Obviously something is wrong with this, but where is the error and why is it one?  $$ \begin{align} \sqrt{-1} &amp;= (-1)^{1/2} \\ &amp;= (-1)^{2/4} \\ &amp;= \sqrt[4]{(-1)^2} \\ &amp;= \sqrt[4]{1} \\ &amp;= 1. \end{align} $$
4300
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}
19786
How do I solve these? $$|-2w-4|&lt;12$$  and  $$(-1/3)|6-5x|+2&lt;1$$  for the first problem i got $-8&lt;w&lt;4$ is that right?
102686
I tried this and that, but could not solve this problem. Here is the code that produces this error:  \begin{equation} o=g(h)=\left\{\begin{array}{rcl}     1 &amp; if &amp; h &lt;   0 \\     0 &amp; if &amp; h \le 0  \end{array}\right \end{equation}   The error is given on the line \end{equation} resides. Thanks.
102266
I am trying to compile a document with equations but whenever I use \begin{eqnarray}..\end{eqnarray} or \begin{align}...\end{align} or anything like this I get the error msg !Missing $ inserted.&lt;inserted text&gt;$. However if I type those equations without using \begin{eqnarray}...\end{eqnarray} or similar things I do not get any error msg and my document runs perfectly.  Even this simple thing does't work....Please help...  \documentclass[a4paper,12pt]{article} \usepackage{amsmath} \usepackage{amssymb}  \begin{document}  \begin{align} $=$  \end{align} \end{document}
74532
I am confused about something: \begin{eqnarray}     (e^{2 i \pi})^{0.5} = (e^{2 i \pi \cdot 0.5})= e^{i \pi}=-1 \end{eqnarray} but \begin{eqnarray}      e^{2 i \pi}=1~ and~ 1^{0.5}=1  \end{eqnarray} where is my mistake??  thanks
19713
Is there any trick to evaluate this or this is an approximation, I mean I am not allowed to use calculator.     $$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$$
55987
Possible Duplicate:          I was thinking on the following line of thoughts: $1 =  \sqrt{1} = \sqrt{-1 \cdot -1} = \sqrt{-1} \cdot \sqrt{-1} = i^2 = -1$  Of course this is not true, but I was wondering which step in this 'line of thoughts' is forbidden to make?   Thanks for the explanation.
41544
Denest $\sqrt {\sqrt[3]{5}-\sqrt[3]{4}}$.   I have tried completing square by several method but all failed. Can anyone help me please? Thank you.  p.s. I'm a poor question-tagger.
51954
I'm currently learning about radicals and simplifying them, and I came across this problem on the internet and tried to solve it:  $$\sqrt{x+5} = x - 1$$  So I used this logic:  $$ \begin{align} \sqrt{x+5} &amp;= x - 1 \\ x + 5 &amp;= (x-1)^2 \\ x + 5 &amp;= (x-1)(x-1) \\ x + 5 &amp;= x^2 - 2x + 1 \\ 0 &amp;= x^2 - 3x - 4 \\ 0 &amp;= (x-4)(x+1) \\ \end{align} $$  Therefore, $x = -1$ and $x = 4$ satisfy the equation $0 = (x-4)(x+1)$.  But then I tried to plug them in the original problem $\sqrt{x+5}=x-1$:  $$ \begin{align} \sqrt{4 + 5} &amp;= 4 - 1 \\ \sqrt{9} &amp;= 3 \\ 3 &amp;= 3 \end{align} $$  So using 4 works as expected, but when using $-1$:  $$ \begin{align} \sqrt{-1 + 5} &amp;= -1 - 1 \\ \sqrt{4} &amp;= -2 \\ 2 &amp;\ne -2 \end{align} $$  At what stage am I going wrong?  And according the , the solution is $x = 4$.
1237
Ok i feel like this is an easy question that I'm making more complicated than it needs to be  $$ \begin{align} 2n+1 &amp;= (2n-1) + 2\\ (2n-1) &amp;= 2(n-1)+1 \end{align} $$  so is it $1$ ?
58567
i have a pretty nasty question. i was glancing through a few olympiad papers and stumbled upon this question:  prove that  $ i = \sqrt {-1}\ $.  i tried the conventional methods namely euler's formula  but could not figure what to do next. how do you actually proceed further? i have checked a few books and a proof is not available.are there any proof available for this
74047
Can you enter the rabbit hole recursively?  If the $ \sqrt{-1} = i $ then, what does $ \sqrt{-i} $ equal?
76152
int t=-1'd1; //trying to assign -1 to t; bit vb[7:0]; // simple declaration  bit matrix[0:1][0:7]='{'{0,1,0,0,0,0,0,0},'{0,0,1,0,0,0,0,0},'{1,0,0,1,0,0,0,0},'{0,0,0,0,1,0,0,0},'{0,0,0,0,0,1,0,1},'{0,0,0,0,0,0,1,0},'{0,0,0,0,1,0,0,1},'{0,0,0,0,0,0,0,0}};// initialising multi-dimensional array.  I am using QuestaSim to compile and simulate. this is the report: # -- Compiling module test # ** Error: (vlog-13032) test.v(8): near &quot;=&quot;: Syntax error. # ** Error: (vlog-13032) test.v(9): near &quot;;&quot;: Syntax error.  I am somewhat New to the Verilog. Please help me in fixing these errors.
41023
Possible Duplicate:             Well, I remembered this after having Algebra II a year ago, is it possible that this is a valid proof that $1 = -1$?  $$ 1 = \sqrt{1} = \sqrt{-1\cdot-1} = \sqrt{-1} \cdot \sqrt{-1} = i \cdot i = i^2 = -1 $$  $$ \therefore 1 = -1 $$  So is this actually fully valid? Or can it be disproved?
99779
The following code works fine  \begin{theorem}[Euler] Blah \end{theorem}   But whenever I try to use square brackets in the description I get an error.  \begin{theorem}[The set $[0, 1]$ is compact] Blah \end{theorem}   Can anybody tell me how to avoid the error, please?
40496
What's the thing with $\sqrt{-1} = i$? Do they really teach this in the US? It makes very little sense, because $-i$ is also a square root of $-1$, and the choice of which root to label as $i$ is arbitrary. So saying $\sqrt{-1} = i$ is plainly false!  So why do people say $\sqrt{-1} = i$? Is this how it's taught in the US?
67427
I came across the following:  $\begin{align} 1 \times 8 + 1 &amp;= 9 \\ 12 \times 8 + 2 &amp; = 98 \\ 123 \times 8 + 3 &amp; = 987 \\ 1234 \times 8 + 4 &amp; = 9876 \\ 12345 \times 8 + 5 &amp; = 98765 \\ 123456 \times 8 + 6 &amp; = 987654 \\ 1234567 \times 8 + 7 &amp; = 9876543 \\ 12345678 \times 8 + 8 &amp; = 98765432 \\ 123456789 \times 8 + 9 &amp; = 987654321. \\ \end{align}$  I'm looking for an explanation for this pattern.  I suspect that there is some connection to the series $\frac{1}{(1 - x)^2} = 1 + 2x + 3x^2 + \cdots$.     asks the same question but has no answers posted.