# .99~ = 1 (revisted)

since the other threads were locked i have to post a new one (much to the anger and rage of many people i'm going to assume having read through the other ones :D please don't kill me)

it's seeming to me that the whole argument stems around a misunderstanding as to the nature of numbers and our decimal notational system.

on one side people are saying .999 contains a limit and therefore it is equal to 1.

on the other side people are saying .999 has no limit and therefore will NEVER be equal to 1.

i am of the latter group of people. If a limit is "implied" by the expression .999 then it SHOULD be written out. not "assumed" in any way shape or form.

if you told someone lim(.999) = 1 no one in their right mind would argue with you.

but .999 as a limitless process drawn to infinity has NO end, and therefore will NEVER equal to 1.

i'll let some people offer their insight to the matter thus far before I continue

Wow, took you a while to post this. Anyway, 0.9... is not a limit. It always has had and always will have an infinite number of nines. There are no more nines being added. 0.9... can be expressed as a sum of a series or as a limit, but this doesnt show that it is not equal to one. 0.9... is simply the terminology used to denote an infinite number of nines after the decimal place. Since it is impossible to actually write an infinite number of nines, we write ... (or correctly a bar which I dont know how to code).

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all non-terminating numbers "do not exist" they are created as the output of a process.

just as .333 is created by dividing one by three. as such they are non-quantifiable numbers (read: irrational) because of the inclusion of infinity in their nature. Just as the digits "never end" the process to create the digits never ends as well.

the sum of that infinite series can NEVER be 1. because no matter how many 9's you calculate out to you'll never get a 10 to complete it all the way back through and make the 1 whole.

the difference of 1 and .999 if beyond human comprehension such that calculus finds it convenient to just approximate the value to be 1, but never forget that it IS and approximation, or a limit, whichever you prefer

Diane_
Homework Helper
.9_ is not a limit

It's a constant, a number. I know we're covering old ground, but try it this way: if two numbers are not the same, then it must be possible to find a number between them. There are no numbers between 1 and .9_, so they must be the same.

Some people try something like .0_1 as a number between .9_ and 1. This is absurd, as there can be no such number. The bar (the underscore here) indicates a digit or a group of digits repeating infinitely. If you put a "1" after it, you are saying that there is a digit 1 at the end of an infinite series of zeros. But if the series of 0's is indeed infinite, then there is no end to it, thus no place to put the 1. If you terminate the zeros at some point, then it ceases to be an infinite series.

.9_ can be expressed as a limit of a sum, which I will do when I'm better at the LATEX thing (translation: when I've played with it enough not to think I'll end up looking foolish). For now, think of it as the sum for i = 1 to infinity of $(9/10)^-^i$.

It seems to me that you're thinking of .9_ as a process of some sort - as something that comes closer and closer to 1 without ever reaching it. This is an error. The value can be expressed as the limit of a sum, but it is a constant, in the same way that the sums of certain infinite series can be constants.

Integral
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Gold Member
ram1024 said:
since the other threads were locked i have to post a new one (much to the anger and rage of many people I'm going to assume having read through the other ones :D please don't kill me)
it's seeming to me that the whole argument stems around a misunderstanding as to the nature of numbers and our decimal notational system.
Perhaps the misunderstanding is yours?
on one side people are saying .999 contains a limit and therefore it is equal to 1.

on the other side people are saying .999 has no limit and therefore will NEVER be equal to 1.
Actually because .999... is limited it is = 1. To say that it has no limit the same as saying that it is not only greater then 1 but greater then any other Real number. Many people confuse the concept of an infinite number of digits with an infinite magnitude. These are completely different concepts and should be kept separate.
i am of the latter group of people. If a limit is "implied" by the expression .999 then it SHOULD be written out. not "assumed" in any way shape or form.

if you told someone lim(.999) = 1 no one in their right mind would argue with you.
The representation of a decimal fraction is given by
$$x = \Sigma _{n=1}^{\infty} D_n 10^{-n}$$
Where the $$D_n$$ represent an integer. So any decimal fraction implies an infinite series. You are talking about something a little bit difference. Let me define a sequence of numbers.
$$x_N = 9\Sigma_{n=1}^N 10^{-n}$$
What you are saying is acceptable.
$$lim_{N \rightarrow \infty} x_N = 1$$
If you think that is true why is it not true that
$$lim_{N \rightarrow \infty} x_N = .999....$$
is true. Just expand the sum that is what you get, how can it not be correct.
but .999 as a limitless process drawn to infinity has NO end, and therefore will NEVER equal to 1.
This is exactly WHY it is equal to 1
I'll let some people offer their insight to the matter thus far before I continue
Gee, Thanks.~^

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the limit of a sum is indeed constant

the NON-limited sum to infinity IS a process. a process such that at any given point "later" in the calculation i have obtained a value closer to 1.

