Is .999 Repeating Equal to 1?: The Proof Behind the Infamous Debate

[jbriggs444]

I think that the stated inequality entails the stated equality only by definition.

By definition of what, exactly? What definition is being invoked? Can you state that definition?

 

[Mark44]

That is, $\forall \epsilon > 0, \exists N : \forall n \ge N, \left(1 - \sum_{i = 1}^n \frac 9 {10^i} \right)< \epsilon$
I would not call this a definition. Instead, it's a proof that .999... is equal to 1.
I would not call this a definition.

I think that the stated inequality entails the stated equality only by definition.

What I wrote above isn't a definition. It is a statement that uses the definition of the limit of a sequence -- a statement that is applying a definition.

 

[sysprog]

What I wrote above isn't a definition. It is a statement that uses the definition of the limit of a sequence -- a statement that is applying a definition.

I didn't say that the inequality that you wrote was a definition; I said that it entails the stated equality, "0.999... = 1", only by definition.

By definition of what, exactly? What definition is being invoked? Can you state that definition?

According to @Mark444, by "the definition of the limit of a sequence"; however, the definition of the density of the reals, or the definition of the infinitesimal as zero, as in the statement "the only non-negative number that is less than any positive real number is zero", which partially defines the reals by exclusion of the hyperreal infinitesimal, would suffice to exemplify the stated equality as by definition.

 

[jbriggs444]

I didn't say that the inequality that you wrote was a definition; I said that it entails the stated equality, "0.999... = 1", only by definition.
According to @Mark444, by "the definition of the limit of a sequence"; however, the definition of the density of the reals, or the definition of the infinitesimal as zero, as in the statement "the only non-negative number that is less than any positive real number is zero", which partially defines the reals by exclusion of the hyperreal infinitesimal, would suffice to exemplify the stated equality as by definition.

One does not normally define the density of the reals. Nor would one assert something about the density of the reals "by definition".

Instead, one would define what it means for one set to be "dense" within another and show from other accepted axioms and theorems how it follows that the set of real numbers is dense within itself. That is, you would prove that between any two distinct real numbers there is yet another real number that lies between the two.

One would not say that there is no strictly positive real number which is less than every other strictly positive real number "by definition". Instead, one might state that fact as a trivial consequence of the fact that the set of real numbers is dense within itself: for any putative minimum positive real number x there is another real number between 0 and x.

If you want to make a statement about the notation "0.999..." and want to make it true "by definition" then you had darned well better have a definition for the notation ready at hand.

You not offered any such definition despite being asked. I assume that you have in mind a definition that the notation "0.999..." denotes the limit of the sequence of partial sums 0.9, 0.99, 0.999, ... If so, one could proceed to prove that this limit is equal to 1.

But that equality is a matter to be proven. Its truth is not a consequence of the definition of 0.999... as the limit of a sequence. Its truth is not "by definition".

 

[sysprog]

But the definition of the limit of a sequence only matters if "0.999..." has a value that is defined as the limit of a sequence. You have not invoked such a definition.

Rather than implicitly using the definition of the limit of an infinite sequence, as @Mark44 did, I used this specific definition: $$(\text{Min.}~ x:\forall(n>0)[0.(9)_n<x]=1)⇒(x=1)~ \text{Def.}$$as an example of a definition by which 0.999... is equal to 1.

Yes, I agree that such is the conventional definition. But it is not the only possible understanding that one can have of the notation.

I think that the inequality written by @mark444 adequately presents the decimal expansion as an infinite summation that converges to 1.

 

[jbriggs444]

Rather than implicitly using the definition of the limit of an infinite sequence, as @Mark44 did, I used this specific definition: $$(\text{Min.}~ x:\forall(n>0)[0.(9)_n<x]=1)⇒(x=1)~ \text{Def.}$$as an example of a definition by which 0.999... is equal to 1.
I think that the inequality written by @mark444 adequately presents the decimal expansion as an infinite summation that converges to 1.

Nowhere in there do I see a definition for "0.999..."

 

[sysprog]

Nowhere in there do I see a definition for "0.999..."

$0.(9)_n=0.999... ~\text{Def.}$

 

[jbriggs444]

$0.(9)_n=0.999... ~\text{Def.}$

OK. We have one notation defined in terms of another. How is the new notation defined? Do we tie it to a value at some point?

