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