.999999999999 repeating = 1

[peoplater]

Take the natural numbers. Then define a set that is what we usually think of as the negative integers. Then the integers is the union of the naturals with the new set. So the naturals are subset of the integers.

If you don't mind, could you provide me with the ordering relation on such a set. The set would be composed of two different types of entities, and I do not see how you could define an ordering relation on such a set.

[peoplater]

Also, if anyone is interested, I found yet another construction of the reals from just the integers. Here is the website http://arxiv.org/PS_cache/math/pdf/0405/0405454.pdf

[LauricAcid]

The set would be composed of two different types of entities, and I do not see how you could define an ordering relation on such a set.

It's straightforward:

Let Q = the rationals (as specially defined in the book)

Let ~ = the set complement

Df. an irrational d is a non-empty bounded downward cut such that d had no greatest member and Q~d has no least member.

(Makes perfect sense when you think of an irrational as something "unreachable" in the rationals. Here d is the cut that is defined by the "convergence" from below and from above.)

Let R = rationals union irrationals.

So Q is the rationals, and R~Q is the irrationals.

Let e = the element relation

Let <q = the ordering on Q (as specially defined in the book)

Df. < = {<x y> | (x,y e Q & x <q y) v

(xeQ & yeR~Q & xey) v

(xeR~Q & yeQ & not yex) v

(x,y e R~Q & x proper subset of y)}

I.e.,

if both x and y are rational, then use the ordering of the rationals.

if x is rational and y is irrational, then x is less than y if x is in the cut that is y.

if x is irrational and y is rational, then x is less than y if y is not in the cut that is x.

if x and y are both irrational, then x is less than y if the cut that is x is a proper subset of the cut that is y.

It's so simple that even I can understand it!

Anyway, I don't know why you think there's anything inherently difficult about ordering relations defined across kinds of things. There's only one kind of thing: sets.

Edited November 14, 2005 by LauricAcid

[LauricAcid]

I found yet another construction of the reals from just the integers. Here is the website http://arxiv.org/PS_cache/math/pdf/0405/0405454.pdf

Thanks for the link. Of course I haven't had time to go over it, but I love math pdf files. I've tucked this one away for future reading.

Pat Corvini gave a fascinating lecture regarding Xeno's paradox and the concept of infinity at last summer's conference in San Diego

What's the gist of her main point?

[peoplater]

Anyway, I don't know why you think there's anything inherently difficult about ordering relations defined across kinds of things. There's only one kind of thing: sets.

I looked at the construction and it is very similar to the way surreal numbers are constructed, but we are still comparing the same type of thing which is an ordered pair <x,y>.

[LauricAcid]

I looked at the construction and it is very similar to the way surreal numbers are constructed, but we are still comparing the same type of thing which is an ordered pair <x,y>.

No, the set of ordered pairs IS the ordering. That's what an ordering is - a set of ordered pairs satisfying certain conditions (depending on the ordering - reflexivity, irreflexivity, anti-symmetry, asymmetry, transitivity, trichotomy, "weak" trichotomy, minimal element, and maybe others). The x's and y's are taken from different kinds of things - rationals and reals. We're not comparing ordered pairs; we're comparing x's and y's. They are compared by whether or not the ordered pair <x y> is in the set of ordered pairs that IS the relation that IS the ordering. Let S = the set of ordered pairs that is the relation that is the ordering. Then x<y <-> <xy> e S. (x is less than y if and only if <x y> is an element of the ordering.) Edited November 15, 2005 by LauricAcid

[peoplater]

No, the set of ordered pairs IS the ordering. That's what an ordering is - a set of ordered pairs satisfying certain conditions (depending on the ordering - reflexivity, irreflexivity, anti-symmetry, asymmetry, transitivity, trichotomy, "weak" trichotomy, minimal element, and maybe others). The x's and y's are taken from different kinds of things - rationals and reals. We're not comparing ordered pairs; we're comparing x's and y's. They are compared by whether or not the ordered pair <x y> is in the set of ordered pairs that IS the relation that IS the ordering. Let S = the set of ordered pairs that is the relation that is the ordering. Then x<y <-> <xy> e S. (x is less than y if and only if <x y> is an element of the ordering.)

right. what did I say?