 

[Mark44]

One problem with the following as a definitiion is that it is too specific.

$(\text{Min.}~ x:\forall(n>0)[0.(9)_n<x]=1)⇒(x=1)~ \text{Def.}$

Would there need to be similar definitions for 1.999..., 2.999..., 3.999..., etc., as well as for .4999..., .24999..., and infinitely more decimal fractions?

 

[Mark44]

$0.(9)_n=0.999... ~\text{Def.}$

Most would interpret the left side to mean some finite number n of 9 digits to the right of the decimal point. The notation 0.999... means an infinite number of 9 digits after the decimal point.

 

[sysprog]

Do we tie it to a value at some point?

That's what the definition

$(\text{Min.}~ x:\forall(n>0)[0.(9)_n<x]=1)⇒(x=1)~ \text{Def.}$

does $-$ it sets the value of 0.999... as equal to 1.

 

[jbriggs444]

That's what the definition
$(\text{Min.}~ x:\forall(n>0)[0.(9)_n<x]=1)⇒(x=1)~ \text{Def.}$
does $-$ it sets the value of 0.999... as equal to 1.

No. It does not. It cannot since it never mentions the text string "0.999...".

As @Mark44 points out, since you use $0.(9)_n$ as a formula with one parameter, it becomes clear that it is not, in fact, identical to the formula $0.999...$ since the latter lacks a free parameter.

 

[sysprog]

Most would interpret the left side to mean some finite number n of 9 digits to the right of the decimal point. The notation 0.999... means an infinite number of 9 digits after the decimal point.

Yes, I would better have said $(\text{Max.}~ n: (0.(9)_n<1)) =0.999...~\text{Def.}$

 

[jbriggs444]

Yes, I would better have said $(\text{Max.}~ n: (0.(9)_n<1)) =0.999...~\text{Def.}$

What does that mean? Is that supposed to denote the maximum element of the set whose elements are given by $0.(9)_n$ for n in $\mathbb{N}$.

If so, I would object that you have failed to define anything. That set has no maximum, though it is bounded above and, accordingly, does have a least upper bound.

 

[sysprog]

What does that mean? Is that supposed to denote the maximum element of the set whose elements are given by $0.(9)_n$ for n in $\mathbb{N}$.

If so, I would object that you have failed to define anything. That set has no maximum, though it is bounded above and, accordingly, does have a least upper bound.

What I meant by it was that for the maximum value of $n$ such that $0.(9)_n$ is less than $1$, $0.(9)_n$ is equal to $0.9...$, but in writing that, I can see that my notation was again faulty $-$ perhaps better would have been $(0.(9)_n:\lnot\exists(n')[n'>n])=0.999...~\text{Def.}$

 

[jbriggs444]

What I meant by it was that for the maximum value of $n$ such that $0.(9)_n$ is less than $1$, $0.(9)_n$ is equal to $0.9...$, but in writing that, I can see that my notation was again faulty $-$ perhaps better would have been $(0.(9)_n:\lnot\exists(n')[n'>n])=0.999...~\text{Def.}$

"The maximum value of $n$ such that $0.(9)_n$ is less than 1"

But there is no such maximum value. For every $n$ in $\mathbb{N}$ that satisfies the condition that $0.(9)_n$ is less than 1 [which is all of them], there is another natural number [e.g. n+1] that is larger. You realize that the notion of the a last item in an infinite sequence is an oxymoron, right?

There are much easier ways to phrase what you should be trying to say. So I worry (possibly without basis) that what you are actually saying reflects an incorrect intuition.

 

[sysprog]

"The maximum value of $n$ such that $0.(9)_n$ is less than 1"

But there is no such maximum value. You realize that the notion of the a last item in an infinite sequence is an oxymoron, right?

The proposed revision doesn't use that:

$(0.(9)_n:\lnot\exists(n')[n'>n])=0.999...~\text{Def.}$

 

[jbriggs444]

The proposed revision doesn't use that:
$(0.(9)_n:\lnot\exists(n')[n'>n])=0.999...~\text{Def.}$

So now you are saying that "0.999..." is defined to be the set containing all numbers of the form $0.(9)_n$ where there is no natural number $n'$ which is greater than n.

Which is to say that "0.999..." is equal to the empty set.

