# .999 does not equal 1 because

1. Nov 6, 2009

### Duncan1

Represent .999... by (1-10-n), limit as n>infinity.
Then
(1-10-n)10n, limit as n>infinity = 1/e (binomial expansion)
(1+10-n)10n, limit as n>infinity = e
(1)10n, limit as n>infinity=1

I think .999... is a transcendental number that can in almost every case be treated as equal to 1. Any solution to a finite algebraic equation that approximates 1 but is less than 1 will always be less than .999...
If .999... is transcendental then so is every other recurring decimal because they are arbitrarily close to a fraction but not equal to that fraction.

The above equation seems to converge with the number of digits of accuracy approximately equal to n.

2. Nov 6, 2009

### epkid08

If you need proof that .999...=1, take a look at the infinite series that represents .999....

3. Nov 6, 2009

### Werg22

If 0.99... is not equal to 1, then there's definitely a number between the two. Can you find one such number?

4. Nov 6, 2009

### Mute

It doesn't work like that. You're saying

$$\mbox{Let}~0.\overline{9} \equiv \lim_{n \rightarrow \infty} (1 - 10^{-n}).$$

First, the limit of that expression is certainly 1 - no doubt about that. It's not the way one tends to define a decimal expansion but let's use it anyways. You then consider

$$\lim_{n \rightarrow \infty} (1 - 10^{-n})^{10^n}$$
which is NOT equal to

$$\lim_{n \rightarrow \infty} (0.\overline{9})^{10^n}.$$

The proper expression given your proposed definition for $0.\overline{9}$ would be

$$\lim_{n \rightarrow \infty} (0.\overline{9})^{10^n} = \lim_{n \rightarrow \infty} (\lim_{m \rightarrow \infty} (1 - 10^{-m}))^{10^n},$$

which is a different thing altogether - there are two limits involved, instead of just the one as in your original post. This makes a very big difference. Note also that you cannot exchange the order of the limits here -that would also be an invalid operation.

5. Nov 6, 2009

6. Nov 6, 2009

### Gerenuk

I agree with Mute. The point is that you cannot make the single limit you had first into a double limit with a second occurence of the same infinity.

7. Nov 6, 2009

8. Nov 6, 2009

### Petek

Correct. If this Wikipedia article doesn't convince you, then probably nothing ever will.

9. Nov 7, 2009

### Duncan1

However,
$$\lim_{n \rightarrow \infty} (1 - 10^{-n})^{10^n}=1/e$$
has a definite value other than 1. Try some values of n in a calculator.
Is this expression in error?
Does this expression not indicate that
$$\lim_{n \rightarrow \infty} (1 - 10^{-n})$$ has a value other than 1 in the above expression?
What then is the value of $$\lim_{n \rightarrow \infty} (1 - 10^{-n})$$ in the above expression?

10. Nov 7, 2009

### Hurkyl

Staff Emeritus
$1^{+\infty}$ is an indeterminate form, just like $0/0$, $(+\infty) / (+\infty)$, and $(+\infty) - (+\infty)$.

If you can prove 0.999... is unequal to 1, then you have found an inconsistency in the arithmetic of integers.

11. Nov 7, 2009

### Redbelly98

Staff Emeritus
To the OP:

What is the decimal representation of 1/3?
Do you accept that 1/3 = 0.333...?

If so, multiply x3 on both sides of that equation and see what you get.

12. Nov 7, 2009

### Count Iblis

If 0.999999... is questioned to be equal to 1, then why would one accept that 0.33333333.... equals 1/3?

I think the answer to ".999... does not equal 1 because..." should be that the number system has not been specified, as pointed out here:

http://arxiv.org/abs/0811.0164

13. Nov 7, 2009

### Redbelly98

Staff Emeritus
That occurred to me. Hence my initial question, "What is the decimal representation of 1/3?" Most people are okay with saying 0.333... in answer to that, and don't question it.
However, the perception that 0.999... looks differently than 1.000... is what distinguishes the two examples in many people's minds.

14. Nov 7, 2009

### Redbelly98

Staff Emeritus
I don't understand -- do we not have an agreed-upon number system?

15. Nov 7, 2009

### Hurkyl

Staff Emeritus
The author of that PDF is writing it pretty much specifically to justify the 0.999... != 1 hypothesis. The main loophole he exploits is that we haven't given a fully precise definition of the number system we are teaching them -- and so he's filling in the missing details in a... peculiar... fashion.

That is not to say there is no merit in non-standard analysis -- but it's like the author is specifically trying to prevent students from understanding what an infinite decimal means.

16. Nov 9, 2009

### Tac-Tics

Most people have a terrible understanding of the real numbers. It isn't usually until college (more likely, never) that people learn about the reals as the completion of the rational number system. The vast majority of people thing an irrational number's defining characteristic is that the decimal expansion doesn't repeat. And they believe that you can add two irrational numbers together in the same way you add rationals, using the algorithm they learned in elementary school -- despite the fact that it obviously doesn't terminate!

The peculiar thing isn't that people don't understand numbers, it's that they so fiercely stick to their broken notions of numbers. They won't even entertain the idea that what they learned was incomplete. After all, business and science are both done with rational numbers! And there's a strange belief (in both math and science) that there's only one correct way these kinds of things can work. But that's the beauty of math! You can make up the rules to be whatever you want as long as you strictly adhere to them.

Last edited by a moderator: Nov 9, 2009
17. Nov 9, 2009

### lurflurf

Most people do not need a good understanding of real numbers. Real numbers are quite obviously a completion of the rationals, they are the numbers between rational like sqrt(2). It is perfectly valid to take nonterminating decimal expansion as a defining characteristic of irrational numbers. Irrationals are aded the same way as rationals, or at least an infinite process analog. I think the trouble is these people (though most of them know .(9)=1 and just want to start stupid threads) do not realize a infinite decimal expansion is inherently a real number. Arguing .(9) is not real is like arguing 17/19 is not rational. One could (and these crack pots do) devise a number system different than the reals (and having horrible properties) but represented by infinite decimal expansions in which two representations that represent the same real number represent different crackpot numbers. The two sides of this "debate" are not even talking about the same thing. We have
Crackpot: In the crackpot number system clearly .(9)!=1.(0)
!Crackpot: In the real number system clearly .(9)=1.(0)
They are both right.

18. Nov 9, 2009

### Mute

No, that expression is perfectly correct.

No, it absolutely does not indicate that. $\lim_{n \rightarrow \infty} (1 - 10^{-n}) = 1$. Your error is the following: consider the expression

$$(1-10^{-n})^{10^m}$$.

If I take the limit $n \rightarrow \infty$ this becomes $(1)^{10^m} = 1$ for any finite value of m. If I instead keep n fixed at some finite value and take $m \rightarrow \infty$ then because $1-10^{-n} < 1$, the expression tends to zero. If I set $m = n$ and then take the limit $n \rightarrow \infty$ I get the value of 1/e. You're mixing up these limiting processes which gives you different results depending on how you take the limit.

19. Nov 9, 2009

### Hurkyl

Staff Emeritus
I'm possibly nitpicking a bit, but:
. an infinite decimal expansion is a real number only because we have chosen to use decimal notation to express real numbers
. we only use the left-finite expansions

20. Nov 10, 2009

### Count Iblis

Most real numbers are not formally describable (the set of formally describable numbers must obviously be countable), so you can question if real numbers really exist.