While working on some field extensions I came across this "proof" that $i = 1$, but I am not sure what is wrong with it:   We know $ i = \sqrt{-1} = (-1)^{1/2}$. Now we have the following equation: $$\begin{align} i \sqrt[4]{-2} =(-1)^{1/2}(-2)^{1/4} &amp;= (-1)^{1/2}(-1)^{1/4}(2)^{1/4}  \\ &amp;= (-1)^{3/4}(2)^{1/4}\\  &amp;= (-1)^{1/4}(2)^{1/4} \\ &amp;= \sqrt[4]{-2} \end{align}$$  Dividing both sides by $\sqrt[4]{-2}$  would yield $i = 1$.
I know there must be something unmathematical in the following but I don't know where it is: \begin{align} \sqrt{-1} &amp;= i \\\\\ \frac1{\sqrt{-1}} &amp;= \frac1i \\\\  \frac{\sqrt1}{\sqrt{-1}} &amp;= \frac1i \\\\  \sqrt{\frac1{-1}} &amp;= \frac1i \\\\  \sqrt{\frac{-1}1} &amp;= \frac1i \\\\  \sqrt{-1} &amp;= \frac1i \\\\  i &amp;= \frac1i \\\\  i^2 &amp;= 1 \\\\  -1 &amp;= 1 \quad !!? \end{align}
proving of inequality $(n!)^2 \leq n^n\cdot n!\leq (2n!)$, where $n\in \mathbb{N}$ How can we prove the inequality $(n!)^2 \leq n^n\cdot n!\leq (2n!)$, where $n\in \mathbb{N}$  $\bf{My \; Try}$::  $1\leq n$  $2\leq n$  $3\leq n$  ....  ....  $n\leq n$  So $1\cdot 2 \cdot 3 ..n \leq n^n$  So $n!\leq n^n\Rightarrow (n!)^2 \leq n^n\cdot n!$  But  did not understand how can we prove $n^n\cdot n! \leq (2n!)$  Help me  Thanks Let $n$ is a natural number. Prove that $$\left\lfloor\frac{n}{1}\right\rfloor+\left\lfloor\frac{n}{2}\right\rfloor+....+\left\lfloor\frac{n}{n}\right\rfloor+\left\lfloor{\sqrt{n}}\right\rfloor$$   is even.   I used induction and the inequality $$x-1&lt;\lfloor{x}\rfloor\le{x}$$ to prove it.  Is there any other, nicer way to do it? Cannot install R-forge package using install.packages Which of the following sentences is/are incorrect? ("Permit" vs. "allow" vs. "let") Which of the following sentences is/are incorrect? Why?   The visa permits you to study for two months. My father would never allow me to study English Let me to go. You're hurting me.   This is question from an English learning book. All the sentences seem right to me. Do you know which one is incorrect and why? Equation of the locus of the foot of perpendicular from any focus upon any tangent to the ellipse ${x^2\over a^2}+{y^2\over b^2}=1$ Find the equation of the locus of the foot of perpendicular from any focus upon any tangent to the ellipse ${x^2\over a^2}+{y^2\over b^2}=1$. will it also be an ellipse? Installing mysql-workbench in Ubuntu 19.04 The error looks like this:  Reading package lists... Done Building dependency tree        Reading state information... Done Package mysql-workbench 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 'mysql-workbench' has no installation candidate   How do I install it? Every time I call \section{**Section Name**}, a number appears to the left of the name of the section.  How do I hide this number? Editing default custom object home screen I would like to add a custom button (URL to Flow) to my default home page for my custom object. Currently, I have a default "New" button to create a new record (see attached screen shot). I'm wanting to add my Flow custom button to this page. Is there a way to do this?  