No. You are indeed working from a faulty intuition about how the infinite works.

There is no maximal element of the natural numbers. There is no sound way of writing down an expression that refers to one by cleverly couching the reference in terms of quantifiers. The natural numbers are an infinite set. The notation "0.999..." does NOT refer to a sequence of the form $0.(9)_n$ for some un-named n. There is no such n. Full stop. The infinite string is different from every finite string.

Every natural number is finite. There are infinitely many of them. None of them are the last one.

 

[sysprog]

So now you are saying that "0.999..." is defined to be the set containing all numbers only that number of the form $0.(9)_n$ where there is no natural number n′ which is greater than n.

I regard that as meaning that the number of 9s to the right of the decimal point is infinite. I think that only an infinite number of 9s satisfies the condition of there being no number of 9s that is greater. In my view, asserting the non-existence of a greater number than a number, avoids the problem with explicitly calling a number the greatest number, which problem arises by virtue of the fact that by definition in the real numbers, the greatest number is always a member of a finite set of numbers.

 

[jbriggs444]

I regard that as meaning that the number of 9s to the right of the decimal point is infinite. I think that only an infinite number of 9s satisfies the condition of there being no number of 9s that is greater. In my view, asserting the non-existence of a greater number than a number, avoids the problem with explicitly calling a number the greatest number, which problem arises by virtue of the fact that by definition in the real numbers, the greatest number is always a member of a finite set of numbers.

No. That is not correct. Infinity is not a natural number. Asserting the non-existence of a thing whose name you are using as if it were a thing is a violation of the rules of reasoned discourse.

Even if you did succeed in somehow naming the non-existent natural number that is greater than all the rest, that does not help if you now call that number n and use it in the notation $0.(9)_n$. We can all guess what that notation means for any finite number n. But what does it mean when n is a non-existent infinite number? You now need a definition for the undefined notation you've just created. You are right back where you are started, needing a definition in order to be able to say "by definition".

Furthermore, you have used the term "by definition" in order to assert that the greatest number is always a member of a finite set of real numbers. Again exactly what definition are you invoking here? And to what purpose?

Edit: You've also mis-stated the result [as I did also for a moment]. The empty set is an example of a finite set of real numbers that has no greatest member.

One can prove that given a non-empty finite set and an order that there is a greatest member. One proof proceeds by mathematical induction and uses the fact that any finite set (*) can be placed in one to one correspondence with an initial subset of the natural numbers. However, this proof is not "by definition". It's a plain old proof by actually doing a proof.

(*) By one definition of finite, anyway.

 

[sysprog]

@jbriggs444, thanks for your patience with me $-$ I'll get back to this later . . .

 

[WWGD]

The axioms of the ( standard) Real numbers (Archimedean Principle) do not allow the existence of indefinitely-small numbers, which are then forced to equal 0. The number 1-0.9999... can be made to be indefinitely-small.

 

[Mark44]

The axioms of the ( standard) Real numbers (Archimedean Principle) do not allow the existence of indefinitely-small numbers, which are then forced to equal 0. The number 1-0.9999... can be made to be indefinitely-small.

First, 1 - 0.999... = 0. The difference is not "getting close to zero" or "approaching zero."
Second, you seem to be contradicting yourself, saying first that the axioms don't allow for indefinitely small numbers, and then saying that 1 - 0.999... can be made indefinitely small.

I would agree if you had said $1 - \sum_{k = 1}^n \frac 9 {10^k}$ can be made as small as one likes, simply by choosing a larger value of n, there is no space between 1 and 0.999...

 

[WWGD]

First, 1 - 0.999... = 0. The difference is not "getting close to zero" or "approaching zero."
Second, you seem to be contradicting yourself, saying first that the axioms don't allow for indefinitely small numbers, and then saying that 1 - 0.999... can be made indefinitely small.

I would agree if you had said $1 - \sum_{k = 1}^n \frac 9 {10^k}$ can be made as small as one likes, simply by choosing a larger value of n, there is no space between 1 and 0.999...

The difference between 1 and 0.999..., aka the partial sums 0.9, 0.99, 0.999,... becomes indefinitely small, so that by the Archimedean principle, it must equal 0. Maybe I used slightly-innacurate shorthand but it stands as explained in this post: as the number of 9s increases, the difference becomes indefinitely small, and so, for the standard reals, is zero.

