# Is There a Rigorous Proof Of 1 = 0.999…?

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Yes.

First, we have not addressed what 0.999… actually means. So it’s best first to describe what on earth the notation $$b_0.b_1b_2b_3…$$ means. The way mathematicians define this thing is

$$b_0.b_1b_2b_3…=\sum_{n=0}^{+\infty}{\frac{b_n}{10^n}}$$

So, in particular, we have that

$$0.999…=\sum_{n=1}^{+\infty}{\frac{9}{10^n}}$$

But all of this doesn’t really make any sense until we define what the right-hand side means. There is an infinite sum there, but what does that mean? Well, we put

$$S_k=\sum_{n=1}^{k}{\frac{9}{10^n}} \ ,$$

then we have a finite sum. So, for example
$$S_1=0.9, \ ~S_2=0.99, \ ~S_3=0.999, \ etc.$$
So, in some way, we want to take the limit of this sequence.

Let’s consider a particularly simple sequence to illustrate the idea behind the definition of a limit of a sequence: 1/2, 1/3, 1/4,… The terms in this sequence get smaller and smaller. You might think that it’s obvious that it goes to 0, or that it’s obvious that a smart mathematician can prove that it goes to 0, but it’s not. It’s impossible to even attempt a proof until we have defined what it means for something to go to 0. So we have to define what the statement “1/2, 1/3, 1/4,… goes to 0” means, before we can attempt to prove that it’s true.

This is the standard definition: “1/n goes to 0” means that “for every positive real number $\epsilon$, there’s a positive integer N, such that for all integers n such that $n\geq N$, we have $|1/n| < \epsilon$”. With this definition in place, it’s quite easy to prove that “1/n goes to 0” is a true statement. What I want you to see here, is that we chose this definition to make sure that this statement would be true. The first mathematicians who thought about how to define the limit of a sequence might have briefly considered definitions that make the statement “1/n goes to 0” false, but they would have dismissed those definitions as irrelevant, because they fail to capture the idea of a limit that they already understood on an intuitive level.

So the real reason why 1/n goes to 0 is that we wanted it to! Similar comments hold for the sequence of partial sums that define 0.999… It goes to 1, because we have defined the concepts “0.999….”, “sum of infinitely many terms”, and “limit of a sequence” in ways that make 0.999…=1. Can we define number systems such that 1=0.999… does not hold? Of course! But these number systems are not as useful, because they don’t conform to our intuition about limits and numbers.

Now that we know what a limit and an infinite sum is, let me give a fully rigourous proof to the equality 1=0.999… This proof is due to Euler and it appears in the 1770’s edition of “Elements of algebra”.
We know that

$$0.999…=\sum_{n=1}^{+\infty}{ \frac{9}{10^n} } = \frac{9}{10} + \frac{9}{10^2} + \frac{9}{10^3} +…$$

This sum is a special kind of sum, namely, it’s a geometric sum. For (infinite) geometric sums, we can find its limit easily:

Let
$$x=\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+…$$
Then
$$9x=0.999…$$

But, we also have $10x=1+\frac{1}{10}+\frac{1}{10^2}+…$, so $10x-x=1$.

This implies that $x=\frac{1}{9}$.

Hence,
$$0.999…=9x=1$$
Does this proof look familiar? It should! It is essentially the same as Proof #2 in the previous post. The only difference is that every step is now justified by operations with limits.

The following forum members have contributed to this FAQ:
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9 replies
1. jedishrfu says:

Great insight, thanks Micro.

2. Josh Meyer says:

Nice!The informal proof I always share with people is that 1/9=0.111…,  2/9=0.222…, 3/9=0.333…, and so on until 9/9=0.999…=1

3. gill1109 says:

Here is a number system in which 1 is not equal to 0.9999…. and it moreover is rather useful in game theory, and some people even imagine it might be useful sometime in the future, in physics: J H Conway's "surreal numbers". https://en.wikipedia.org/wiki/Surreal_number

4. WWGD says:

You can also use, though not as nice,  the perspective of the Reals as a metric space, together with the Archimedean Principle:  then d(x,y)=0 iff x=y. Let then ## x=1 , y=0.9999….  ##Then ##d(x,y)=|x-y| ## can be made indefinitely small ( by going farther along the string of 9's ), forcing ## |x-y|=0## , forcing  ##x=y ##. More formally, for any ##\epsilon >0 ##, we can make ##|x-y|< \epsilon ##

5. alva says:

Letx=110+1102+1103+…Then9x=0.999…But, we also have 10x=1+110+1102+…, so 10x−x=1.This implies that x=19.Hence,0.999…=9x=1

6. DMartin says:

I can show that whatever the meaning of the number 0.0000000 => 00001, with an infinite number of zeros, it is different from 0. This means that other similar numbers are probably the same, though it does depend on the context. The method involves showing that the two numbers 0.0000000 => 00001 and 0 give completely different output numbers when put into an equation.To show that 0.0000000 => 00001 does not equal 0.Take the equation θ/(sin θ) , where θ is an angle in degrees.For θ = 0.0000000 => 00001, it gives θ/(sin θ) = 180/π = 57.295779513082320876798154814105this is known because a series of increasingly small angles such as 0.0001, 0.0000000001, 0.0000000000000001 etc.give numbers that approach 180/π.But for θ = 0, it gives θ/(sin θ) = 0Therefore 0.0000000 => 00001 does not equal 0.Any thoughts would be appreciated, thanks.

7. DMartin says:

Well, that may be so, but this thread is a discussion on the basis that such numbers are worth talking about, so we're assuming they have some meaning before we start. If you say 'there is no such number', then presumably you think this whole thread is pointless.

8. DMartin says:

So you have a rule that 'each numerical digit must have it's concrete position'. I suppose you know the positions of all the 9s in 0.99999….. then. But even if you argued that their positions are more concrete than the 1 in the number I used, it's not clear where the rule came from.what I've shown is a series that converges on, or approaches, a number at infinity, and the point is, whatever the existence status of that number at infinity, it isn't zero. And surely whatever its existence status, it's similar to the existence status of the numbers you're talking about.

9. DMartin says:

I don't know what you mean by existence, when you say you can prove the existence of your number. But it seems clear enough that if we can talk about numbers with an infinite number of decimal places that all make a difference, then my number and yours are very similar. And my number and yours add up to one, which gives them more common ground.About the position of the 1 in my number, out of all those 9s of yours, there must be one that corresponds to my 1. So whatever problems I have with my number (and God knows it's hard to keep them all in line), you must have the same problems with yours.

10. DMartin says:

If you say that one number exists and another doesn't, you need to say what you mean by exists. If you mean exists within mathematics, Gödel showed that mathematics isn't necessarily a self-consistent system, so existing within mathematics isn't necessarily a meaningful concept.If you're not keen on how my number is expressed, perhaps you'd prefer it if I said:an angle θ such that θ/(sin θ) = exactly 180/π. I can prove that this angle is not zero, because θ = 0 gives a different result.