.999... does not equal 1 because

**Duncan1** · Nov6-09, 03:50 PM

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.

**epkid08** · Nov6-09, 03:58 PM

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

**Werg22** · Nov6-09, 04:36 PM

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

**Mute** · Nov6-09, 06:25 PM

Originally Posted by 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.

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

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

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

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

$LaTeX Code: <BR>\\lim_{n \\rightarrow \\infty} (0.\\overline{9})^{10^n}.$

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

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

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.

**Count Iblis** · Nov6-09, 07:12 PM

http://arxiv.org/abs/0811.0164

**Gerenuk** · Nov6-09, 08:15 PM

Originally Posted by Mute

The proper expression given your proposed definition for $LaTeX Code: 0.\\overline{9}$ would be
$LaTeX Code: \\lim_{n \\rightarrow \\infty} (0.\\overline{9})^{10^n} = \\lim_{n \\rightarrow \\infty} (\\lim_{m \\rightarrow \\infty} (1 - 10^{-m}))^{10^n},$

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.

**rochfor1** · Nov6-09, 10:19 PM

This has been done to death. Google is your friend.

**Petek** · Nov6-09, 10:40 PM

Originally Posted by rochfor1

This has been done to death. Google is your friend.

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

**Duncan1** · Nov7-09, 07:36 AM

Originally Posted by Mute

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

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

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

$LaTeX Code: \\lim_{n \\rightarrow \\infty} (1 - 10^{-n})^{10^n}$

which is NOT equal to

$LaTeX Code: <BR>\\lim_{n \\rightarrow \\infty} (0.\\overline{9})^{10^n}.$

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

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

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.

However,
$LaTeX Code: \\lim_{n \\rightarrow \\infty} (1 - 10^{-n})^{10^n}=1/e$
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
$LaTeX Code: \\lim_{n \\rightarrow \\infty} (1 - 10^{-n})$ has a value other than 1 in the above expression?
What then is the value of $LaTeX Code: \\lim_{n \\rightarrow \\infty} (1 - 10^{-n})$ in the above expression?

**Hurkyl** · Nov7-09, 07:54 AM

$LaTeX Code: 1^{+\\infty}$ is an indeterminate form, just like

, $LaTeX Code: (+\\infty) / (+\\infty)$ , and $LaTeX Code: (+\\infty) - (+\\infty)$ .

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

**Redbelly98** · Nov7-09, 08:46 AM

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.

**Count Iblis** · Nov7-09, 09:17 AM

Originally Posted by Redbelly98

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.

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

So long as the number system has not been specified, the students' hunch that .999... can fall infinitesimally short of 1, can be justified in a mathematically rigorous fashion.

**Redbelly98** · Nov7-09, 09:37 AM

Originally Posted by Count Iblis

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

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.

**Redbelly98** · Nov7-09, 09:44 AM

So long as the number system has not been specified, the students' hunch that .999... can fall infinitesimally short of 1, can be justified in a mathematically rigorous fashion.

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

**Hurkyl** · Nov7-09, 10:07 AM

Originally Posted by Redbelly98

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

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.

**Tac-Tics** · Nov9-09, 09:39 AM

Originally Posted by Redbelly98

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

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.