0.999 repeating

mackdaddy

Junior Member
Joined
Mar 6, 2013
Messages
58
I have a question regarding the claim that 0.999 repeating is equal to 1 the proof is as following
ps. I'm going to use ... as repeating as i donot know how to make the bar on my computer

let 0.999...=x

then

10x = 9.999...
-x -0.999...

9x = 9

then x=1 or 0.999...

But.

Wouldn't there always be an ever so small difference between the two numbers, and if not how could one say that a function whose asymptotes are the x and y axis, doesn't include zero in the Range or Domain of the function, even though the function becomes infinitely closer to the line? Is this not the same idea?
 
Yes, there is a difference. For example:
Suppose that you are going to define an expression for the field of rational numbers.
Suppose you want to use repeating decimals for that purpose.

Notice that there are two candidate expressions for the multiplicative identity.
\(\displaystyle \frac{1}{1}=1. \bar{0}\)
and possibly
\(\displaystyle \frac{1}{1}=0. \bar{9}\)

While all other rational numbers have a unique expression, for example
\(\displaystyle \frac{1}{3}=0. \bar{3}\)

This problem is avoided by definition.
 
I have a question regarding the claim that 0.999 repeating is equal to 1 the proof is as following
ps. I'm going to use ... as repeating as i donot know how to make the bar on my computer

let 0.999...=x

then

10x = 9.999...
-x -0.999...

9x = 9

then x=1 or 0.999...

But.

Wouldn't there always be an ever so small difference between the two numbers, and if not how could one say that a function whose asymptotes are the x and y axis, doesn't include zero in the Range or Domain of the function, even though the function becomes infinitely closer to the line? Is this not the same idea?
.
\(\displaystyle \displaystyle 0.999...\ = \sum_{i=1}^\infty\left(9* 10^{- i} \right) \implies\)
.
\(\displaystyle \displaystyle 10 * 0.999...\ = 10 * \sum_{i=1}^\infty\left(9* 10^{- i}\right ) = \sum_{i=1}^\infty \left(9 * 10^{(1 – i)}\right) = (9 * 10^0)+ \sum_{i= 2}^\infty\left(9 * 10^{(1-i)}\right).\) OK so far?
.\(\displaystyle j = i - 1 \implies j = 1\ if\ i = 2\ and\ - j =1 - i.\)
.
\(\displaystyle So\ \displaystyle 10 * 0.999...\ = 9 + \sum_{j=1}^\infty\left(9* 10^{- j}\right) \implies\)
.
\(\displaystyle \displaystyle (10 * 0.999...) - (0.999) = (0.999...)(10- 1) = 9 * (0.999...) = 9 + \sum_{j=1}^\infty\left(9 * 10^{- j}\right) -\sum_ {i=1}^\infty\left(9* 10^{- i}\right).\)
.
But the two sums on the right are exactly equal in EVERY term so their difference is exactly zero. Nothing left over.
.
\(\displaystyle So\ 9 * (0.999...) = 9 + 0 = 9 \implies 0.999...\ = 1.\)
 

I just thought of a silly way create \(\displaystyle 0.9999\cdots\)


Suppose we ignore the fact that \(\displaystyle 10 \div 1 \,=\,10\)

Instead, we think that 10 divided by 1 is 9 with remainder 1.

The division looks like this:

Code:
          0.9 9 9 9 ...
         --------------
      1 ) 1.0 0 0 0 ...
            9
          ---
            1 0
              9
            ---
              1 0
                9
              ---
                1 0
                  9
                ---
 
Wouldn't there always be an ever so small difference between the two numbers, and if not how could one say that a function whose asymptotes are the x and y axis, doesn't include zero in the Range or Domain of the function, even though the function becomes infinitely closer to the line? Is this not the same idea?

Your argument is important.
There is a big difference between an expression and a limit.
The infinite ordered set of elements in a convergent series, does NOT include the limit.

Jeff's post is correct if you define 0.9999... to mean the value of the limit S = {0.9, 0.99, 0.999, ...}
However, 1 is not in S.

The question becomes less clear when you talk about a set of finite expressions.
My post was restricted to discuss the set of finite expressions for the rationals using a Vinculum. As you pointed out, the set of those expressions is not well defined until you assign the value 1 to two different expressions.
 
Last edited:
then x=1 or 0.999...

