Why doesn't .9999~ = 1?

[Exapno_Mapcase]

*Originally posted by criminalcatalog *
**I don’t claim to be a mathematician, or that I have extensive education on the subject. But I have realized that mathematics is our most useful tool for understanding the world we live in. Unfortunately, a very nasty side effect is the way a mathematic mind tends to limit itself to reason built from systems that are too rigid, formulaic, and limited by “boundaries” (ah? the post recirculates itself!) which reflect an unopen mind.

I don’t mean to insult; this is just how I feel. **

This is why I asked the question about the psychology of those who are denying the obvious truth that .9999… = 1 (and also that .3333… = 1/3, etc.)

Note that everyone who has posted to this thread who does claim to be a mathematician or has extensive education on the subject all line up on one side. Those having no education or understanding line up on the other.

So why would anyone voluntarily come in late to this discussion and put their weight on the side of the uneducated? Rather than make a judgment, I’d really like to understand what would make anyone act like this?

P.S. Phage said:

How about pi? The ending point of pi has not been found, and is projected to be unending. Pi does not have a limit such as 1!

How about pi = 4 (1 - 1/3 + 15 - 1/7 + 1/9 - 1/11 …)?

What does that come to? The number closest to pi but smaller than it? What is that number, if so?

[Exapno_Mapcase]

Bah.

pi = 4 (1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 …)

[Joe_Random]

*Originally posted by Phage *
In order for (0.999…) to be accurately expressed in decimal form, you must have an infinite number of 9s on the number. The “…” refers to the abstract idea of infinity, and so is not decimal notation.

I’m not following. Are you saying that it is only valid if we actually write an infinite number of 9’s rather than saying that there are an infinite number of 9’s?

**Infinity must be part of the real number system. How many numbers are there in total? **

Infinity need not be part of the real number system. Consider for a moment a new number system that I’m about to invent. We’ll call them “Joe Numbers”. The set of Joe Numbers consists of “0, 1, 2, 3, 4”.

There are five Joe Numbers, yet “5” is not a Joe Number.

There are infinite real numbers, yet infinity is not a real number.

[Lemur866]

OK, I had another thought. Suppose that 0.9999… !=1. That means that 0.9999… is the real number closest to 1 but is not 1. According to Math-L, this cannot be. But suppose it is true, under Math-P. Every number would have a number that is closest to it, right? 4.99999…is the closest number to 5. And 0.499999…is the closest number to 0.5. And so on.

So, what number is the closest to 0.99999…, differing from it by only an infintesimal? How would you write such a number? And what would it be? It isn’t 0.999…98, because that implies a terminating decimal. 0.999…998 would be closer, right? Such a number would have to exist in Math-P, but it cannot be expressed. And there would have to be a number larger than 1, but closest to it also. 1.000…1. But what about 1.000…01? Is there a difference between those two numbers? No? Then what exactly is an infintesimal? If we claim in Math-P that there is a difference, what IS the difference?

It could be that people are just getting hung up by decimal notation. As ultrafilter points out, 1 is just the short way to write 1. You could alternatively write it …0000000001.00000000… An infinite number of zeros on either side. But according to Math-P there would be some difference between 1 and 1.000…, although what it would be I don’t know. Odd system, that Math-P.

[Race_Bannon]

*Originally posted by Exapno Mapcase *
**So why would anyone voluntarily come in late to this discussion and put their weight on the side of the uneducated? Rather than make a judgment, I’d really like to understand what would make anyone act like this?

**

I’m with you Exapno, but I don’t have a good answer. The shame is that the folks on the 0.999… !=1 side have some really good questions, and the proper sense of wonder, which are both good prerequisites for a learning experience. But, somehow, they stubbornly cling to their opinions as if we were arguing about a political issue or something.

It must have something to do with the lack of consequences. If an untrained person were faced with the controls of an airplane, or were getting ready to ski a double black diamond run, they’d probably think twice about proceeding against the advice of experts.

But here, talk is cheap.

[RealityChuck]

*Originally posted by Exapno Mapcase *
**Note that everyone who has posted to this thread who does claim to be a mathematician or has extensive education on the subject all line up on one side. **

FWIW, I don’t claim to be mathematician, and I don’t have extensive education on the subject. However, the proof that 0.9999… = 1 only requires basic middle school algebra. To prove 0.3333… = 1/3 only requires grade school math.

