leo@philmds.UUCP (Leo de Wit) (06/27/88)
In article <3637@teklds.TEK.COM> daniels@teklds.UUCP (Scott Daniels) writes: [stuff about binary floating point representation deleted] >Ok, here we go: > To represent 2.0 exactly, we could use 2/.1000, but that represents the >interval 1.8750:2.1250. Now, there is a tighter specification which is >entirely within that interval: 1/.1111 (which represents 1.8125:1.9375), so >we should use that tighter interval since no poin inside it is any further >from the desired value 2.0 than the range that 2/.1000 gives. Hence the >besty representation (the tightest) for 2.0 is an interval which does not >even include 2.0! Now I may've got this wrong, but 2.0 - 1.8125 = 0.1875 (the most far point). 2.1250 - 2.0 = 0.1250 (upper far point of interval) 2.0 - 1.8750 = 0.1250 (lower far point of interval) And 0.1875 > both other distances. So your theorem does not hold (yet). Leo (Q.E.D. Quick End Dirty, sorry for the misspelling 8-)