# 0.9999 = 1

1. May 6, 2010

### losin

0.9999.... = 1

i want to show 0.9999....=1

by showing 0.9999... is a supremum of numbers smaller than 1,

i would be able to prove it.

how can i show 0.999.... is a supremum of numbers smaller than 1?

2. May 6, 2010

### Tac-Tics

Re: 0.9999....=1?

This is a common question on the forums. Do a search.

If you want to create a hard proof of the statement: 0.999.... = 1, you first need to give a rigorous definition for a repeating decimal. The usual definition is it's the LIMIT of the sequence created by appending digits one at a time:

So 0.9999..... is the limit of the sequence 0, 0.9, 0.99, 0.999, 0.9999, ...

Using this definition, the proof is simply to show that that sequence converges to 1.

3. May 6, 2010

### losin

Re: 0.9999....=1?

i want to show 0.9999... is strictly same as 1,

not sequence 0.9, 0.99, 0.999, ... converges to 1.

4. May 6, 2010

### Dickfore

Re: 0.9999....=1?

1.00..01 > 0.999.... > 0.99...9 for all finite decimals. Both the limits tend to 1. By the Dedekind's cut principle, 0.999... = 1. Q.E.D.

5. May 6, 2010

### Tac-Tics

Re: 0.9999....=1?

Yes, but what is 0.999....?

It's the limit of the sequence 0.9, 0.99, 0.999, .... .

Therefore, showing that sequence converges to 1 is the proof.

6. May 6, 2010

### losin

Re: 0.9999....=1?

can i show 0.9999....=sup{x|real number less than 1} ?

7. May 6, 2010

### Dickfore

Re: 0.9999....=1?

See my above post.

8. May 7, 2010

### Landau

Re: 0.9999....=1?

What could you possibly mean by this?

9. May 7, 2010

### Dickfore

Re: 0.9999....=1?

$$1 + \frac{1}{10^{n}}$$

10. May 7, 2010

### Anonymous217

Re: 0.9999....=1?

1/3 = .3333333...
Multiply both sides by 3 and see what you get.

By the way, this is assuming that you have established that 1/3 = .33333... in the first place.

11. May 7, 2010

### Dickfore

Re: 0.9999....=1?

And also, $3 \times 0.333... = 0.999...$

12. May 7, 2010

### Anonymous217

Re: 0.9999....=1?

^I don't get the point of your post, but okay.

13. May 7, 2010

### Dickfore

Re: 0.9999....=1?

We start multiplying from the rightmost digit. What is the rightmost digit of 0.333...?

14. May 7, 2010

### Martin Rattigan

Re: 0.9999....=1?

You only start from the rightmost digit because it's convenient not to have the carries messing up what you've already written down. I quite often multiply from left to right.

But really its just the result that if $\Sigma a_n$ converges, $\Sigma ka_n$ converges to $k\Sigma a_n$.

15. May 7, 2010

### Dickfore

Re: 0.9999....=1?

lol.

16. May 7, 2010

### Martin Rattigan

Re: 0.9999....=1?

It saves working out how much space to leave before you start writing it down. Mind you I do quite often get it wrong, but then that applies if I go the other way as well.

17. May 7, 2010

### Dickfore

Re: 0.9999....=1?

18. May 7, 2010

### Martin Rattigan

Re: 0.9999....=1?

Sorry.

The comment about convergence is a proof. That is $0.\={3}$ is defined as $\sum_{0}^{\infty}3\times 10^{-r}$, so if you've proved this converges to $\frac{1}{3}$ you can say $\sum_{0}^{\infty}3\times 3\times 10^{-r}=\sum_{0}^{\infty}9\times 10^{-r}=0.\={9}$ (by definition) converges to $3\times \frac{1}{3}=1$.

This doesn't mean that showing $0.\={3}=\frac{1}{3}$ is any easier than showing $0.\={9}=1$ in the first place, so I wouldn't personally recommend this route.

Last edited: May 8, 2010
19. May 8, 2010

### Rats_N_Cats

Re: 0.9999....=1?

Is this proof acceptable?
$$0.\={9} = 0.999... = 0.9 + 0.09 + 0.009 + ... = \sum_{k=1}^{\infty} \frac{9}{10^k} = \frac{9/10}{1-1/10} = 1$$
Using the infinite geometric series formula, of course.

20. May 8, 2010

### Martin Rattigan

Re: 0.9999....=1?

That would obviously depend on losin.

The formula for the sum of an infinite gp depends on some basic results about limits, e.g. $|x|<1$ implies $\lim_{n\rightarrow \infty}x^n=0$.

It sounds to me as if losin may be working from axioms for the reals at a stage before much in that way has been proved, in which case he may prefer a direct proof from the axioms using the fact that the $n^{th}$ partial sum for $0.\={9}$ is $1-\frac{1}{10^n}$. This seems to fit in more with his posted ideas.

When I said that showing $0.\={3}=\frac{1}{3}$ is no easier than showing $0.\={9}=1$, I didn't in any way mean to imply either was particularly hard, so I'm sure losin will soon settle on a proof he finds satisfactory even if it doesn't necessarily correspond with any of the suggestions on the forum.