 

[Mark44]

The difference between 1 and 0.999..., aka the partial sums 0.9, 0.99, 0.999,... becomes indefinitely small, so that by the Archimedean principle, it must equal 0. Maybe I used slightly-innacurate shorthand but it stands as explained in this post: as the number of 9s increases, the difference becomes indefinitely small, and so, for the standard reals, is zero.

But 0.999... is not AKA the partial sums 0.9, 0.99, 0.999. The number 0.999... is the limit of the sequence of partial sums.

 

[WWGD]

But 0.999... is not AKA the partial sums 0.9, 0.99, 0.999. The number 0.999... is the limit of the sequence of partial sums.

Others have given formal, rigorous answers so , to complement said answers, I am trying to to give something more informal which I believe is still clear. If you want more rigor:
$Lim_{n \rightarrow \infty} ( 1- \Sigma _ {k=1}^{n} 9( 10)^{-n}) =0$. The partial differences can be made indefinitelly-small, which forces equality. Therefore, $0.99999...=1$.

 

[Mark44]

My complaint was that you equated ("aka") 0.999... with the partial sums 0.9, 0.99, and so on.

With regard to providing something more informal, "keep things as simple as possible, but no simpler."

 

[Frodo]

The simple "high school proof" takes only two lines.

divide 3 into 1 to get 1/3 = 0.333333...
multiply both sides of the equation by 3 to get
1 = 0.999999...

or divide 9 into 1 to get 1/9 = 0.111111...
multiply both sides of the equation by 9 to get
1 = 0.9999999...

 

[jbriggs444]

The simple "high school proof" takes only two lines.

divide 3 into 1 to get 1/3 = 0.333333...
multiply both sides of the equation by 3 to get
1 = 0.999999...

or divide 9 into 1 to get 1/9 = 0.111111...
multiply both sides of the equation by 9 to get
1 = 0.9999999...

Someone who is not convinced that 0.333... is equal to 1/3 but who thinks that it "is always a little bit less" can then reasonably contend that 3 * 0.333... = 0.999... which "is always a little bit less" than 1.

There is a reasonable sense in which such a contention is correct.

The notation 0.333... is a finite shorthand which we could take to denote the infinite string where the digits to the right of the decimal point are indexed by the natural numbers and all of them are 3.

The notation 0.999... is a finite shorthand which we could take to denote the infinite string where the digits to the right of the decimal point are indexed by the natural numbers and all of them are 9.

As long as we treat these strings as character strings, there is no difficulty in looking at "1.000...", comparing it to "0.999..." and seeing that they are distinct. One can apply a simple sort order so that it is clear that "0.999..." is strictly "less than" "1.000...".

The difficulty is trying to put together a consistent arithmetic using one-way infinite decimal strings as numbers. It does not work as easily as one might hope. One ends up having to compromise on some useful property or other.

The standard compromise that is made is to preserve the property of closure under addition, subtraction, multiplication and division (by non-zero divisors) but to give up the property that representations are unique. The result is the set of real numbers.

If our arithmetic is to make sense, we insist that 0.999... is a notation for a real number rather than a notation for an infinite character string.

 

[jack action]

I've been following this conversation with interest and here's what I understand:

The question is «prove that $0.\overline{9} = 1$.» And the answer should be based on the use of a geometric series where you can supposedly get the equivalent fraction of a number with a repeating decimal.

But from the same Wikipedia page where it is stated that $0.\overline{9} = 1$, it is also mentioned that a geometric series gives a proof of convergence. Proof of convergence is not a proof of equality. And the actual equivalent fraction for $0.\overline{9}$ would be (where $a$ is the first term of the geometric series and $r$ is the common ratio):
$$\lim_{n \rightarrow \infty} \frac{\frac{a}{r}(1-r^n)}{\frac{1}{r} - 1}$$
Or:
$$0.\overline{9} = \lim_{n \rightarrow \infty} \frac{\frac{0.9}{0.1}(1-0.1^n)}{\frac{1}{0.1}-1} = \lim_{n \rightarrow \infty} (1-0.1^n) \neq 1$$
Since $\frac{a}{r} = \frac{1}{r} - 1$, it can be rewritten - to find the actual fraction - as:
$$\lim_{n \rightarrow \infty} \frac{\frac{1}{r^n}}{\frac{1}{r^n}}(1-r^n) = \lim_{n \rightarrow \infty} \frac{(\frac{1}{r^n}-1)}{\frac{1}{r^n}}$$
Or:
$$0.\overline{9} = \lim_{n \rightarrow \infty} \frac{(10^n-1)}{10^n} \neq \frac{10^n}{10^n}$$
It makes sense that the equivalent fraction of a number with an infinite number of decimals has the concept of a limit approaching infinity describing it.

