One equals point nine repeating

created by dpride
(idea) by dpride (5.4 mon) (print)   (I like it!) Tue Apr 25 2000 at 4:20:08
this is something I learned in high school, though I didn't believe until college. It absolutely stumps some people, but to others, it is pretty much common sense.

one equals point nine repeating

The simplest explanation is that .999... goes on forever so there is no number between .999... and 1 therefore they are effectively the same number.

There are other, much more complicated proofs, so you are free to type one up...

(idea) by N-Wing (3.2 d) (print)   (I like it!) 1 C! Tue Apr 25 2000 at 4:38:47

This is based on the method used to convert repeating decimals into fractions:

You can also think of it this way:
0.1111... = 1/9
0.2222... = 2/9
      etc.
0.8888... = 8/9
0.9999... = 9/9 = 1

(idea) by pfft (1.5 wk) (print)   (I like it!) Tue Oct 03 2000 at 17:50:22

I think the problem many people (both "believers" and "nonbelievers") have with this is simply one of notation.

A decimal number like 0.312 is defined as the sum 3*10-1 + 1*10-2 + 2*10-3.

"0.999..." is defined similarily as the infinite series 9*10-1 + 9*10-2 + 9*10-3 + ... = sumi=1 to infinity 9*10-i. Such a sum is in turn defined as limn->infinity sumi=1 to n 9*10-i. It is clearly believable that that limit is 1, and it is not difficult to prove. (update: 10998521 gives a proof below).

Now, if there are any great deeply philosophical lessons to be learnt from this, I do not know. I think it is enough to note that if we allow "..." notation to indicate series, there can be several different decimal representations of a given number.

PS. On second thought, it is perhaps not completely a matter of arbitrary definitions. If you want the string "0.9999..." to have the value of a real number r such that r≤1 and r≥1-10-n for arbitrarily large n, then the only value r can have is 1. Similar to how there are no infinitesimals, completeness of the reals means there there are "not enough" real numbers to fit between 1 and the increasing sequence of point nines. To find such a rare beast you have to move to a bigger number system, like in non-standard analysis.

(idea) by 10998521 (2.7 wk) (print)   (I like it!) 1 C! Sat May 05 2001 at 19:37:51
If we let S(n) = .9999...9 (with n 9s)

S(n) = sigmar = 1 to n 9 * 10-r (From the definition of a base 10 number)

10S(n) = sigmar = 0 to n-1 9 * 10-r (Multiply each term of the sum by 10)

9S(n) = 10S(n) - S(n) = sigmar = 0 to n-1 9 * 10-r - sigmar = 1 to n 9 * 10-r = 9 - 9 * 10-n

S(n) = 9S(n)/9 = 1 - 10-n

limn goes to infinity S(n) = 1 - limn goes to infinity 10-n = 1 - 0 = 1

QED

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