# 0.9999..=1 proof

• Apr 22nd 2012, 02:43 PM
rt20
0.9999..=1 proof
Hire some DIFIRENT numbers:
1
0.9...9=x
0.9...8
0.9...7
.......
0.9...1
0.9...0
0.9..-1
.........
0.0...1=y
0
x=0.9.. / lets *10
0.9..+0.9..+0.9.. ten times =(1-y)+(1-y).. ten times SO
10x= 10-10y / -0.9...
10x-x=(10-10y)-(1-y)
9x=9-9y
9x=9-9y
x=(9-9y)/9 and this is NOT 9 !
x=9(1-y)/9
x=0.9....
:)
This last one clearly shows that0.9...9 is not some "magic" number its number like anyother. This 0.00...1 is more "special" but is stillordinary number.
• Apr 22nd 2012, 03:07 PM
HallsofIvy
Re: 0.9999..=1 proof
Quote:

Originally Posted by rt20
Hire some DIFIRENT numbers:
1
0.9...9=x
0.9...8
0.9...7
.......
0.9...1
0.9...0
0.9..-1
.........
0.0...1=y

I have no idea what any of these mean. I know that "0.999..." means the decimal number whose only digits are "9". But I don't know what you mean by "0...9". How many digits does that "..." skip over? It can't be an infinite number since there is no "last" digit in that case.

Quote:

0
x=0.9.. / lets *10
0.9..+0.9..+0.9.. ten times =(1-y)+(1-y).. ten times SO
"ten times"? Do you mean 0.9+ 0.9+ 0.9+ 0.9+ 0.9+ 0.9+ 0.9+ 0.9+ 0.9= 10(0.9)= 9? I don't know what you mean by 1- y because I don't know what you mean by "0.0...1" which is apparently how you have defined y
And so nothing beyond here makes sense.

Quote:

10x= 10-10y / -0.9...
10x-x=(10-10y)-(1-y)
9x=9-9y
9x=9-9y
x=(9-9y)/9 and this is NOT 9 !
x=9(1-y)/9
x=0.9....
:)
This last one clearly shows that0.9...9 is not some "magic" number its number like anyother. This 0.00...1 is more "special" but is stillordinary number.
Should I conclude that you do not understand how "decimal notation" works?

If you really want a proof that 0.999...= 1, do this: 0.999...= 0.9+ 0.09+ 0.009+ ...= 0.9+ 0.9(.1)+ 0.9(.01)+ 0.9(.001)+ ... In other words, it is a geometric series, of the form $\sum ar^n[$ with a= 0.9 and r= 0.1. It is easy to show that the sum of such a geometric series is $\frac{a}{1- r}$ which, here, is $\frac{0.9}{1- 0.1}= \frac{0.9}{0.9}= 1$.
• Apr 22nd 2012, 04:22 PM
rt20
Re: 0.9999..=1 proof
"It can't be an infinite number since there is no "last" digit in that case."
Well there is. I just write it. In mathematics you can have any number you can think of! Why not ?

Saying that that there is not "last digit" is like saying 0=1 witch is not true.
0= 0 + 0 + 0 + ...
= (1 - 1) + (1 - 1) + (1 - 1) + ...
= 1 - 1 + 1 - 1 + 1 - 1 + ...
= 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + ...
= 1 + 0 + 0 + 0 + ...
= 1

I am saying that 0.0..1 is the closest to zero number. By definition. I don't have to prove it - i just to define it : )
0.9.. is the closest to 1, with make 1-0.0...1 =0.9.... Its very simple.