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# 3 proofs that 0.999... is the same as 1

First of all writing the number 0.999... isn't not valid but for the sake of this post let's allow it.

Proof 1:

This is probably the most easy to understand.

Let's assume:
x = 0.999...
Next we'll multiply both sides by 10, so we have:
10 x = 9.999...
Let's subtract
x
on both sides.
10 x - x = 9.999... - x
Since
x = 0.999...
we get:
9 x = 9.999... - 0.999... = 9
We now can divide each side by 9 because it's not equal to zero and we get:
x = 1

Because we assumed that
x = 0.999...
and we have that
x = 1
that means that
1 = 0.999...

Proof 2:

Because the real numbers are dense, that means that you can find always a real number between two real numbers. For example we can use the arithmetic mean:
(a + b) / 2
Let's assume that
0.999... ≠ 1
. That means that with the arithmetic mean we can find a number in between, with that we would get
0.999...5
? Which isn't a real number, there is nothing like a "last digit". And even it there was, that number wouldn't be in between 0.999... and 1. That means that our initial assumption is wrong thus 0.999... is equal to 1.

Proof 3:

This is basically the actual proof because it's per definition the same. Let's look at what we mean when we write 0.999... it means that ever digit after the 0 is 9. Let's rewrite that.
0.999... = 0.9 + 0.09 + 0.009 + ...
We can rewrite that further to
= 9 * 0.1 + 9 * 0.01 + 9 * 0.001 + ...
We can add "0" since it doesn't change the value, or more specifically let's add 9 and subtract 9. With that we get:
= 9 * 1 + 9 * 0.1 + 9 * 0.01 + 9 * 0.001 + ... - 9
Let's rewrite it again
= 9 * (1/1) + 9 * (1/10) + 9 * (1/100) + 9 * (1/1000) + ... + 9
= 9 * (1/10)^0 9 * (1/10)^1 + 9 * (1/10)^2 + 9 * (1/10)^3 ... + 9
We can pull 9 outside of the sum to get:
= 9 * [(1/10)^0 + (1/10)^2 + (1/10)^3 + ...] - 9
To write it a bit more abstract we get:
That this is 9 times the sum from i = 0 to infinity over (1/10)^i and from that we subtract 9 again.
This sort of sum is called geometric series, it's of the form:
The sum from i = 0 to infinity over q^i. There's a formula for the result of the sum if |q| < 1:
1/(1-q)
In our case q = 1/10 which is smaller than one, so we can use that formula here and we get:
= 9 * [1 / (1 - q)] - 9
with q = 1/10 we have:
= 9 * [1 / (1 - 1/10)] - 9
When we simplify it we get
= 9 * [1 / (9/10)] - 9 = 9 * [10 / 9] - 9 = 10 - 9 = 1
Thus by writing 0.999... it's the same as 1. And because that's ambiguous, it's not allowed to write it so that every number written as digits is unique.
DrWatson · 70-79, M
Absolutely!

Here's another argument. If someone is willing to accept that

1/3 = 0.333333.....

Then multiply both sides by 3.
Luke73 · 22-25, M
@DrWatson Yes that's one too. But it's more complicated because you first have to show that 1/3 = 0.333... and that 0.333... * 3 is 0.999...
helenS · 36-40, F
@DrWatson You would have to prove that 0.3333... = 1/3; the proof would be exactly the same as for 0.9999... = 1.
1/3 is the limit of the infinite series of which 0.3333... is a shorthand notation.
DrWatson · 70-79, M
@helenS @Luke73 Absolutely. All of these arguments, if they are to be made rigorous, rely on defining a decimal as a series.

That is why I used the word "argument" rather than "proof." Many people are willing to accept the representation of 1/3 but are skeptical about the representation of 1. To such a person, we can argue , "The truth of the second equality follows from your own acceptance of the first."

I have found this to be rhetorically persuasive even if it is not mathematically rigorous. And after someone is convinced by this, they might then be more open to seeing a real proof.
helenS · 36-40, F
... for some reason my reply got lost. Here it goes again:

First you will have to define the exact meaning of the vague expression "0.999..."

The only sense I can make of it is an infinite series, like this:
0.999... = 9 Σ_{n=1}^∞ 10^{–n}

It can easily be shown that the above infinite series has a limit L=1, using the usual L±ε approach. The longer the series will run up to a given n, the nearer its sum σ_n will approach to L such that by taking the value of n to be sufficiently large, the difference ε=L − σ can be made arbitrarily small. If this be the case, this value L is said to be the limit of the infinite series. This approach goes back to Bolzano, Cauchy and Weierstrass, and it eliminates every possible mysticism about 0.999... and similar expressions. (Learning about ε came as a revelation to me...)

Therefore it would be much better, in my opinion, to write
lim 0.9999... = 1, instead of
0.9999... = 1.

The latter expression make no sense (to me, at least) because the infinite (= endless) series will never "reach" its end, but it's limit is well-defined.

hS
Luke73 · 22-25, M
That's what I've done in my third proof. The convergence of the series is rather simple with the formular for the geometric series aswell as the limit of it.

I would disagree with your last part, because it would be the same as saying that the limit of 3.14... is pi, 3.14... is the number pi.
helenS · 36-40, F
helenS · 36-40, F
@Luke73 There is really only potential infinity, in which an algorithm generates a sequence with no terminating element, so each individual step is achieved in a finite number of steps. That's the meaning of an "infinite" series – it simply has no end.
In many favorable cases it's possible to assign a limit L to an infinite series, and to prove that L is the limit, by using only a finite sequence of steps.
In this sense, the number 1 is the limit of the series "0.9; 0.99; 0.999; (...)"
You may also think of a single real number as the limit of an ever-decreasing interval around that number. It's not necessary for the sequence to "reach" the limit; in fact the limit will never be reached. It cannot be reached because an infinite number of steps would be necessary to reach it.

PS I sort of remember that "actual" infinity plays an important rule in set theory, of which I have no clue.
Luke73 · 22-25, M
@helenS I agree with you that 1 is the limit of the sequence 0.9; 0.99; 0.999... But I disagree that a number is the same as a series.

I'm not 100% sure if it's completely right but if you just look at the interval [0;1] and you assume that the digits each numbers represent are a limit to that, then they would be countable because you would just reiterate over every digit at each decimal place. That's not possible though because the real numbers are not countable.
Heartlander · 80-89, M
0.9999… isn’t a number. It’s an expression to demonstrate that it will never be one, an unreachable limit, much like infinity is unreachable. The best you can do is treat it as one and accept that it’s close enough to one that treating it as one is good enough.

To say that it is actually one is to say that there is an end to smallness.
Luke73 · 22-25, M
@Heartlander Then where is the flaw in all three proofs?