[23skidoo]

*Originally posted by Lemur866 *
So, what number is the closest to 0.99999…, differing from it by only an infintesimal? How would you write such a number? And what would it be? It isn’t 0.999…98, because that implies a terminating decimal. 0.999…998 would be closer, right?

Aren’t both of those decimals the same: a decimal followed by an infinite number of 9’s, followed by an 8?

Here’s a slight hijack: Is that even a number? I mean, if there’s an infinite number of 9’s, would you ever “get to” the 8?

[Q.E.D]

*Originally posted by 23skidoo *
**Here’s a slight hijack: Is that even a number? I mean, if there’s an infinite number of 9’s, would you ever “get to” the 8? **

No, .999…8 is meaningless, for the reason you suggested. And the poster you were quoting knew this. That was his point. It would be like saying ¥ - 1.

[netscape_6]

Lemur866, could I sum up your posts uo with mathmatics just work better and don’t break down as much if we assume .999,=1?

If so that helps me out

[Donut]

Originally posted by criminalcatalog
Unfortunately, a very nasty side effect is the way a mathematic mind tends to limit itself to reason built from systems that are too rigid, formulaic, and limited by “boundaries” (ah? the post recirculates itself!) which reflect an unopen mind.

See, this bothers me. Not because it’s not true–it is, sort of. It bothers me because I don’t see how it could be any other way. I mean, what do you want? You want mathematicians to just make up anything that agrees with their intuition, and say “Hey, I like that. Let’s say it’s true”? No, mathematics is rigid. If you want to claim something is true you have to prove it. The kind of mathematics you want, “math without boudaries”, is completely useless. When you get right down to it this kind of thinking-imaginatively-but-staying-inside-the-box is hard, far harder than just thinking up any damn thing you please. It is also, however, exponentially more useful.

.

Originally posted by criminalcatalogNote that, according to the link provided, infinity holds a quality of “having no boundaries or limits”.

This is just down right dishonest. As pointed out, pulling out a dictionary during a technical argument is useless, but if your going to do it you could at least take one of the definitions provided that is specifically mentioned as being used in mathematics.

Originally posted by netscape 6
Lemur866, could I sum up your posts uo with mathmatics just work better and don’t break down as much if we assume .999,=1?

You can and you can’t. We could work out a set of assumptions that include 0.999… = 1, make that set the basis of our math and it would work, BUT we don’t. What we do instead is formulate a set of assumptions that are far more intuitive and obvious, things that I doubt even criminalcatalog would take issue with, and then prove that 0.999… = 1

[spanna]

Just do the Math using MS Calc … Problem solved

[kabbes]

Can I just direct your attention to this rather useful page? It contains the definition of a real number. In particular, note alternative definition 2:

Equivalence classes of decimal sequences (sequences consisting of natural numbers between 0 and 9, and a single decimal point), where two decimal sequences are equivalent if they are identical, or if one has an infinite tail of 9’s, the other has an infinite tail of 0’s, and the leading portion of the first sequence is one lower than the leading portion of the second.

In other words, 0.999… = 1 is one of the definitions of real numbers!.

Another point from that page: real numbers are an “ordered field”, which means that they have to obey certain rules. This means that infinity is not a real number: it does not obey these rules.

Honestly folks. To people that have actually studied mathematics at any kind of serious post-high-school level this is trivial stuff. It’s not one of the great debates of our time - that’s why it stays in GQ!

Our difficulty in explaining it lies in the fact that to get to the point where the consequence is trivial to you too, we’ll have to give you a twelve week lecture course, detailing things like Cauchy sequences, equivalence classes, fields and, in particular, Dedekind Cuts. That ain’t gonna happen.

But the beauty of this bit of mathematics is that it can be shown to a non-mathematician, just not in a completely intuitive way. Hence the “10x - x = 9” ‘proof’. Hence the “what lies between?” argument. Hence the “limit of the sequence” discussion. They’re all ways of trying to get you to understand something without having to go through that rigorous training. In return, we’re just asking that where you aren’t convinced, you consider that maybe the fault lies in the gap in your knowledge rather than the tenets of all of modern mathematics!

But if you do want to get a bit more rigorous with your understanding, you’re going to have to understand arguments like this:

Let D be any dense subring of the rational numbers. That is, D is any subring of the rational numbers other than the ring of integers. We have in mind the ring of decimal fractions, those rational numbers that can be expressed with denominator a power of 10. A Dedekind cut in D may be defined as a nonempty proper subset S of D such that if x < y, and y is in S, then x is in S.

