0.999... =1 Proofs?

Question

Does anyone know how many proofs there are for 0.999999...continued =1 and what they are.

Answer 1

Let x = .999999999 ...

Then 10x = 9.9999999999 ...

Now subtract:
10x = 9.999999999 ...
- x= .999999999 ...
________________
9x = 9
x = 1

Answer 2

If they are different you should be able to find a number between them which isn't zero. call this x i.e. 0.99999
call this number so x = 0.999......+ e where e is non zero.

For any number you come up for e with all the 9's will carry one to the next decimal place from the position of the first non zero number. This will come up with a number 1.00000 + decimal places to the right of the non zero number of e. this is inconsistent with the assumption that there is a difference.

Therefore 0.9999 and 1 are the same number (proof by contradiction). Ok this could be made more formal but you get the idea.

Answer 3

I suppose there are lots of proofs, the easiest one is the way you convert repeating decimals to fractions. you know, the one that goes like this:

1 K = .9999...
10K = 9.9999
subtract:
10 K = 9.9999...
- 1 K = .9999...
---------------------
9K = 9
K = 1

A more elegant proof is by summation of the series
9/10 + 9/100 + 9/1000 + ... , that is
9summation(1/10^n), which is a geometric sequence with r = 1/10, which sums to 9 * 1/(10 - 1) = 9 * 1/9 = 1

Answer 4

0.9999 is 9 times the infinite sum:, S where

S = sum k=1 to inf { (1/10)^k }

you can show that this sum is equal to

S = 1/[1-I1/10)] - 1 = 1/9 and so 9S = 9*[1/9] = 1.

The problem that people have is that 0.99999... looks like it's just a little short of 1 but that's because they fail to understand that this IS an infinite sum.

Answer 5

its a rule that after a decimal point any number which exceeds 5 then its previous number can be added with 1.so if v consider
x = 1/0.9999 then,
x = 1.00000001
how will u consider this ?
x = 1 (right...)

Answer 6

0.333.... = 1/3
3 * 0.333... = 3 * 1/3 = 3/3
0.999... = 1

There are more but this is just a very basic one