But.

Wouldn't there always be an ever so small difference between the two numbers...?
The point of the "..." is that the 9's continue "forever". For there to be a difference, small or otherwise, between 0.999... and 1, there would have to be some number that could fit "between" the two numbers.

What would you propose that "between" number to be? ;)
 
I have a question regarding the claim that 0.999 repeating is equal to 1 the proof is as following
ps. I'm going to use ... as repeating as i donot know how to make the bar on my computer

let 0.999...=x

then

10x = 9.999...
-x -0.999...

9x = 9

then x=1 or 0.999...

But.

Wouldn't there always be an ever so small difference between the two numbers, and if not how could one say that a function whose asymptotes are the x and y axis, doesn't include zero in the Range or Domain of the function, even though the function becomes infinitely closer to the line? Is this not the same idea?
A function is not a number! You are not talking about a function that takes on different values for different x and so "goes" to some limit, you are talking about a specific number that has a single specific value. There is no "always" about it.
 
Having seen Bob Brown's post as well as stapel's and having discussed Bob's with him, I'd like to clarify my post a bit in light of their excellent posts. It is usual to teach elementary algebra using the real (or complex) numbers, which strictly exclude transfinite numbers. The sums that I said were equal are so only because they are sums of an infinite number of terms. If they were sums of a finite number of terms, there would be a small difference between the sums as you pointed out and Bob confirmed. After all, 0.999,999,999 is not equal to 1. So my demonstration does not work if you exclude the concept of infinity from your tool kit (as it is excluded from the tool kit of elementary algebra). The demonstration would then have to involve the concept of limit, which is usually introduced in calculus. However, the notation 0.999... implies infinity rather than limit, as staple took the trouble to explain whereas I failed to do so. Consequently, excluding infinity from the explanation of something that implies infinity seems to me contradictory, and I felt free to use sums with an infinite number of terms.
 
Yes, there is a difference. For example:
Suppose that you are going to define an expression for the field of rational numbers.
Suppose you want to use repeating decimals for that purpose.

Notice that there are two candidate expressions for the multiplicative identity.
\(\displaystyle \frac{1}{1}=1. \bar{0}\)
and possibly
\(\displaystyle \frac{1}{1}=0. \bar{9}\)

While all other rational numbers have a unique expression, for example
\(\displaystyle \frac{1}{3}=0. \bar{3}\)

This problem is avoided by definition.

There are a countable number of rationals having two different expansions.

\(\displaystyle \dfrac{1}{2} = .4999999...\)
\(\displaystyle \dfrac{2}{25} = .07999999...\)

edit: I think the biggest misconception about repeating decimals is the gerund that sits in its name. The number .9999... isn't doing anything, it isn't repeating, it isn't continuing, it isn't approaching anything. It is a fixed number.
 
Last edited:
edit: I think the biggest misconception about repeating decimals is the gerund that sits in its name. The number .9999... isn't doing anything, it isn't repeating, it isn't continuing, it isn't approaching anything. It is a fixed number.

The adjective repeating is very descriptive. The biggest misconception about repeating decimals is the missing nouned-verb at he end of the name.

Correct, unabbreviated name: repeating decimal expression.

These are NOT numbers fixed in a continuum. They are expressions that are given meaning by definitions. Once definitions are in place, then a notion of their meaning can emerge. I have suggested that a useful and brief set of definitions could leverage from the definitions of rational numbers . Edit: {Then, as danon points out, the most reasonable notion, is a fixed number representing a fixed location in a continuum. Further, you may conclude, that there is really no distinction to be preserve by defining a new field -- and simply apply the definition of "numeral" to these objects, as Hallsofivy did.}

The advantage of a one-to-one pairing of repeating decimal expressions to the set of rational numbers is this: A definition of repeating decimal arithmetic would be unnecessary.

