# Q: Is 0.9999… repeating really equal to 1?

Mathematician: Usually, in math, there are lots of ways of writing the same thing. For instance:

$\frac{1}{4}$ = $0.25$ = $\frac{1}{\frac{1}{1/4}}$ = $\frac{73}{292}$ = $(\int_{0}^{\infty} \frac{\sin(x)}{\pi x} dx)^2$

As it so happens, 0.9999… repeating is just another way of writing one. A slick way to see this is to use:

$0.9999... = (9*0.9999...) / 9 = ((10-1) 0.9999...) / 9$

$= (10*0.9999... - 0.9999...) / 9$

$= (9.9999... - 0.9999...) / 9$

$= (9 + 0.9999... - 0.9999...) / 9 = 9 / 9 = 1$

One.

Another approach, that makes it a bit clearer what is going on, is to consider limits. Let’s define:

$p_{1} = 0.9$
$p_{2} = 0.99$
$p_{3} = 0.999$
$p_{4} = 0.9999$

and so on.

Now, our number $0.9999...$ is bigger than $p_{n}$ for every n, since our number has an infinite number of 9’s, whereas $p_{n}$ always has a finite number, so we can write:

$p_{n} < 0.9999... \le 1$ for all n.

Taking 1 and subtracting all parts of the equation from it gives:

$1-p_{n} > 1-0.9999... \ge 0$

Then, we observe that:

$1 - p_{n} = 1 - 0.99...999 = 0.00...001 = \frac{1}{10^n}$
and hence

$\frac{1}{10^n} > 1-0.9999... \ge 0$.

But we can make the left hand side into as small a positive number as we like by making n sufficiently large. That implies that 1-0.9999… must be smaller than every positive number. At the same time though, it must also be at least as big as zero, since 0.9999… is clearly not bigger than 1. Hence, the only possibility is that

$1-0.9999... = 0$

and therefore that

$0.9999... = 1$.

What we see here is that 0.9999… is closer to 1 than any real number (since we showed that 1-0.9999… must be smaller than every positive number). This is intuitive given the infinitely repeating 9’s. But since there aren’t any numbers “between” 1 and all the real numbers less than 1, that means that 0.9999… can’t help but being exactly 1.

Update: As one commenter pointed out, I am assuming in this article certain things about 0.9999…. In particular, I am assuming that you already believe that it is a real number (or, if you like, that it has a few properties that we generally assume that numbers have). If you don’t believe this about 0.9999… or would like to see a discussion of these issues, you can check this out.

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### 58 Responses to Q: Is 0.9999… repeating really equal to 1?

1. mathman says:

lol. You need to know what each “number” is before you can take the difference. Of course the difference is zero, because you say that they are the same before you subtract, assuming you are subtracting “numbers”

2. George says:

While the mathematical argument presented here is correct, I have to say that limit and/or mathematical infinity introduced created confusions for many. What I wanted to say is that math infinity is really a math trick. It cannot and should not be real in the real physical world. Today this infinity plagues physics and its calculations. Maybe we need something new…

3. EMINEM says:

go to this https://www.youtube.com/watch?v=wsOXvQn3JuE site and get a mathematical proof that 0.999… doesnot equal to 1….

4. . says:

EMINEM, the video was published on April fool’s day for a reason!

5. Josh says:

The simplest explanation I’ve seen for this is:

1/3 = 0.333…
(1/3)3=1 = (0.333…)3=0.999…
1 = 0.999…

I don’t see how anyone can come to any conclusion other than 1=0.999… after understanding this.

6. mathman says:

1 / 3 = ??
Use long division. There are 2 parts: the quotient and the remainder. For every iteration.
First iteration is 0 R 1/3
Second is 0.3 R 1/30
Third is 0.33 R 1/300
Etc etc.

Where does your remainder go, Josh?
You don’t account for it anywhere.

The remainder is never zero.

How many 9’s are there in your 0.999… ?

7. Josh says:

Hey mathman,

There are infinitely many 9s in my “0.999…”. If it were anything other than infinitely many, then it would be less than 1.

And as far as 1/3, your “etc. etc.” is an infinitely repeating “0.333…”. I’m failing to see how anything you’ve presented has disproved what I said.

8. Angel says:

@mathman:

The remainder does not need to be accounted for. The division 1/3 is never formally complete if there is a remainder. Therefore, one must keep iterating the operations in long division. As the number of iterations n approaches infinity, the remainder does approach zero. That is what matters here. The number of 3s in the decimal expansion of 1/3 is infinite, and so is the number of 9s in the figure 0.999…