Re: Problems I have with 1.999...=2

stephen@xxxxxxxxxx wrote:
Kirby Cook <kwmcook@xxxxxxxxxxx> wrote:

: Let me try it another way. My assertion might be stated (I hope) as : follows. Given the set whose elements are nine tenths, nine tenths plus : nine hundredths, nine tenths plus nine hundredths plus nine thousandths, : etc., the least upper bound of the set is one, and one is not a member : of the set.

That is true. 1 is not a member of the set { .9, .99, .999, ... }. But if .999.... is a set, then it is not a number,
and the question does 1=.999.... does not make much sense.
If you interpret .999.... as a number, which is what most people
do, then you really do not have a lot of choices about what number it is.

Stephen

Since the preceding post could only be described as inattentive, I was going to ignore it. But I won't, after all.
My point, which I consider sufficient, is that .999... is a member of the described set, and 1 isn't.
.

Relevant Pages

  • Re: Problems I have with 1.999...=2
    ... Given the set whose elements are nine tenths, nine tenths plus nine hundredths, nine tenths plus nine hundredths plus nine thousandths, etc., the least upper bound of the set is one, and one is not a member of the set. ...
    (sci.math)
  • Re: Problems I have with 1.999...=2
    ... My assertion might be stated as ... Given the set whose elements are nine tenths, nine tenths plus ... >nine hundredths, nine tenths plus nine hundredths plus nine thousandths, ... means that least upper bound, ...
    (sci.math)
  • Re: Problems I have with 1.999...=2
    ... nine tenths plus nine hundredths plus nine ... The preceding post is perfectly attentive. ... If you are here to understand math - then try to understand what ... >member of the described set, ...
    (sci.math)
  • Re: Decimals
    ... Mark Brader: ... twenty-eight plus two tenths plus three hundreds plus seven thousandths ...
    (alt.usage.english)