[LauricAcid]

right. what did I say?

I thought your point was that this ordering does NOT mix "kinds" of numbers. In the cases of all reals being equivalence classes of Cauchy sequences or all reals being Dedekind cuts, there is only one "kind" of number, so it's not surprising that we can specify the ordering. But you wondered how we could specify an ordering for a set of which not all the numbers are of a single "kind". And I showed an ordering in which the numbers are not of a single "kind". Then I understood your repsonse to be there is not something special about that ordering since it only compares ordered pairs anyway. But the the ordering I showed IS special in the sense of being one of different "kinds" of numbers and it does not compare ordered pairs of real numbers; it compares real numbers. Of course, there is only one "kind" of thing - sets, but in the sense you had mentioned of "kinds" ('types' was your word, but I'm using 'kinds' since I wish to steer clear of confusion with 'type' in the sense of 'type theory'), this is an ordering of different kinds of numbers. On the other hand, If I misunderstood your remark about ordered pairs, then I did not mean to, and I note that, happily, we're on the same page here after all. Edited November 15, 2005 by LauricAcid

[bisket530]

How do you know that 9(.999...) = 8.999...? That seems just as dubious a step as saying that 9.999... - .999... = 9.

Actaully, Bryan, it is wrong because we are dividing by zero.

because we set a equal to b, dividing by (a- would mean we would have to divide by zero, which cannot happen.

Nice try though

[bisket530]

Another way to think about this problem.

This problem deals with elements similar to limits and infinate sequences.

Lets say we are in a 100 meter race. I want to get to the finish line but would have to go the half mark, which is the 50 meter mark first. But, even before this, I would have to go to the 25 meter mark. But before this, 12.5 meter mark. Then 6.25, then 3.125... you get the point.

As many of you know, we would have to infinatley divide the distances. And of course, we cannot because we will have to keep diving. It will never end.

But first, lets try to add all of the half distances. 50+25+12.5+6.25+3.125.... we will have to add until the last digit will become zero, which cannot happen. So then this means that we will never be able to reach the finish line. That is, we will be infinately close to the finish line, but we will never reach it.

But, of course, we always can reach the finish line. Why? because the number infinately close to 100 meters is 100 meters. If this werent true, then we would never be able to finish races. This idea is one of Xeno's paradoxes; check it out sometimes.

Bringing this idea to .9999999....=1. Something infinately close to something is the same thing.

It seems logical, rebelious, and even cool to think that .9999... does not equal 1, but it's is true.

Hopefully, the race analogy helped some of you.

[Robert J. Kolker]

But, of course, we always can reach the finish line. Why? because the number infinately close to 100 meters is 100 meters. If this werent true, then we would never be able to finish races. This idea is one of Xeno's paradoxes; check it out sometimes.

There is a branch of mathematics in which this is not the case. It is called non-standard analysis or the theory of hyper-real numbers. See -Non Standard Analysis- by Allain M. Robert for a gentle introduction to the subject. This book is now available from Dover so it is not too expensive. There are more complicated treatises based on the theory of ultra-filters, but for people who are not professional mathematicians or graduate students in mathematics, Allain Robert's book is just fine.

When Isaac Newton invented calculus he based his results on the notion of infinitely small quantities that are not zero. Bishop Berkeley has a good time hitting on this idea. The Bishop referred to the infinitesimals used by Newton and Leibniz as "the ghosts of departed quantities". It turns out that Berkeley's criticism of the basis of calculus was right on the mark, but it was not until the 19th century when the limit concept was introduced to banish this embarrassment. But that is not the end of the matter. In the 1960's, the mathematician Abraham Robinson found a way of rigorously and logically justifying the concept of the infinitesimal and he invented a new branch of mathematics (theory of hyper-reals) to do this. It turns out that the reasoning of Newton and Leibniz based on infinitesimal quantities was really right, after all. And now you know the REST of the story.

