le c_1\tag5$$ $$\mu_t(K^c)\le k_1(\varepsilon)\tag6$$ $$(\mu_t\ast\nu_t)(K)&lt;c_1\nu_t(K-K)+c_2k_1(\varepsilon)\tag6$$ $$\mu_t(E)\ge\alpha_1\tag7$$ $$(\mu_t\ast\nu_t)(E)\le c_3\tag8$$ $$\nu_t(E)\le\frac{c_3}{\alpha_1}=:c_2\tag9$$ $$(\mu_t\ast\nu_t)(E)\ge\alpha_3\tag{10}$$ $$\nu_t(E)\ge\frac{\alpha_3}{c_1}\tag{11}$$ $$(\mu_t\ast\nu_t)(K^c)&lt;k_3(\varepsilon)$$ The assumptions that I'm imposing is that $(\mu_t)$ and $(\mu_t\ast\nu_t)$ are bounded above (by $c_1$ and $c_3$) and below (by $\alpha_1$ and $\alpha_3$) in total variation norm and are tight (so that we can find $K$ for arbitrary $k_1(\varepsilon)$ and $k_3(\varepsilon)$). In order to obtain the claim we need to choose $k_1(\varepsilon)$ and $k_3(\varepsilon)$ such that $(4)$ holds. From everything above, we obtain $$\nu_t(E)-\nu_t(K-K)&lt;\frac{c_3}{\alpha_1}-\frac{\alpha_3}{c_1}+\frac{c_3}{\alpha_1c_1}k_1(\varepsilon)+\frac1{c_1}k_3.(\varepsilon)\tag{12}$$ Even if we simplify this by assuming $\alpha_1=\alpha_3$, $c_1=c_3$ and $k_1(\varepsilon)=k_3(\varepsilon)$, the equation $$\frac{c_3}{\alpha_1}-\frac{\alpha_3}{c_1}+\frac{c_3}{\alpha_1c_1}k_1(\varepsilon)+\frac1{c_1}k_3.(\varepsilon)=\varepsilon\tag{13}$$ doesn't seem to admit a solution $(k_1(\varepsilon),k_3(\varepsilon))$ with positive $k_1(\varepsilon)$ and $k_3(\varepsilon)$; unless $\frac{c_3}{\alpha_1}-\frac{\alpha_3}{c_1}\ge0$; which clearly doesn't need to be the case (it is clearly the case for probability measures though ...)  $^1$ Remember that a family $\mathcal F$ of finite signed measures on $\mathcal B(E)$ is called tight if for all $\varepsilon&gt;0$, there is a compact $K\subseteq E$ such that $$\sup_{\mu\in\mathcal F}|\mu|(K^c)&lt;\varepsilon.$$ Why is the plural of "basis" "bases" and not "baseis"? Looking at the , it indicates that the plural is either bases or baseis.  It looks like the rare  comes from the Greek, but the common  just refers back to basis and that does not explain the plural etymology.  Is there a standard rule at work here, or is this just an exceptional case that arose over time by convention? Write this equation I want to write this equation in this way down but I have been unsuccessful. How can I improve this codes in order to get the equation below?  \[ DeltaC{t} = \left\{ \begin{array}{ll}  -1.40- 0.588w_{t-1} - 0.5351\DeltaC_{t-1}+ 0.169\DeltaBC_{t-1} &amp; \mbox{ if $w_{t-1}&lt;- 0.62$} \\ -1.40 - 0.588w_{t-1} - 0.5351\DeltaC_{t-1} + 0.169\DeltaBC_{t-1} &amp; \mbox{$w_{t-1}&gt;0.62$} \end{array} \right. \] Add new alias to Gmail without SMTP (forwarding-only address) I used to be able to go to Accounts and Import → Add another email address you own and add an email account that's only a forwarding address. That is, it's not a mailbox and I don't have SMTP access to it.  Gmail would send me a confirmation email to that address, I'd confirm it's mine, and done. Now it only gives me the option to add SMTP credentials.  Did they stop supporting the previous method? Is there any other way I can get that option back? Proving that $(3n)!$ is divisible by $n! \times (n + 1)! \times (n + 2)!$ if $n$ is greater than 2 Prove that:  If $n$ is greater than 2, then $(3n)!$ is divisible by $n! \times  (n + 1)! \times  (n + 2)!$  From Barnard &amp; Child's "Higher Algebra".  I know that the highest power of a prime $p$ contained in $N$! is  $$ \left\lfloor{\frac{N}{p}}\right\rfloor + \left\lfloor{\frac{N}{p^2}}\right\rfloor + \left\lfloor{\frac{N}{p^3}}\right\rfloor ... $$  I'm unable to show that the formula above gives a higher value for $N = 3n$ than for the sum of its values when $N$ = $n$ , $n + 1$ and $n + 2$, considering the condition that $n &gt; 2$. In my experimentation, I've used only BJTs as switches (for turning on and off things like LEDs and such) for my MCU outputs. I've been repeatedly told, however, that N-channel enhancement-mode MOSFETs are a better choice for switches (see  and , for examples), but I'm not sure I understand why. I do know that a MOSFET wastes no current on the gate, where a BJT's base does, but this is not an issue for me, as I'm not running on batteries. A MOSFET also requires no resistor in series with the gate, but generally DOES require a pull-down resistor so the gate doesn't float when the MCU is rebooted (right?). No reduction in parts count, then. There doesn't seem to be a great surplus of logic-level MOSFETs that can switch the current that cheap BJTs can (~600-800mA for a 2N2222, for example), and the ones that do exist (TN0702, for example) are hard to find and significantly more expensive. When is a MOSFET more appropriate than a BJT? Why am I continually being told that I should be using MOSFETs? How to make a curve path from scratch given a list of (x, y, z) points? How to compute $m$ value from RSA if $phi(n)$ is not relative prime with the $e$? Here is some information we got : We know the value of $n$, with size $1043$. We know the value of $p$ (size $20$) and $q$ (size $1023$) as the factors. $e = 65537.