At first one might think that just calling \(\displaystyle 0. \bar{9}\) undefined might solve everything. However, later, when trying to define arithmetic operations with these objects you may want \(\displaystyle 0. \bar{9}\) back. Example: \(\displaystyle 0. \bar{3}\) + \(\displaystyle 0. \bar{6}\) = \(\displaystyle 0. \bar{9}\). My solution is to have just one equality rule and define \(\displaystyle 1. \bar{0}\) = \(\displaystyle 0. \bar{9}\) as the multiplicative identity. Edit: {It is important to me that this object be able to stand alone as a field, because I have an interest in extending/understanding repeating decimals as a subset of the p-adic number field. It is sufficient to have a simple conversion rule to define repeating decimal expressions as alternative rational numerals (then always convert and use the +, * arithmetic and field postulates for [n,d] ). However, without thinking, most people believe that long division, multiplication, addition and subtraction works for the decimals -- well it does -- but not without thinking (quite complicated). This is the area that I am thinking about. Any help is extremely welcome! -- Responses should be made in the new thread }

I believe that is what you (daon2) are suggesting also.
 
Last edited:
I haven't quite seen my favorite argument.

"Wouldn't there always be an ever so small difference between the two numbers"

If \(\displaystyle 1 \ne 0.99\overline{9}\), then \(\displaystyle 1 - 0.99\overline{9} > 0\) or \(\displaystyle 1 - 0.99\overline{9} = \epsilon > 0\), where \(\displaystyle \epsilon\) is the "ever so small difference" the OP is talking about. Unfortunately for our little friend \(\displaystyle \epsilon\), we cannot pin down just how small it is. ANY positive value assigned can be shown to be too big simply by writing enough 9s.

In words, if 1 is not EXACTLY the same as 0.9999...., then you had better tell me how different the two are. As soon as you pick a value, I can show you that you have chosen a value that is greater then the actual difference. Since you can pick any value you wish, we have established EQUALITY.
 
What do you mean it is not fixed?

A definition is not needed to require that 1=.999.... Two real numbers are distinct if and only if some real number lies between them. It is a result of the axioms of the real numbers that the equality is true; either 1) density or 2) the uniqueness of a least upper bound will do.

edit: tkhunny beat me by .999... minutes ;)
 
My opinion: 0.99999..... is a fictitious number - there is physical process to get there other than limit consideration. 0.333... can be achieved with 1/3 and all other numbers have fractional equivalence - and a physical way to get there (e.g. 0.7272... = 72/99 = 8/11).
 
What do you mean it is not fixed?

A definition is not needed to require that 1=.999.... Two real numbers are distinct if and only if some real number lies between them. It is a result of the axioms of the real numbers that the equality is true; either 1) density or 2) the uniqueness of a least upper bound will do.

edit: tkhunny beat me by .999... minutes ;)

Sorry danon2, I do not understand your post. We are talking about expressions, not real numbers. These expressions can be useful to express the notion of rational numbers one-to-one (with the one notable exception being discussed).
 
.333...=1/3
3*.333...=3*(1/3) = > > 3*1/(1/3) < <
.999...=1
It should be "3*1/3" or "(3*1)/3," otherwise your last expression equals 9.


.111=1/9
9*.111...=9*(1/9)= > > 9*1/(1/9) < <
.999...=1

Likewise, this should be "9*1/9" or "(9*1)/9," otherwise your last expression equals 81.


Both equations are valid due to the Transitive Property [If you don' know the Transitive property,
I will state it... Transitive Property- If a=b and b=c then a=c]
......................................................................................................................................

...
 
Last edited:
There are a countable number of rationals having two different expansions.

\(\displaystyle \dfrac{1}{2} = .4999999...\)
\(\displaystyle \dfrac{2}{25} = .07999999...\)

edit: I think the biggest misconception about repeating decimals is the gerund that sits in its name. The number .9999... isn't doing anything, it isn't repeating, it isn't continuing, it isn't approaching anything. It is a fixed number.

My description of the one-to-one nature of this notation does require a canonical form for repeating decimals. I thought it was obvious -- but as you illustrate, a lack of rigor can be confusing. This is an important exercise though, because it underscores the fact that we are talking about an expression representing what you call "a fixed number". {Edit: I agree.} However, It is NOT a point on a number line, it represents a point on a number line.

Edit: I agree That, "The number .9999... isn't doing anything, it isn't repeating, it isn't continuing, it isn't approaching anything." The adjective repeating, inspires those notions. Those notions are interesting. But as you point out, they generate a lot of diversion.
Below is a response to your valid statement, "There are a countable number of rationals having two different expansions." It shows a lack of rigor (on my part) to claim that there is only one exception. Once the notation that I suggest is accepted, then rule #3) can be used to resolve all of those cases. Rule #3) is stating that 0.999.... = 1.