Bob Kolker

[punk]

There is a branch of mathematics in which this is not the case. It is called non-standard analysis or the theory of hyper-real numbers. See -Non Standard Analysis- by Allain M. Robert for a gentle introduction to the subject. This book is now available from Dover so it is not too expensive. There are more complicated treatises based on the theory of ultra-filters, but for people who are not professional mathematicians or graduate students in mathematics, Allain Robert's book is just fine.

When Isaac Newton invented calculus he based his results on the notion of infinitely small quantities that are not zero. Bishop Berkeley has a good time hitting on this idea. The Bishop referred to the infinitesimals used by Newton and Leibniz as "the ghosts of departed quantities". It turns out that Berkeley's criticism of the basis of calculus was right on the mark, but it was not until the 19th century when the limit concept was introduced to banish this embarrassment. But that is not the end of the matter. In the 1960's, the mathematician Abraham Robinson found a way of rigorously and logically justifying the concept of the infinitesimal and he invented a new branch of mathematics (theory of hyper-reals) to do this. It turns out that the reasoning of Newton and Leibniz based on infinitesimal quantities was really right, after all. And now you know the REST of the story.

Bob Kolker

We must be clear that standard analysis ("normal" reals) and non-standard analysis (hyper-reals) sort of exist in parallel to each other. The mathematician can choose to work within standard analysis (then 0.99999... = 1.000...) or the mathematician can choose to work within non-standard analysis (so 1.0000... - 0.9999... = d, where d is an infinitesimal). It is simply a matter of the axiom system that is chosen.

How does one choose an axiom system? Well if you are a pure mathematician, basically because you think it is interesting. If you are anyone else, you choose it because it is deemed appropriate to the problem you are working on.

In a more modern viewpoint, one doesn't choose an axiom system as one chooses a topos. All topoi are equally "fundamental", and they all satisfy certain properties that all topoi have, but one topos or another might be better to work in given the problem.

So does 0.9999... = 1.0000... or does 1.000... - 0.999... = d?

That depends on the problem you are working on, and what works best for you.

Edited September 29, 2007 by punk

[ebmusicman84]

This may have been pointed out, I only read about five pages of this topic, but I would like to state this:

Many of the proofs I read in this thread contained the idea of limits. I would like to point out that it is because of the idea that 0.999~ = 1 that limits exist. An infinite number of nines after a decimal point is impossible to achieve. Limits communicate the idea that, were an infinite amount of nines attainable, stick a zero and a decimal in front of them and you get 1.

[LIMIT(n--->infinity)0.999..repeating n times] = 1 (to borrow from shakthig) says: The limit of [0.999~n times] as n approaches infinity equals one, or the more nines you put after 0. the closer you get to one. So yes, 0.999~ = 1 because this communicates that there is an infinite amount of nines after the 0. which indeed is equal to one.

Edited November 21, 2007 by ebmusicman84

[Marty McFly]

It's a limit, which is a concept that is used in calculus all the time. As the 9's after the decimal point approach infinity the value of the decimal aproaches 1. The limit therefore equals 1.

The problem is that the limit will NEVER equal 1. it will just keep approacing it. because there is no limit. it's infinity. this is what sucks about these decimal #s with the bar on top (namely repeating decimals)

it's likeyou're ALMOST there... ALMOST there... it's right there right out of reach, getting closer ALMOST THERE but never quite there.

Edited November 21, 2007 by Marty McFly

[John McVey]

it's like you're ALMOST there... ALMOST there...

And then some cloned prat comes from behind, or your missile just impacts on the surface. That sucks, too.