$ $\varphi(n)$ = $(q-1)(p-1)$ When I calculated $\gcd$ and $\text{modinv}$, I got : $\gcd(e,\varphi(n)) = 65537$ $modinv(e,\varphi(n)) = 1 $ So we can tell that they are not relatively prime. So, how to compute the d, and get the value of m? I'm not that good with math, so I cant understanding well the theory. so can anyone please make an example implementation or write a clear formula? Boot failure : No DEFAULT or UI configuration directive found! Split string on spaces in Java, except if between quotes (i.e. treat \"hello world\" as one token) Is there a way to see the history of questions protected by me or another user? Having a record of all actions taken on a site, as is provided by the &quot;activity&quot; and &quot;votes&quot; tabs of our profile, comes in handy very often. Perhaps the most common case is being able to find a question I've interacted with in the past, to see if anything has changed about it. This applies to questions a user has protected as much as anything else. Users are theoretically encouraged to unprotect questions when it's no longer necessary, as emphasized , but in practice they often do not. In many cases it's hard to blame them: unless they specifically marked it as a favourite, the site doesn't keep any record of the question to allow them to find it again. We should help users more easily review their own protections in the future.    For personal review, the votes tab is adequate, but there may also be value in displaying this publicly in the activity tab. There may be users with a history of repeatedly protecting questions over-eagerly. It would be useful to be able to review other users' protection histories to see if this is the case. Even if this is not available to regular users, it may be helpful to moderators, unless their user activity views already allow them to filter down to see this. How to prove that the sum and product of two algebraic numbers is algebraic? Suppose $E/F$ is a field extension and $\alpha, \beta \in E$ are algebraic over $F$. Then it is not too hard to see that when $\alpha$ is nonzero, $1/\alpha$ is also algebraic. If $a_0 + a_1\alpha + \cdots + a_n \alpha^n = 0$, then dividing by $\alpha^{n}$ gives $$a_0\frac{1}{\alpha^n} + a_1\frac{1}{\alpha^{n-1}} + \cdots + a_n = 0.$$     Is there a similar elementary way to show that $\alpha + \beta$ and $\alpha \beta$ are also algebraic (i.e. finding an explicit formula for a polynomial that has $\alpha + \beta$ or $\alpha\beta$ as its root)?    The only proof I know for this fact is the one where you show that $F(\alpha, \beta) / F$ is a finite field extension and thus an algebraic extension. How to get text in between two sections? I want to get text in between START &amp; STOP. Remaining text should be deleted, like some_text_1 and some_text_2.  some_text_1 some_text_1 . . . START --- Values that I want to get ... STOP some_text_2 some_text_2   From my findings, sed does similar job, but only oposite, like: sed '/START/,/STOP/d' file which would give me everything else then content in between START &amp; STOP.  Any idea? What is the " := " operator in (VB).NET or what is it good for? How to prove $a^2 + b^2 + c^2 \ge ab + bc + ca$? How can the following inequation be proven?  $$a^2 + b^2 + c^2 \ge ab + bc + ca$$ I'm trying to prove that $e^{\sqrt 2}$ is irrational. My approach: $$ e^{\sqrt 2}+e^{-\sqrt 2}=2\sum_{k=0}^{\infty}\frac{2^k}{(2k)!}=:2s $$ Define $s_n:=\sum_{k=0}^{n}\frac{2^k}{(2k)!}$, then: $$ s-s_n=\sum_{k=n+1}^{\infty}\frac{2^k}{(2k)!}=\frac{2^{n+1}}{(2n+2)!}\sum_{k=0}^{\infty}\frac{2^k}{\prod_{k=1}^{2k}(2n+2+k)}\\&lt;\frac{2^{n+1}}{(2n+2)!}\sum_{k=0}^{\infty}\frac{2^k}{(2n+3)^{2k}}=\frac{2^{n+1}}{(2n+2)!}\frac{(2n+3)^2}{(2n+3)^2-2} $$ Now assume $s=\frac{p}{q}$ for $p,q\in\mathbb{N}$. This implies: $$ 0&lt;\frac{p}{q}-s_n&lt;\frac{2^{n+1}}{(2n+2)!}\frac{(2n+3)^2}{(2n+3)^2-2}\iff\\ 0&lt;p\frac{(2n)!}{2^n}-qs_n\frac{(2n)!}{2^n}&lt;\frac{2}{(2n+1)(2n+2)}\frac{(2n+3)^2}{(2n+3)^2-2} $$ But $\left(p\frac{(2n)!}{2^n}-qs_n\frac{(2n)!}{2^n}\right)\in\mathbb{N}$ which is a contradiction for large $n$. Thus $s$ is irrational. Can we somehow use this to prove $e^\sqrt{2}$ is irrational? Ubuntu 20.04 graphics driver problem I have upgraded from Ubuntu 19.10 to 20.04, and now I have a problem with my graphics rendering (strange squares on background and font problems). I use the i915 driver. Graphics card: Skylake GT2 HD Graphics 520. I tried to reinstall GNOME shell and Xorg server. Nothing helped. Also Chrome browser breaks on most of pages. Calculate last day of month If you provide 0 as the dayValue in Date.setFullYear you get the last day of the previous month:  d = new Date(); d.setFullYear(2008, 11, 0); //  Sun Nov 30 2008   There is reference to this behaviour at . Is this a reliable cross-browser feature or should I look at alternative methods?