# Math Question

This one has been bothering me for a week or two. How do we express the number that is equal to
1 minus .9999… ?
Is it zero? If not, then how do we explain that (1/3)3 doesn’t equal 1?

.9999 does not equal 1…

.9999 is an approximation of 1/3+2/3 as expressed in decimals…

-Imp

If you draw it on a graph you simply get an ever-decreasing gap. It’s like actually trying to measure the circumference of a circle

Yes. It is zero. But we do wonder why because we commonly think that reals have exact decimal representation or correspondence, when this is not the case. You can call it approximation, as Imp says. And you can choose to be precise up to a certain digit, with 1 as limiting value.

There is one and only one intelligible meaning of .999… in our present mathematical framework, and that is that .999… = 1.

Hi to All:

Imp claimed that 1-.999… was not equal to 0. Just for fun I would like to offer a proof that the assertion is false.

1. Assume that 1-.999… is not equal to 0.
2. Let d = 1-.999…
3. Then d is not equal to 0
4. Let Sn = 1-.9(Sum for i=0 to n of (.1^i))
5. Then Sn =1-.9[size=75]1[/size]9[size=75]2[/size]9[size=75]3[/size]…9[size=75]n[/size] for all n
6. Lim as n goes to [size=100]infinity[/size] of Sn = 1-.9(lim as n goes to infinity of Sum for i=0 to n of (.1^i))
7. In general lim as n [size=100]goes[/size] to [size=100]infinity[/size] Sum for i=0 to n of r^i = 1/(1-r) where r < 1. High School Math.
“8)” By substituting we get Lim as n goes to infinity of Sn = 1-.9(lim as n goes to infinity of Sum for i=0 to n of (.1^i)) = 1-.9(1/(1-.1)) = 1-.9/.9 = 0
8. By definition the lim as n goes to infinity of Sn exists if and only if for every e there exists an N such that if n > N then |Sn - lim| < e. Notice that lim = 0 in this case.
9. Select e=d/2
10. Then if n>N Sn<d/2
11. Since Sn = 1-.999… For all n we must have that 1-.999… <d/2
12. Now 1-.999… = d and 1-.999… is <d/2. Since d is not equal to 0 this is a contradiction.
13. Therefore the assumption is false
14. Thus 1-.999… is equal to 0

I can not seem to substitute my item number 8 for my smiley face. Oh well .