So, for my posts assume the following canonical form for repeating decimals (base 10) in pseudo- scientific notation to be as follows.

Example : \(\displaystyle \frac{230}{7}\text{ = 3.}\overline{285714}\) E 1

1) Digits under the vinculum represent no more than one cycle of repetition.
2) Decimal point is always located just before the first digit under the vinculum.
3) \(\displaystyle \overline{9}\) is replaced by \(\displaystyle \overline{0}\) and add 1 to the mantissa.
4) E n represents that the resulting number is multiplied by 10^n
5) Represent the additive identity as \(\displaystyle \text{ = 0.}\overline{0}\) E 0

Edit: Analogous example for repeating integer digits
{ Example : \(\displaystyle \frac{230}{7}\text{ = }\overline{571428}\text{.9}\) E 2 } used in future posts.

Now there is only one way to express a rational number using this repeating decimal notation.
The discussion brought by the OP is that he finds it unclear what is meant when someone writes the expression 0.999... We have in this thread demonstrated that lack of clarity (as is true in message boards across the web). I submit that the root of this confusion comes from not calling 0.9999... an expression, and not considering what is the most useful (while consistent) definition.

My vote: Let 0.999... be a non-standard way of writing
\(\displaystyle \frac{1}{1}=1. \bar{0}\) or perhaps \(\displaystyle \frac{1}{1}=1. \bar{0}\) E 0
 
Last edited:
Just for fun

This thread has appeared in every math message board that I have visited.
However it reminds me of an OP question that I had never seen before.

Repeating Integers (thread)

It is an apparent attempt by some sadistic grade school teacher to expose the class to base 10 p-adic integers :)

\(\displaystyle \frac{1}{7}\text{ = }\overline{285714}3.0\)

I'm not sure how rigorous it is, but after playing with this notation, I have been able to express all of the rationals (positive and negative) with this (unsigned) notation!

BUT, I have to define \(\displaystyle \text{-1 = }\overline{9}.0\) :) :) :(
 
Last edited:
Well I appreciate the feedback on this particular post, but... I would like propose one more idea

say you 0.9 the difference from 1 is 0.1

then you took 0.99 then the difference is 0.01, and even if you kept doing this an infinite amount of times would you not always have a decimal 0.0...01?

what this proposes is that you don't work a terminating process on an infinitely repeating decimal, but instead work an infinitely repeating process on infinite terminating decimals.
 
Well I appreciate the feedback on this particular post, but... I would like propose one more idea

say you 0.9 the difference from 1 is 0.1

then you took 0.99 then the difference is 0.01, and even if you kept doing this an infinite amount of times would you not always have a decimal 0.0...01?

what this proposes is that you don't work a terminating process on an infinitely repeating decimal, but instead work an infinitely repeating process on infinite terminating decimals.
The limit of that process is zero. If you look at my first post, you will see sums with infinite numbers of terms: there is no terminating implied. Personally, I think this is a spurious question. It arises by trying to think in physical or procedural terms about the real numbers, which are free creations of the human mind and have never been observed in the physical universe. In a physical sense, the real number system is fictitious; no one has ever measured anything that was not a rational number. In any case, in terms of the theory of real numbers, 0.999... is not a process, but a specific real number. There is no mystery that the same number can be expressed in multiple ways: positive square root of 16, 10 minus six, and 3 plus 1 all mean 4. You do not doubt that 1/3 = 0.333.... Nor do you doubt that 3 * (1 /3) = 1, and 3 * 0.333... self-evidently equals 0.999.... . As daon has now explained several times, the real number system is dense, and two real numbers are the same by definition if no real number intervenes between them. Try figuring out what number could intervene between 1.000... and 0.999... You may now understand Kronecker's comment that God invented the integers, but the rest are the work of man.
 
And sorry but one more question regarding my question about the functions.

A function whose asymptotes are the x and y axes, such as f(x)=1/x has a Domain of -infinity>0<infinity and the same for range, and is always approaching the axes, would cross every single x coordinate and y coordinate except zero. But what about f(x)=1/infinity? I don't think infinity is used in functions and this proposed value for "f(x)" is impossible to calculate but it is interesting to think about.

And if one says that this is still never equal zero, would that not be the same as saying 0.999... could never equal 1, or 1-0.999... could never equal 0?
 
Top