Therefore the question is misleading, since there is no proof of equality, only convergence.

 

[WWGD]

I've been following this conversation with interest and here's what I understand:

The question is «prove that $0.\overline{9} = 1$.» And the answer should be based on the use of a geometric series where you can supposedly get the equivalent fraction of a number with a repeating decimal.

But from the same Wikipedia page where it is stated that $0.\overline{9} = 1$, it is also mentioned that a geometric series gives a proof of convergence. Proof of convergence is not a proof of equality. And the actual equivalent fraction for $0.\overline{9}$ would be (where $a$ is the first term of the geometric series and $r$ is the common ratio):
$$\lim_{n \rightarrow \infty} \frac{\frac{a}{r}(1-r^n)}{\frac{1}{r} - 1}$$
Or:
$$0.\overline{9} = \lim_{n \rightarrow \infty} \frac{\frac{0.9}{0.1}(1-0.1^n)}{\frac{1}{0.1}-1} = \lim_{n \rightarrow \infty} (1-0.1^n) \neq 1$$
Since $\frac{a}{r} = \frac{1}{r} - 1$, it can be rewritten - to find the actual fraction - as:
$$\lim_{n \rightarrow \infty} \frac{\frac{1}{r^n}}{\frac{1}{r^n}}(1-r^n) = \lim_{n \rightarrow \infty} \frac{(\frac{1}{r^n}-1)}{\frac{1}{r^n}}$$
Or:
$$0.\overline{9} = \lim_{n \rightarrow \infty} \frac{(10^n-1)}{10^n} \neq \frac{10^n}{10^n}$$
It makes sense that the equivalent fraction of a number with an infinite number of decimals has the concept of a limit approaching infinity describing it.

Therefore the question is misleading, since there is no proof of equality, only convergence.

But convergence of a decimal expansion is considered equality, e.g., the famous $\pi$s defined as the limit of its decimal expansion $3.141592...$, aka, the limit of the decimal expansion. This us what $\pi$ is.

 

[jack action]

But convergence of a decimal expansion is considered equality, e.g., the famous $\pi$s defined as the limit of its decimal expansion $3.141592...$, aka, the limit of the decimal expansion. This us what $\pi$ is.

But $1$ is not defined as the limit of the decimal expansion of $0.999...$

 

[WWGD]

But $1$ is not defined as the limit of the decimal expansion of $0.999...$

Well, it gets kind of technical and confusing that some reals will have more than one expansion. Any Real with a finite expansion $a.a_1...a_n$ will have an equally-valid expansion $a.a_1...(a_n-1)99999...$ So decimal expansions are not always unique.

 

[Frodo]

Someone who is not convinced that 0.333... is equal to 1/3 but who thinks that it "is always a little bit less" can then reasonably contend that 3 * 0.333... = 0.999... which "is always a little bit less" than 1.

That is why the proof starts by "divide 3 into 1" instead of saying "consider 0.33333 ...".

Any person who can divide by 3 should be able see that 1 divided by 3 is exactly equal 0.333333..., going on to infinity (my ... says goes to infinity), and that the only digit which can appear is 3.

If they are then able to multiply by three they can see that the left side becomes 1; and the right side must become 0.999999..., going on to infinity, and that the only digit which can appear is 9.

 

[jack action]

@WWGD , are you saying that 'converges' is the same as 'equals'? To me converges means tends to a limit, without reaching it. So if we accept $0.999...$ as the result of a geometric series , the sum cannot be $1$, it must be assumed to be less. If we assume $0.999... = 1$, then it cannot be represented as a geometric series (that is proven to always be convergent).

$\frac{1}{\infty}$ cannot be equal to $0$. It is close, but not equal. The $1$ on the numerator proves it.