If D is the ring of decimal fractions, then each decimal number u gives rise to a Dedekind cut

{x in D : x < u or x = u}

in D. Note that this cut contains 0. Conversely, any Dedekind cut S in the ring of decimal fractions, that contains 0, is associated with a unique decimal number u as follows. For fixed m, the largest element of S that can be written as a fraction with denominator the m-th power of 10 gives the digits in u up to the m-th place to the right of the decimal point. The nonterminating decimal number 3.1415926535… corresponds to the cut consisting of all those decimal fractions r such that r < 3, or r < 3.1, or r < 3.14, or r < 3.141, or … .

Let cut D denote the set of all Dedekind cuts in D. Define the sum of two cuts in the usual way

u + v = {x + y : x is in u and y is in v}.

It is easily shown that the commutative and associate laws hold. The additive identity is the principal cut 0 = {x in D : x < 0 or x = 0}. The elements of D, in the guise of principal cuts, form a subgroup of cut D. In fact D consists precisely of the (additively) cancellable elements of cut D. This is because

u + d = {x + d : x is in u}

while if 1¯ = {x in D : x < 1}, then u + 1¯ = u + 1 for any cut u that is not principal. However, the nonprincipal cuts are cancellable among themselves, and are closed under addition, so they also form a subgroup of the monoid, cut D. This group may be identified with the traditional real numbers, as Rudin does with cuts in the rational numbers. Recall that any traditional positive real number has a unique nonterminating decimal expansion. Note that 0¯ = {x in D : x < 0} is the identity element of the group of nonprincipal cuts.

The order on cut D is given by inclusion of cuts. The weakly positive cuts are those that contain the rational number 0. These correspond exactly to the decimal numbers if D is the ring of decimal fractions. The product of two weakly positive cuts u and v is defined to be {st : s is in u and t is in v}. This multiplication on weakly positive cuts shows how to multiply any two decimal numbers. It’s straightforward to show that the associative, commutative and distributive laws hold. So the decimal numbers form a positive, totally ordered, commutative semiring.

The picture here is the traditional real numbers, in the form of nonprincipal cuts, living uneasily together with the ring D, in the form of principal cuts. For each element d of D, there is a traditional real number d¯ just below it, and u + d¯ = u + d for each traditional real number u. That, for traditionalists, is a complete description of the additive structure of cut D. Note that d¯ = d + 0¯.

Clearly 0.9* = 1 + 0¯, so 0¯ is a sort of negative infinitesimal. On the other hand, you can’t solve the equation 0.9* + X = 1 because, in cut D, the sum of a traditional real with any real is a traditional real.

Now - are you going to study until you understand that argument or are you just going to take our word for it?!

pan

[Apollon]

What does 0.999… look like in other systems than the decimal?
How about base 9 or base 8? Will it still require an infinite number of digits to be represented?

[kabbes]

Well the snappy answer is that it will look like 1.

In fact, since we are taking about the infinite-tail decimal expansion that is equal to 1.0000… then the answer is that it will be:

0.111… in base 2
0.222… in base 3
0.333… in base 4

and so on.

pan

[MC_Master_of_Ceremonies]

Apollon if you use nonimal fractions instead of decimal fractions:

you get by working from the first decimal places:

9/10 = 8/9 r 1/90

9/100 = 729/8100; + 1/90 = 819/8100 = 8/81 r 19/8100

9/1000 =…etc.

I don’t know if it’s clear from that but using base-9 fractions instead of base-10 fractions you get 0.88888…

[kabbes]

Same answer, different route…

[Orbifold]

*Originally posted by Phage *
The moment that you attempt to use .999… in an equation it must be converted into a limit. Otherwise it would be impossible to actually finish said equation. The necessity of such a conversion leads to the introduction of an infinitely small margin of error.

There is no margin of error. It is entirely possible to compute certain limits exactly. The limit of the sequence 0.9, 0.99, 0.999, 0.9999, …, in one of those limits. The limit of that sequence can be computed exactly, by standard rules for limits, and is exactly one. 0.999… is exactly one.

