How can we prove 0.9999... is equal to 1?

[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?

 

[Tac-Tics]

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.

 

[losin]

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

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

 

[Dickfore]

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.

 

[Tac-Tics]

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

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

 

[losin]

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

 

[Dickfore]

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

See my above post.

 

[Landau]

1.00..01

What could you possibly mean by this?

 

[Dickfore]

What could you possibly mean by this?

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

 

[Anonymous217]

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.

 

[Dickfore]

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...

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

 

[Anonymous217]

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

 

[Dickfore]

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

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

 

[Martin Rattigan]

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 $k\Sigma a_n$.

 

[Dickfore]

I quite often multiply from left to right.

lol.

 

[Martin Rattigan]

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.

 

[Dickfore]

Please refrain from off-topic posting. The OP wanted a rigorous proof and you are talking about your arithmetical adventures.

 

[Martin Rattigan]

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.

 

[Rats_N_Cats]

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.

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.

 

[Martin Rattigan]

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.