From: utzoo!decvax!harpo!ihnp4!ihuxf!jhillis Newsgroups: net.math Title: equality for all Article-I.D.: ihuxf.198 Posted: Tue Jan 25 13:45:08 1983 Received: Fri Jan 28 05:52:46 1983 Reply-To: jhillis@ihuxf.UUCP (Jeffrey Hillis) As was pointed out last time this was argued: .9999... = .9+.09+.009+.0009+... = sum (.9)*(.01)**n where n=0,1,2,3... = (.9)*[1/(1-.01)] = (.9)*(1/.9) = 1