**
You do not seem to understand that there does not need to be a number between .999… and 1. The fact is .999… is the number that is closest to 1 yet still less than 1. It IS THE CLOSEST! There is no number closer yet still less than 1! Is that so hard to understand?**

You don’t seem to understand that this is a fundamental property of the real numbers: between any two distinct real numbers, there is always a third real number. Period. There is no such thing as “the closest real number to one which is still less than one”.

You said to Dr. Matrix that (1+0.999…)/2 “cannot be computed in finite time”, or something like that. This is complete nonsense. Time has nothing to do with any of this. Just as the limit of the sequence 0.9, 0.99, … can be computed to be exactly 1, so can (1+0.999…)/2 be computed exactly. But even that doesn’t matter; we don’t have to compute it. All that matters is that such a number exists.

(1+0.999…)/2 exists, and must be less than or equal to 1, and greater than or equal to 0.999… . And IF 0.999… < 1, as you seem to wish to assert, then by the usual rules of algebra we must have 0.999… < (1+0.999…)/2 < 1. Which, aside from leading to another contradiction as I outlined in a previous post, demonstrates that there’s no such thing as “the number closest to one which is still less than one”.

**
.999… is not, by your definition, a real number. Why do you insist that it is?
**

First, when in this thread have I defined real numbers? Second, that’s wrong. 0.999… is a real number.

To sum up:[list=1]
[li]The limit of the sequence 0.9, 0.99, 0.999, … can be computed exactly, and that limit is one.[/li][li]0.999… is a real number, and equals one, as has been proved multiple times in this thread.[/li][li]For any two distinct real numbers x and y, (x+y)/2 always exists and is strictly between x and y. This is a property of the real numbers.[/li][li]There is no such thing as “the closest real number to one which is still less than one”, because of point #3.[/li][li]You are completely wrong in denying any of the above.[/li][/list=1]

[The_Controvert]

Just chiming in on a fascinating thread. Loved the arguments and counterarguments, especially RealityChuck’s simple proof.

The conclusion I have come to is that for some people, it is impossible to reach through the internet and open their eyes. Like a sleeping human trapped in a virtual reality world (a “matrix” if you will) they must choose their reality. Some people cannot accept your reality, so you just have to let them cling to their delusions.

[Phage]

Orbifold,
Originally, your definition of a “real number” involved the necessity of finding a number between it and another number. When you could not find such a number you pointed out that this seemed to contradict the statement that 0.999… was a real number, and used that as a basis for disbelieving it was real. My inquiry about your belief that 0.999… was a real number was completely independent of stating the actual definition of a real number.
As proved by kabbes excellent link, 0.999… is included in the definition of a real number. 0.999… does not conform to your definition that 2 real numbers must have a number between them. Therefore, we must conclude that your definition of a real number is false.

Now to address the heart of the issue:

1. The limit of the sequence 0.9, 0.99, 0.999, … can be computed exactly, and that limit is one.

This is false. A limit cannot be “computed”; if it could be, there would be no reason for it to exist. Such a sequence never ends, so it cannot EVER be computed. The limit that is derived from such a sequence is an approximation of a more math-friendly value that the sequence comes close to.

2. 0.999… is a real number, and equals one, as has been proved multiple times in this thread.

First four words are true, but the rest of this statement is false. 0.999… does not equal 1. This should not be included in a summation, as it has not been proven.

3. For any two distinct real numbers x and y, (x+y)/2 always exists and is strictly between x and y. This is a property of the real numbers.

(0.999…) is a real number, and so it can be said that (x+y)/2 does exist for (0.999…). However, in order to actually compute the answer math operations must be performed on a number that has no end. It would take an infinite amount of time to complete such an operation, and since “after an infinite amount of time” is nonsensical it can never happen.

4. There is no such thing as “the closest real number to one which is still less than one”, because of point #3.

I am afraid that I do not follow your logic in this statement.

5. You are completely wrong in denying any of the above.

As a summation of your opinion, I suppose that this is legal to include. I, on the other hand, believe that most of what you have said above is false and I intend to prove it to you.

All
For those that think 0.999… equals 1, at what point do you make such a jump? If 0.999… is followed by a finite number of nines then you agree that it is not equal to 1, correct? From what I can observe from the rest of this thread, you are saying that it will equal 1 only after an infinite period of time! As stated above, this is an illogical way of thinking.

[MC_Master_of_Ceremonies]

No, because we don’t need to take the time to count them, we have already defined that there infinite 9’s, so we can evaluate the number exactly in a finite time period.