There are no numbers between 1 and .9_, so they must be the same.
there are no KNOWN or Human Comprehensible numbers between them. Which doesn't mean they're the same, it just means you can't understand the difference.

krab
Diane_ said:
.9_ can be expressed as a limit of a sum, which I will do when I'm better at the LATEX thing (translation: when I've played with it enough not to think I'll end up looking foolish). For now, think of it as the sum for i = 1 to infinity of $(9/10)^-^i$.
I'll help

$$.9999...=\sum_{i=1}^\infty 9(10)^{-i}$$

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Diane_
Homework Helper
Thank you, Krab!

A gentleman indeed.

Latex is cool

so is this latex thing creating the images from another site, or is there a latex image generator on this site.

it's very spiff

but .999 as a limitless process drawn to infinity has NO end, and therefore will NEVER equal to 1.
This is exactly WHY it is equal to 1
1. if you stop the process at any decimal digit then it doesn't equal 1
2. if you take a limit on the process as a convergence then it DOES equal 1
3. if you DON'T stop the process and let it run forever then it doesn't equal 1

Integral
Staff Emeritus
Gold Member
click on the equations to see the code used to generate the equation. See the link at the top of the General Physics forum to get more specific instructions.

god that is too cool :D

much thanks Integral

Hurkyl
Staff Emeritus
Gold Member
The technical definition of a decimal number is that it is a function from the integers into a set of digits (e.g. {0, 1, 2, ..., 9}) (and a few other details).

This function is, intuitively, a list for the digits. If a function a is a decimal number, then we say a(n) (or more commonly, $a_n$) is the n-th digit. To say it another way, if n is nonnegative, then a(n) is the (n+1)-th digit to the left of the decimal place, and if n is negative, a(n) is the n-th digit to the right of the decimal place.

Boy, that was a mouthful! An example is in order:

The decimal number 12.34 is really the function f where:

f(n) = 0 whenever n > 1
f(1) = 1
f(0) = 2
f(-1) = 3
f(-2) = 4
f(n) = 0
whenever n < -2

This may seem a strange and mysterious way to define things, but if you think about it, it makes sense; all we're doing when we write a decimal number is listing digits and their positions, so it makes sense that the technical representation would be a function that says what digit is in what position.

Anyways, now that we have seen (part of) the actual definition of a natural number, we can clear up some things. When we write $0.\bar{9}$, it is the function f where:

f(n) = 0 for n >= 0
f(n) = 9 for n < 0

Whether you want to think of a function as a generator, a process, a lookup table, a set of ordered pairs, or anything else, the point is that, in all technicality, a decimal number is the function, not the digits it spits out.

My favorite way to actually carry out the construction of the decimal numbers involves doing all of the work on these functions and never saying what real number they're supposed to represent. This approach simply defines $0.\bar{9} = 1$ (and all similar situations); in other words, it's true simply because that's the way we've defined equality of decimal numbers.

It's late so I'm gonna stop here. I need a sleepy smiley.

Integral
Staff Emeritus
Gold Member
ram1024 said:
1. if you stop the process at any decimal digit then it doesn't equal 1
2. if you take a limit on the process as a convergence then it DOES equal 1
3. if you DON'T stop the process and let it run forever then it doesn't equal 1

A number is not something that moves. An infinite number of 9s represents a fixed point on the number line, It is NOT a changing thing, that is what you are implying. There must be an infinite number of 9s to reach 1, any finite number does not make it. You seem to be thinking that infinity is simple a large finite number which keeps moving. That is not the way to think of it. It is not a large finite number, it does not move.

An infinite number of 9s represents a fixed point on the number line, It is NOT a changing thing
an infinite number of decimal sequential digits of 9 even if it WERE to simply EXIST, it is a unique number and quite simply NOT equal to 1. it is certainly NOT rational whereas 1 is

Ram a repeating infinite decimal is rational. It can be expressed as a fraction, and so is rational. The definition of a rational number includes repeating infinite decimals. On this point you are simply wrong.

infinitely repeating decimals can only be approximated as fractions because of the inclusion of infinity in their nature.

i'm not budging on this one

http://www.math.fau.edu/Richman/html/999.htm [Broken]

Perhaps the situation is that some real numbers can only be approximated, like the square root of 2, whereas others, like 1, can be written exactly, but can also be approximated. So 0.999... is a series that approximates the exact number 1. Of course this dichotomy depends on what we allow for approximations. For some purposes we might allow any rational number, but for our present discussion the terminating decimals---the decimal fractions---are the natural candidates. These can only approximate 1/3, for example, so we don't have an exact expression for 1/3.
pure theory as is everything else with this imaginary number, but at least someone else "gets" what i'm alluding to.