Cheers, Sean. I understand there is a method for solving simultaneous modular equations. For example; $$x = 2 \mod{3}$$ $$x = 3 \mod{5}$$ $$x = 2 \mod{7}$$ We find numbers equal to the product of every given modulo except one of them - giving $5 \cdot 7$, $3 \cdot 7$ and $3 \cdot 5$. We then find the multiplicative inverses of these numbers with modulo equal to the number missing from the product. The numbers found are then 2, 1 and 1 in this case. The value of x is then given by: $$x = 2 \cdot (5 \cdot 7) \cdot 2 + 3 \cdot (3 \cdot 7) \cdot 1 + 2 \cdot (3 \cdot 5) \cdot 1 = 233 = 23 \mod{3\cdot5\cdot7}$$  But I do not understand how this method correctly gives the value of $x$. I understand that the Chinese remainder theorem proves that there is a unique value of $0\le x \lt 3\cdot5\cdot7 \mod{3\cdot5\cdot7}$ but can someone please explain why this method finds this value of x? Closed form for $ \int_0^\infty {\frac{{{x^n}}}{{1 + {x^m}}}dx }$ I've been looking at  $$\int\limits_0^\infty  {\frac{{{x^n}}}{{1 + {x^m}}}dx }$$  It seems that it always evaluates in terms of $\sin X$ and $\pi$, where $X$ is to be determined. For example:  $$\displaystyle \int\limits_0^\infty  {\frac{{{x^1}}}{{1 + {x^3}}}dx = } \frac{\pi }{3}\frac{1}{{\sin \frac{\pi }{3}}} = \frac{{2\pi }}{{3\sqrt 3 }}$$  $$\int\limits_0^\infty  {\frac{{{x^1}}}{{1 + {x^4}}}dx = } \frac{\pi }{4}$$  $$\int\limits_0^\infty  {\frac{{{x^2}}}{{1 + {x^5}}}dx = } \frac{\pi }{5}\frac{1}{{\sin \frac{{2\pi }}{5}}}$$  So I guess there must be a closed form - the use of $\Gamma(x)\Gamma(1-x)$ first comess to my mind because of the $\dfrac{{\pi x}}{{\sin \pi x}}$ appearing. Note that the arguments are always the ratio of the exponents, like $\dfrac{1}{4}$, $\dfrac{1}{3}$ and $\dfrac{2}{5}$. Is there any way of finding it? I'll work on it and update with any ideas.    UPDATE:  The integral reduces to finding   $$\int\limits_{ - \infty }^\infty  {\frac{{{e^{a t}}}}{{{e^t} + 1}}dt} $$  With $a =\dfrac{n+1}{m}$ which converges only if   $$0 &lt; a &lt; 1$$  Using series I find the solution is     $$\sum\limits_{k =  - \infty }^\infty  {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} $$ Noun-adjective reversal - was it ever in use in plain speech? If we disbelieve $H_0$, why quote a p value calculated assuming $H_0$ was true? How can I increase my disk space when Ubuntu is installed inside Windows? The sign in the formula for the voltage across the inductor This question is about voltage induced in a inductance.  I know the rate of change of current induces EMF in the inductor. The direction of the induced EMF will be opposite to the change that is causing it. The EMF induced is given by the following formula.  $$V_l = -L \frac{dI}{dt}$$  The negative sign indicates direction of induced emf across the coil is opposite the current change.     In the picture I have drawn current direction and \$V_l\$ direction.  If \$\dfrac{dI}{dt}\$ is postive, \$V_l\$ will be negative according to formula right? Or does the direction of \$V_l\$ has to be reversed?  In many Internet references I found \$V_l = L \dfrac{dI}{dt}\$. Which one is correct?  I hope they reversed \$V_l\$ reference points and removed the negative sign in the formula. Please clarify. Will using 'var' affect performance? Make bibliography appear in toc as a chapter, not section Are Android OS updates region specific? Well, I have a Galaxy Ace with 2.3.3, build XWKPY and Samsung Kies doesn't show an update to 2.3.4 yet.  I'm thinking of manually updating it through Odin and I need to know if I can just download any Gingerbread 2.3.4 update file or is there anything else that I need to be worried about? Quoting within $(command substitution) in Bash Prove that equation has exactly 2 solutions If $p,q$ are positive quantities and $0 \leq m\leq 1$ then Prove that $$(p+q)^m \leq p^m+q^m$$    Trial: For $m=0$, $(p+q)^0=1 &lt; 2= p^0+q^0$  and for $m=1$, $(p+q)^1=p+q =p^1+q^1$.   So, For $m=0,1$ the inequality is true.How I show that the inequality is also true for $0 &lt; m &lt; 1$.   Please help.