JJM

[Amaroq]

Jesus, this is pissing me off.

I'm in a debate with someone over this very thing.

I've proved that 1 != .999... with two different approaches already.

Divide 1 / 1. What do you get? 1.

If you try to do long division, the only way to get anything else is to say that zero 1's fit into 1. Then you get 0 with a remainder of 1. Pull down the next decimal place, and choose to only fit 9 of your 1's into the resulting 10. Once again, you get a remainder of 10.

The only way to arrive at .999... by division of 1/1 is to be dishonest, frankly.

For the other proof, basically tried to argue that .999... + .000...1 = 1. He says .000...1 does not exist.

If you infinitely divide 1 by 10 over and over again, adding more and more zeroes all with a 1 at the end, what happens to the 1 when you reach infinity? Does it just drop off and disappear?

How much more of an evasive moron can you be?

Edited October 18, 2009 by Amaroq

[JeffS]

This is really interesting. I came across this problem in one of my kid's math. I was going to ask his teacher, but there's a lot of smart people here so maybe they can answer it for me.

Here's the problem:

Turn 5.999999.... into a fraction.

The only method of solving this I know of is to call your fraction "x" to formulate this equation: x=5.999... . Multiply each side by 10 so you get 10x=59.999... . Subtract x from the left side and 5.999... from the right to eliminate the repeating digits. You get 10x-x = 59.999... - 5.999... . This resolves to 9x=54. Divide both sides by 9 to arrive at x=6. So, when you put x back into the original equation you get 6=5.999..., which is clearly incorrect.

Can anyone explain this to me?

[JeffS]

Sorry, ignore that last question. I'm going to go over the last 10 pages as it seems my question is answered there. Mea Culpa.

[Amaroq]

I apologize to anyone I may have offended. The insult is not directed at anyone here. Though I really have to ask:

What is wrong with what I've given as proof? Either manipulating the rules of division is acceptable, or there's a contradiction between simple division and the advanced "proofs" in favor of this thing. Can someone disprove my logic regarding the simple division, and maybe tell me why you apparently aren't allowed to have .000...1?

[Eiuol]

1 / 3rd is .33333 repeating

(1/3) * 3 = 1.

(1/3) * 3 = 1 is the same as

(1/3) + (1/3) + (1/3) which would logically be expressed as .999 repeating too though

The problem to me is talking about fractions. To express a fraction as a decimal isn't exactly a fun thing to do and I'm not sure it can ever be 100% perfect (like Pi). I'm no (so please correct anything I just said if you are) math major but .999 repeating at best is an approximation, so it wouldn't be saying A = !A. 1 is the same as .9999999repeating, just like 1/3 is the same as .333333repeating. .999repeating itself cannot actually exist and be concrete, I believe, but conceptually it exists and it is the same as 1.

Edited October 18, 2009 by Eiuol

[Amaroq]

Yeah, that was one of the proofs I saw in favor of 1 = .999...

I'm no math major either. Never really liked math. But I had another thought on this that I'd like to pose here.

Let's examine why 1/3 = .333...

You've divided one by three. Three doesn't fit, so you bring down the next decimal, etc. There's always a remainder left over that forces you to keep adding threes.

I think the problem is trying to express a fraction in base 10 that can't be evenly distributed over ten numerical symbols.

Think of it like grouping ten marbles into three groups. There's no way to evenly do that. The ten marbles would be the ten possible numbers from 1-10. (Or 0-9, whatever.) The 1/3 is just one of the three "slices" you're taking out of that 10.

If you want to take a more precise slice, you split each marble into smaller pieces. But each piece must be split into the same number of smaller pieces. No matter how many pieces you split them into, 10, 100, 1000, you're never going to take a completely even portion.

You obviously take as even portions as possible. The more precise your portion, the more repeating decimals there are.