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Consider my good friend Zeno (all hail)

he once defined a problem to where motion could not exist because in order to traverse a distance you must first travel half that distance then half the remaining distance then half of THAT remaining distance ad infinitum

this is unequivocally true. sure it's provably false that motion doesn't exist, but within the confines of the problem it can be concluded that given INFINITE number of half distances you could never reach your destination.

the LIMIT of that expression is of course the total distance, the sum is undefined or indeterminable but CONVERGES to the total distance.

so apply what we can from this example to what the situation of $$.99\bar{9}$$ and we can conclude that it is a very good approximation of 1 but not quite 1. the difference being indeterminate, yet real.

that's my thoughts on the matter at least :(

stupid spellings

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Integral
Staff Emeritus
Gold Member
On most web sites on the INTERNET, you can have long arguments about this. However this is not MOST websites. We have active mentors who are chosen because we KNOW WHAT WE ARE TALKING about. There will be no protracted argument about this fact.

You need to change modes. Get out of the "I think I know what I am talking about" mode and get into the learning mode. You may actually learn something you did not know about the the Real Number line if you read and attempt to understand what you are being told. You have been given good information by several others in this thread. I posted a link in the similar thread in Logic on this topic go read that pdf and think about it.

You need to modify your concept of infinity. Zeno did not know about convergent infinite sequences, with that knowledge there is no paradox.

matt grime
Homework Helper
Will it never end? Ram, please look up the definitions of the real numbers. They have nothing to do with what you can write down as the result of some process like this. I dont' know why but this is the single biggest point of misapprehension amongst the, erm, keen but ill-educated, perhaps. (And I mean mathematically, I don't mean intellectually.) Constructibility, Turing machines and fear of the infinite are uniting themes in the crank world and do not affect in the slightest the truth of the statement 0.99... =1 because that truth is inherent in the meaning of all those symbols and the under lying assumption that we are working in base 10 decimal expansions of R.

Integral
Staff Emeritus
Gold Member
Matt,
It goes beyond base 10, since .111.... (binary)=1 and .222... (base 3)=1 etc. this is more another facet of the wonders of the largest digit in what ever base you choose.

Infinite representations are just a fact of life in the real number system. Think about this. .1(base 10) = .0001100110011... (base 2) now is .1 a fixed definite point on the number or is it not!

Ram,
You really should not put so much faith in a single quote, why not make an effort to learn more about the construction of the Real number line before being so argumentative.

Zurtex
Homework Helper
ram1024 said:
Consider my good friend Zeno (all hail)

he once defined a problem to where motion could not exist because in order to traverse a distance you must first travel half that distance then half the remaining distance then half of THAT remaining distance ad infinitum

this is unequivocally true. sure it's provably false that motion doesn't exist, but within the confines of the problem it can be concluded that given INFINITE number of half distances you could never reach your destination.

the LIMIT of that expression is of course the total distance, the sum is undefined or indeterminable but CONVERGES to the total distance.

so apply what we can from this example to what the situation of $$.99\bar{9}$$ and we can conclude that it is a very good approximation of 1 but not quite 1. the difference being indeterminate, yet real.

that's my thoughts on the matter at least :(

stupid spellings
You assume the universe is continuous?

As you have said it would take an infinite number of half distances to get there and that is similar to $$.999...$$ reaching 1, but there are an infinite number of 9s in $$.\bar{9}$$

HallsofIvy
Homework Helper
ram1024: would you please state clearly what DEFINITION of "real number" you are using? I know several definitions that could be used- they all lead immediately to the conclusion that 0.9999... is exactly equal to 1.

Hurkyl
Staff Emeritus
Gold Member
an infinite number of decimal sequential digits of 9 even if it WERE to simply EXIST, it is a unique number and quite simply NOT equal to 1.
As strings of digits they would be unequal. Why do you think they would be unequal as numbers?

i'm not budging on this one
Then you have already defeated yourself.

pure theory as is everything else with this imaginary number, but at least someone else "gets" what i'm alluding to.
So, how much further than this sentence did you read? Did you catch, for instance,

So we see that in the traditional definition of the real numbers, the equation 0.9* = 1 is built in at the beginning.
?

but within the confines of the problem it can be concluded that given INFINITE number of half distances you could never reach your destination.
No; it can be concluded that when considering successive half-distances that reaching the desintation will not be included in the consideration.

so apply what we can from this example to what the situation of $.9\bar{9}$ and we can conclude that it is a very good approximation of 1 but not quite 1. the difference being indeterminate, yet real.
Ok, then, what is the number halfway between $.9\bar{9}$ and 1? It exists because we can compute it via ($.9\bar{9}$ + 1)/2.

matt grime
Homework Helper
Integral said:
Matt,
It goes beyond base 10, since .111.... (binary)=1 and .222... (base 3)=1 etc. this is more another facet of the wonders of the largest digit in what ever base you choose.

Infinite representations are just a fact of life in the real number system. Think about this. .1(base 10) = .0001100110011... (base 2) now is .1 a fixed definite point on the number or is it not!

I will bite back my natural inclination and point out that 0.999... is not equal to one in base 11. I am perfectly well aware of all that.