# Is that true?

is 2,99999....... = 3 ?

What is the limit of 2.9..., or what is the limit of 3 - 1/(10^m), where m is a positive integer?

Otherwise stated: Does 1/10^m = 0 ?

If the 9s continue indefinitely, then it is equivalent. You can show this by showing that there is no real number between 2.999... and 3, which is the same as evaluating the infinite series represented by the decimal 2.999... .

If the 9s continue indefinitely, then it is equivalent. You can show this by showing that there is no real number between 2.999... and 3, which is the same as evaluating the infinite series represented by the decimal 2.999... .
You seem to be satisfied if there is no real number between 2.999... and 3.

What is a real number?

Mentallic
Homework Helper
is 2,99999....... = 3 ?
Any reason why you chose to use 2.99..=3 rather than the usual 0.999..=1?

You seem to be satisfied if there is no real number between 2.999... and 3.

What is a real number?
Someone asking the question in the original post is probably not familiar with extending the real numbers or with alternative formulations of analysis. They have also either not understood or have never seen a proper definition of the set of real numbers, using which, they would immediately see the truth of the statement. The purpose is to lead them towards a motivation of the definition.

Last edited:
The standard proof is the following (which you undoubtetly have already seen):

Let x=2.9999....
Then 10x=29.9999....
Thus 10x-x=27
Which yields x=3.

So, we have 3=2.999...

HallsofIvy
Homework Helper
In my opinion, a slightly more rigorous proof:
$$2.999...= 2+ 0.9+ 0.09+ 0.009+ \cdot= 2+ \frac{9}{10}+ \frac{9}{100}+ \frac{9}{1000}+ \cdot\cdot\cdot$$
$$= 2+ 0.9(1+ 0.1+ 0.01+ \cdot\cdot\cdot)$$

$1+ 0.1+ 0.01+ \cdot\cdot cdot$ is a geometric series, $\sum_{n=0}^\infty ar^n$ with a= 1, r= 0.1. The sum of such a serie is
$$\frac{a}{1- r}= \frac{1}{1- .1}= \frac{1}{0.9}$$

So the series is
$$2+ .9\left(\frac{1}{.9}\right)= 2+ 1= 3$$

Most are okay with x + 1/3=x.333... because that's what their calculator spits out, just think of any x.999... = x + 1 as an extension of that. The only difference is that the calculator simplifies it to x + 1, since, unlike 1/3 and its repeating decimal causing multiples, x + 1 is an integer.

In terms of equalities:
2 + 1/3 = 2 + .333... - The original problem
1/3 = .333... - Subtract 2
1 = .999... - Multiply by 3

There are many much more rigorous proofs than this, but IMO, this is the most accessible.

Mentallic
Homework Helper
Perhaps an attempt to avoid inevitable locking of the thread?
Perhaps if I created a thread asking if x+0.99...=x+1 for all x then we would have a general case and then there's no excuse in making more of these threads.

Perhaps debating x + .999... = x + 1 is fun, and an excuse in itself.

gb7nash
Homework Helper
Perhaps debating x + .999... = x + 1 is fun, and an excuse in itself.
What's there to prove here though? Assuming x is a real number, this is equivalent to subtracting x on both sides and proving .999... = 1.

HallsofIvy
Homework Helper
What he meant was that it is like rattling the bars at the monkey house- you get lots of commotion from the monkeys.

gb7nash
Homework Helper
What he meant was that it is like rattling the bars at the monkey house- you get lots of commotion from the monkeys.
Whoops, I'm looking far too into this.

how we can say if two numbers are equals?...
a friend told me than if a=b, then 0<|a-b|<E*. So, if we gues than a is not iqual b, then |a-b| is not equal 0. ok... the geat an E=|a-b|/2...the definition say 0<|a-b|<E=|a-b|/2. and that's not true. so a=b.
2,9999...->3
=>3-2,99....->0
the diference is so small.

It is very difficult to tell what you are trying to say.

It is true that if a=b then a-b=0.
It is NOT true a=b implies 0<|a-b|<|a-b|/2
It is true if a=b then 0 ≤|a-b|≤|a-b|/2, because if a=b then |a-b| =0 so |a-b|/2 = 0 which gives us 0 ≤|a-b|≤0. Which implies |a-b| = 0, i.e. a-b=0.

I understand the notion you’re grasping at. You feel like 0 < 3 - 2.9999….. This isn’t true. Your problem is you are thinking of 2.999… as some progression or something that is growing. 2.9999… is just a symbol representation of a number, it is a fixed thing that doesn’t entail any type of change.

Last edited:
gb7nash
Homework Helper
how we can say if two numbers are equals?...
a friend told me than if a=b, then 0<|a-b|<E*
Is your friend a math major? Has he taken real analysis?

If a=b, then |a-b|= 0, it isn't greater than 0.

I think little ant means the following:

$$a=b~\text{if and only if}~\forall \epsilon>0:~|a-b|<\epsilon$$

This is indeed true, and show why 3=2.999....

Borek
Mentor
I have a feeling some think not about the number 2.999... but about sequence of numbers 2.9, 2.99, 2.999 ...

I have a feeling some think not about the number 2.999... but about sequence of numbers 2.9, 2.99, 2.999 ...
I don't think they are that wrong in doing so. The Cauchy construction of the reals says that 2.999... is the equivalence class of the fundamental sequence (2.9, 2.99, 2.999, ...). But that very construction also says that 3 is the equivalence class of (2.9, 2.99, 2.999, ...). Thus 3=2.999...
So I have no problem with people that see real numbers as sequences. I have problems with people who work with sequences incorrectly. Specifically, saying that $$x_n<N$$ implies that $$\lim_{n\rightarrow +\infty}{x_n}< N$$. It is this mistake that leads many people astray...