The reason I think 1 is different than .999 is because the same unresolved issue in the repeating 3's is here too. There's something forcing you to continue to add 9's to the end. If you take 3/3, then you've taken the whole. You don't have any repeating-ness to resolve unless you take an infinitely small portion of the cake away.

I've kind of lost my train of thought, but what I'm saying is, the issue that forces you to repeat forever is due to trying to distribute fractions over a base (thirds over the decimal system, or base 10, for example) that doesn't fit. I think that once you take the whole, all pieces of the fraction, the issue is resolved, no longer a problem.

So 1 isn't .999...

Multipying .333... by 3 just... skips over .999... completely and gives you 1.

It sounds completely silly, but that's the only way it makes sense to me.

EDIT: Added to the marble analogy.

Edited October 18, 2009 by Amaroq

[aleph_0]

No, the mathematician's claim is not that .999... = 1 is an approximation. The mathematician claims that .999... is identical to 1. As someone else explained before, this is a naming issue. Mohammad Ali = Cassius Clay and likewise .999... = 1. They are two names for the same thing. Eiuol's proof is, to me, the most obvious proof.

As for your argument, Amaroq, long-division isn't a tool the pure mathematician uses. It is a method for calculating things near enough to the answer. Sometimes it even gets you exactly the right answer, but not always (for instance, divide pi by any rational and you'll never reach the end), and so it is not acceptable in a mathematical proof.

You can use heuristic arguments if you like, in order to feel more at ease with the notion that .999... = 1, but the fact is that this identity is true merely as a product of the formalization we give to mathematics. To take another instance, we could define a homomorphic operation * over the integers, such that 0*0 = 0, 0*1 = 1, 1*0 = 1, 1*1 = 0. Then 2 = 2*0 = 1*1*0 = 1*1 = 0, so 2 = 0. But this is just a result of some formalism that we chose. (In fact, it is the familiar formalism of the integers mod 2. That is, you repeat your numbers up to 2. For the integers mod 3, you only deal with the numbers 0, 1, 2. So 0+1 = 1, 1+1 = 2, 3 = 1+2 = 0 [mod 3], and 2+2 = 1 [mod 3]. Does this mean, in any metaphysical way, that 2 = 0 or 3 = 0? No. It just means that we picked a formalism that is useful for some task and live with the deductive consequences of the formalism.)

[TuringAI]

If you infinitely divide 1 by 10 over and over again, adding more and more zeroes all with a 1 at the end, what happens to the 1 when you reach infinity? Does it just drop off and disappear?

How much more of an evasive moron can you be?

It does, because the 1 would be out of scope. There is no 'infinitesimal' or 'infinite' place in the decimal system. The decimal system is a method for representing the real numbers, and in any (integer) base, the number of the place must be an integer, therefore the size of the amount of places is aleph-null.

A real number isn't the same thing as a computation of that real number. All computations of real numbers involve successively applied algebraic operations, and so a computation of 0.9999999~ is dependent upon how long the computation is. But the computation is just the representation of the real number for a finite number of steps, not the real number itself.

So the real number 0.999~ is 0 with a decimal point followed by an infinite number of 9s. That doesn't mean there is a PLACE HOLDER called the infinitieth place to the right of the decimal point, even though HOW MANY 9s there are in the representation ARE infinite.

It is not in general true that infinity - infinity = 0. It does not follow the algebraic rules of real numbers, and this is because it is not an algebraic number. So infinity + 1 = infinity. This means that multiplying 0.999999 by 10 does not increase how many 9s there are to the right of the decimal place. You cannot increase or decrease infinity to get a new number, it remains infinity. The only thing that changes by multiplying 0.999999~ by 10 is that there is now a 9 to the left of the decimal representation of 0.999999~ multiplied by 10. So 0.999999~ * 10 = 9.999999~ where ~ indicates an infinite number of 9s to the right of the decimal place.

