Prove that 0.999999999(9)=1

[Math Tutor]

Prove that
0.999999999(9)=1

[Guest]

[tex]0.999999999(9) = 0.(9)=0.9+0.09+0.009+......=0.9+0.9*10+0.9*10^{2}+......[/tex] Right?
So [tex]0.(9)[/tex]- Indefinitely - decreasing geometric fraction
But we know formula for Indefinitely - decreasing geometric fraction,[tex]S=\frac{b_1}{1-q }[/tex] , where q = 0.1
[tex]S=\frac{0.9}{1-0.1 } = \frac{0.9}{0.9 } = 1.[/tex]

[Guest]

Interesting solution but I think that this should be proved using limes.

[shemet]

Formula for Indefinitely - decreasing geometric fraction is proven by limes!!!

[Guest]

We will prove that 0,(9)=1
0,(9)=0,9(9)=x
So 10x=x+9 so x=1

[Guest]

Hey guest that was interesting solution but this can be also solved through limit and will be more easy by using scientific notation calculator it can be done in a second.

[Math Tutor]

I am not sure and it has not to be calculated and proved.

[Alicelewis11]

Sorry, I have also no idea about it.

[leesajohnson]

I don't think that there is any real prove of this problem. It is just assumed equal to 1 or we can say that to round off the fraction digit we write 0.999999 equal to 1.

[Guest]

It is already proved(the second post) using geometric progression.

[Guest]

It is already proved(the second post) using geometric progression......Query......

Are you sure about the second post......is it not just saying it converges on 1 because it is converging.......but never gets there...??????

[Guest]

If one item has a value of 0.99999999999999......
Then 10 items have a total value of 9.9999999999999999.......
Subtract the one item from ten items gives nine items have total value of 9.000000000000..... or just 9.0
Divide value of nine items by nine to get value of one item, this gives value of 1.0 as the value of one item.

So original given value of one item as 0.99999999999......is same as calculated value of one item equals 1.0000000000 OR just 1.

[Guest]

Subtract the one item from ten items gives nine items have total value of 9.000000000000..... or just 9.0

Sorry...I meant to say last post..........

Subtract the value of one item (0.999999999....) from the value of ten items gives the total value of nine items as 9.000000000000..... or just 9.0

....................................

[Guest]

0.9999.... means [tex]9/10+ 9/100+ 9/1000+ 9/1000+ …=[/tex][tex](9/10)(1+ 1/10+ 1/100+ 1/1000+ …)= (9/10)(1+ (1/10)^1+ (1/10)^2+ …)[/tex]

That last is a geometric sequence, [tex]1+ r+ r^2+ …[/tex] with r= 1/10.

It is easy to show that such a geometric sequence converges to [tex]\frac{1}{1- r}[/tex]:

The sum of an infinite series like that is the limit of the "partial sums"- the sums up to a finite power. Let [tex]S_n= 1+ r+ r^2+ …+ r^n[/tex]. Then [tex]S_n- 1= r+ r^2+ r^3+ …+ r^n= r(1+ r+ r^2+ …+r^{n-1})= r(1+ r+ r^2+ r^{n-1}+ r^n- r^n)[/tex][tex]= r(S_n- r^n)= rS_n- r^{n+1}[/tex]. So [tex]S_n- rS_n= (1- r)S_n= 1- r^{n+1}[/tex] and then [tex]S_n= \frac{1- r^{n+1}}{1- r}[/tex]. Now, take the limit as n goes to infinity. That limit exists only if |r|< 1 and, in that case [tex]r^{n+1}[/tex] goes to 0 so that limit is [tex]\frac{1}{1- r}[/tex].

In the case that [tex]r= \frac{1}{10}[/tex], [tex]\frac{1}{1- r}= \frac{1}{1- \frac{1}{10}}= \frac{1}{\frac{9}{10}}= \frac{10}{9}[/tex].

So 0.99999... is equal to [tex](9/10)(10/9)= 1[/tex].

[Guest]

It is already proved(the second post) using geometric progression......Query......

Are you sure about the second post......is it not just saying it converges on 1 because it is converging.......but never gets there...??????

Then you are saying that you do not know what a "limit" is. The sequence "converges on 1" and the limit of that sequence is 1.

[Guest]

This is sad. There are a number of people, including one who uses the user-name "math tutor" responding to this who clearly have no idea what "limits" are!

"Math tutor", who started this thread, responded, "I am not sure and it has not to be calculated and proved." I don't even know what that means except that the first part, "I am not sure" should have prevented him from responding at all! And why would he have asked if he belies "it has not to be calculated and proved"? The proof given in the first response, that this is a geometric sequence that converges to 1, was completely correct.

Saying "it should be proved using limes" ignores the fact that the convergence of geometric sequences is proved using limits. No, it is not necessary to repeat that unless this is a homework exercise that explicitly stated that. The person who wrote "is it not just saying it converges on 1 because it is converging.......but never gets there" is just saying that he does not know what the "limit of a sequence" means and does not know what and infinite decimal is!