# .99 = 1?

1. Jun 13, 2008

### Ed Aboud

Hi all.
I can't remember where I saw this but it really confuses me.

If

x = .999999999.....
10x = 9.9999999......
10x - x = 9x

x = 1?

So does that mean if that .9 recurring is 1 exactly?
Thanks for any help.

2. Jun 13, 2008

### Staff: Mentor

Yes, by definition and derivation. You got it.

3. Jun 13, 2008

### nicksauce

Yes it does mean that. .99999.... is an alternate representation for the number 1.

Another way of looking at it: Suppose a and b are real numbers with b>a. There there exists another number, c, such that a < c < b. Suppose we have the hypothesis that .9999.... is not equal to 1. Can you find a number between .999999.... and 1?

4. Jun 13, 2008

### Diffy

5. Jun 13, 2008

### arildno

6. Jun 13, 2008

### maze

Assuming .99999... is actually represents a number, it can't be bigger than 1. It also can't be smaller than 1 because of Nicksauce's reasoning. Therefore the only option is for .9999.. to be 1.

The question of whether .999999... is even a real number at all - the question if the sequence .9 + .09 + .009 + ... converges - is a bit more complicated and requires some basic real analysis.

7. Jun 13, 2008

### arildno

Really?

From what I know, a usual axiom appended to the real number field is that every bounded, increasing sequence converges..
Not too complicated to show that the condition holds in this case..

8. Jun 13, 2008

### Diffy

9. Jun 13, 2008

### Ed Aboud

Ok, thanks very much for the help guys.
And I am sorry if it was posted before, I'm relatively new to PF.

10. Jun 13, 2008

### GRB 080319B

11. Jun 13, 2008

### maze

Yeah, I mean it's not really too complicated, its just probably a little beyond a standard highschool level.

The possible "problem" would be if the real numbers weren't complete (they are complete, so it's not a problem). The analogy would be, the rational numbers are not complete, so the rational sequence 3, 3.1, 3.14, 3.1415, ..., a bounded increasing sequence, doesn't actually converge in the rational numbers. If the real numbers were not complete, .9, .99, .999, ... might converge to some non-real "infintessimal" sort of number, in the same way that 3.14, 3.141, ... converges to a non-rational number.

Last edited: Jun 13, 2008
12. Jun 13, 2008

### LukeD

The form of that axiom that I've always seen is that the real numbers are complete (i.e. every Cauchy sequence of real numbers has a limit in the real numbers). A bounded, increasing sequence is a Cauchy sequence though, so it has a limit.

13. Jun 14, 2008

### D H

Staff Emeritus
This is true, but be fair to the young'uns. This is far beyond high school mathematics. It is even beyond the typical Calc 1/2/3/... sequence required of all science and engineering undergraduates in most colleges. Science majors do study advanced math, but it is stuff like Green's theorem, complex analysis, numerical methods, ... That $$0.\bar{9}=1$$ is even beyond a lot of high school math teachers. I know, because long, long ago I saw the simple 9*0.111... "proof" and pointed it out to a high school math teacher who insisted I was wrong. I didn't learn that I was right until I went to college and took a real algebra course. Bottom line: We should go a bit easy on the kind of people who typically ask this question.

What we need is a standard answer to this question, and now we have the vehicle for this answer: the PF library. Unfortunately, I will be very busy for the next few weeks. Any takers on writing this up? Then when the question comes up again we can simply point to the library entry.

14. Jun 14, 2008

### cristo

Staff Emeritus
That's a good idea that perhaps someone could undertake.

There are also hundreds of threads on this topic that one can find by searching. I'm thus locking this thread.