However, here's the thing. If EVERY number to the right of the decimal place in the decimal on the left is equal to EVERY number to the right of the decimal place in the decimal on the right, then no matter how far you go, there is no 9 that is not matched to another 9. Since 9 is finite, 9 - 9 = 0. So 9.999999~ - 0.999999~ = 9.000000~, which is the same as 9. So that X=0.999999~ and 10X=9.999999~ method is actually being used correctly. So 0.999999~ = 1, simply put. Or not so simply put, as the case may be.

[Rockefeller]

I'm no math major but .999 repeating at best is an approximation

...

1 is the same as .9999999repeating, just like 1/3 is the same as .333333repeating. .999repeating itself cannot actually exist and be concrete, I believe, but conceptually it exists and it is the same as 1.

Conceptually, 0.99 repeating is not the same as 1, because it combines a concept of method (repetition) with the concept of decimal numbers (which have existential referents). However, as soon as you specify how many times 9 is being repeated, it becomes a decimal number.

Only when it is expressed as a decimal number, you can strictly compare it with 1.0 (a decimal number). Now, you can easily see 0.99 (repeating 9 say 50 times) refers to less number of referents than 1.0.

Attempting to compare 0.99-repeating and 1.0 is analogous to comparing density and volume (two non-commensurable characteristics).

Also, there is a difference in case of fractions 1/3 and 1/1 - which are symbols for concepts "one part in three" and (say) "three parts in three". In order to express the concept 1/3 as a decimal number, you need an approximation; but in order to express the concept of unity, you don't need an approximation - the denotation 1.0 (a decimal number) accurately refers to desired number of referents. It's analogous to the fact that the English word "Book" can accurately be expressed by the Hindi word "Kitaab", but the English word "Train" can only be approximately expressed in Hindi as "Gaadi" (which actually means anything with more than 2 wheels).

[Jake_Ellison]

Jesus, this is pissing me off.

I'm in a debate with someone over this very thing.

I've proved that 1 != .999... with two different approaches already.

I doubt you have. 0.(9), or 0.999..., if you prefer, equals one by definition.

He says .000...1 does not exist.

You might wanna start listening to this guy, instead of arguing, whoever he is.

Conceptually, 0.99 repeating is not the same as 1, because it combines a concept of method (repetition) with the concept of decimal numbers (which have existential referents). However, as soon as you specify how many times 9 is being repeated, it becomes a decimal number.

First off, 0.99 repeating is not the same as 0.(9). 0.99 repeating is nothing (unless you care to mention how many times it's repeating, then it will be a number--a different number than 0.(9) ), while 0.(9) is a different notation for the number 1. So it's also a number. But a number is not a concept, it is an instance of a concept.

The notation itself is a concept, it is a notation that is used to represent fractions in the x.y(z) format, rather than in the x/y format.

By definition, x.y(z) equals a fraction, defined by a somewhat complex equasion that will give you 1/3 for 0.(3), and 1 for 0.(9). That is the definition of the concept, there really is nothing at all to argue about. 0.(9)=1 follows directly from the definition of the mathematical notation, the bracket, used.

The only reason you're arguing is because you don't know that definition.

It's analogous to the fact that the English word "Book" can accurately be expressed by the Hindi word "Kitaab", but the English word "Train" can only be approximately expressed in Hindi as "Gaadi" (which actually means anything with more than 2 wheels).

That's absurd. Those are all concepts. Numbers are not even concepts, math is not a language, and 0.(9) is most definitely not the English to the Hindi of 1.

There nothing even remotely similar about the two examples.

Attempting to compare 0.99-repeating and 1.0 is analogous to comparing density and volume (two non-commensurable characteristics).

Jesus Christ, with the "analogous". The definition of decimals and fractions is that they are different representations of the same thing: rational real numbers.

On a separate note, how do you feel about this statement, by the creator of C++:

"Proof by analogy is fraud. --Bjarne Stroustrup"?

Edited October 18, 2009 by Jake_Ellison