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 2009-09-01, 20:22 #23 mart_r     Dec 2008 you know...around... 3·191 Posts $666*777=517482$ and $666/777=.857142...$ $49899989999999^4=6200144830041037000103201165060058012000400400040000001$
 2009-09-26, 04:45 #24 Damian     May 2005 Argentina 101110102 Posts $1+2+3+4+5+...+\infty = -\frac{1}{12}$
 2009-09-26, 06:16 #25 paleseptember     Jun 2008 Wollongong, .au 3·61 Posts $\mbox{The Area of an Ellipse is }\pi*a*b$
 2011-10-09, 18:38 #26 Dubslow Basketry That Evening!     "Bunslow the Bold" Jun 2011 40
 2011-10-09, 19:56 #27 ccorn     Apr 2010 100100112 Posts $G_6(\frac{a+i}{N})=0$ iff $a^2\equiv-1\quad\pmod{N}$ where $i^2=-1$ and [TEX]G_6(\tau)=\sum_{(m,n)\in\mathbb{Z}^2\backslash\{(0,0)\}}\frac{1}{(m+n\tau)^6}[/TEX] ($N$ must be representable as $N=c^2+d^2$ with $c, d$ coprime integers) Last fiddled with by ccorn on 2011-10-09 at 20:45 Reason: added condition on N
 2011-10-09, 22:22 #28 EdH     "Ed Hall" Dec 2009 Adirondack Mtns 3·5·223 Posts $-1 = -1$ $\frac{1}{-1} = \frac{-1}{1}$ $\sqrt{\frac{1}{-1}} =\sqrt{ \frac{-1}{1}}$ $\frac{\sqrt{1}}{\sqrt{-1}} =\frac{\sqrt{ -1}}{\sqrt{1}}$ $(\sqrt{1})^2 = (\sqrt{-1})^2$ $1 = -1$
 2011-10-10, 09:34 #29 xilman Bamboozled!     "πΊππ·π·π­" May 2003 Down not across 2×32×569 Posts $i^2 = j^2 = k^2 = ijk = -1$

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