ijk (01/26/83)
I just couldn't resist adding my 1.999... cents worth. This proof comes from 11th grade math class - a long time ago. Let x = .999... (remember that ... stands for '9' ad infinitum) then 10 * x = 9.999... subtract first line from second and we get simply: 9x = 9. (!exactly) or x = 1. IT is true. As people have pointed out, it takes an understanding of limits to really see what is going on, but it is not necessary for the proof. Just another example of the mysteries of infinity (what has more points in it: a line an inch long or a line a mile long?? Ans: they have the same).
bryan (01/26/83)
I must admit that .999... is equal to 1. But i still defend .000...1 is a number; for example consider the following series, (.1 * .01 * .001 * .0001 ...). If .999 is equal to 1 then .000...1 is equal to 0. If you agree that the above is correct then the following is correct. (.000...1) + (.999...) = 1 since 0 + 1 = 1. Also another proof that .999... = 1 . 1/3 + 2/3 =3/3 1/3 = .333... 2/3 = .666... .333... + .666... = .999... 3/3 = .999... 1 = .999... !ihlpb!bryan Bryan DeLaney
kenchan (01/28/83)
I presume most of you are using Dedekind Cuts or Cauchy sequences for the definition of the real numbers. Can anyone out there prove .99...=1. (or even give a definition of .999...) using J.H. Conway's definition of numbers (J.H. Conway, "On Numbers and Games", Academic Press)? Ken Chan American Bell, Lincroft (houxq!kenchan)
ech (01/29/83)
#R:ihlpb:-27100:whuxlb:7200014:000:1034 whuxlb!ech Jan 26 17:19:00 1983 There is a subtle distinction here that I haven't communicated, between .999... carried to n places and .999... carried to an infinite number of places. The latter is the LIMIT of the sequence .9, .99, .999,... and is equal to 1, but does not appear anywhere IN the sequence. Similarly, we can define .1, .01, .001,... which converges to 0 but never reaches it, only the limit being 0. Now, I will not dispute that all those numbers in the second sequence are numbers; however, the notion of putting the 1 after INFINITELY many 0's is simply nonsense: the limit of the second sequence is .000... carried to an infinite number of places, which is quite clearly 0. Another, quite simple proof, that .999... is 1 is from the schoolboy method for resolving repeated fractions into rational number (p/q) form: x = 0.999... 10x = 9.999... subracting, 9x = 9.0 x = 1 The above proof relies only on the idea that .999... must be equal to itself, which had better be self-evident even in the weird world of real numbers. =Ned=