I would like to argue about .999

[uart]

actually no. please do not assert that I've said things that i have not.

Actually yes. I quoted you verbatim and you said "decimals of fractions never accurately equal those fractions". If you are referring to only recurring decimal then use the word "recurring". And you've got the nerve to say that others here are poor communicators.

you could tell clearly from the pattern that i had set up with my earlier argument about .111... and it's relation to .999... that my argument only dealt with fractions who are represented by infinitely repeating digits after a decimal.

Yes that's true. That's why I was trying to help you word your question more correctly ok.

and as for your argument that there is a trap. 1-.999... = 1-.999... since there is no other way to write it that i know of.

Ok you didn't get as far as I expected. Most "layman" when confronted with this usually go something like,

1 - 0.9 = 0.1
1 - 0.99 = 0.01
1 - 0.999 = 0.001

therefore 1 - 0.999... = 0.00...1
where the "..." means repeats forever.

Usually at this point even the layperson can see the absurdity of what they've just written in 0.00...1. Essentially this is saying that there is a decimal point followed by a never ending number of zeros then at the end of this never ending string of zeros we put a one.

Clearly that one on the end is redundant so therefore 0.00...1 = 0.00... which is just an inefficient way of writing zero.

The difference between 1 and 0.999... is zero and therefore they are equal.

 

[Mark44]

But the decimal .333... is also a decimal number, the ellipses "..." is just shorthand notation for the infinite string that constitutes this decimal number. Once you are talking of an infinite string of 3's there is no sense in asking of when does the sum reach 1/3 since it has by then reach 1/3. Any additional 3 in the string would be only adding a zero to the sum since n/[infinity] where n is a finite number = 0. Where are these additional 3's that you're talking about that don't contribute to the sum?

Infinity is not a number in the real number system, so you can't divide by it or otherwise do arithmetic with it. Limits are what you need to be working with if a variable gets large without bound.

That is a property of infinity. I see what you are saying but as long as you are talking in terms of decimal places of an infinite string any "additional decimal places" would effectively merely be adding zero to the sum. Also in reality you can not put a number on the number of decimal places to infinity or ever reach it to add to it. If you could you wouldn't be at infinity.

 

[Fredrik]

1) You don't know how to communicate

Are you serious? What exactly makes you say that? I really want to know.

2) the easiest way to say that .999... equals one is to say that the space after the decimal is infinite in size but is being filled with and infinite amount of 9's therefore the requirement needed for it to be "bumped" up to 1 is met. Also, if 1/9 equals .111... then 1/1 should equal .999... it's the same concept but with a different appearance.

This doesn't make any sense. The actual reason why 0.999...=1 is explained in post #35.

For every r>0, all but a finite number of members of the sequence 0.9, 0.99, 0.999,... are between 1-r and 1+r.​

See #35 for a more thorough explanation.

3) if the reaction to my original assertion had been strictly productive instead of egotistical then this thread may well have never occurred which links us back to the 1st item on this list. you need to learn to communicate more concisely and without an attitude (you being certain members that have responded).

You have displayed an attitude problem in just about every one of your posts. The problem is with you, not with the people who have tried to help you. No one gave you "attitude" before you started insulting everyone.

and does not the ... mean the 9's continue on into infinity?

Yes, but what does that mean? You clearly don't know the answer to that, and as long as you don't, there's no way you can understand this problem. This is why you should read post #35, where it's explained.

also, are there any cases where time can somehow be applied to this idea of ... so that at a certain point you have not actually reached 1?

Not in mathematics. If you consider the problem of physically writing the decimal expansion on paper, then time is obviously an issue, but then we are no longer talking about mathematics. Time has absolutely nothing to do with the issue of why 0.999...=1.

 

[Calrid]

Infinity is hard to comprehend, so some people are tragically incapable of accepting that at infinity anything equals anything. No one can really visualise an infinity but what they can do is accept that as a limit it makes sense that certain values converge to an exact form.

If you can't accept that the infinite is basically an indefinite value that is without bound, then you are looking at it finitely and hence you are not going to grasp the implication of what infinity actually means. If you were to count to infinity it would take you forever, literally and if I stopped you in ten years you still would be infinitely far away.

Perhaps if you could put 1 billion 9s after the decimal and tried to do an equation you'd see an incredibly small amount of deviation from the exact value, now imagine that there are an infinite amount of these .9s how small is it now?

 

[MathematicalPhysicist]

As homage to this thread or any such thread I believe we should post a thread along the lines:"I would like to kill (you) about .999..."

Any in favour?

 

[ramsey2879]

Infinity is not a number in the real number system, so you can't divide by it or otherwise do arithmetic with it. Limits are what you need to be working with if a variable gets large without bound.

Where in my post did I say to multiply or do any arithmetic with infinity? In fact I actually said the contrary, to wit infinity was boundless. I was responding to the Op's argument that fractions such as .333... or .999... never add up to 1/3 or 1 becauase if they did the concept of infinity would be contradicted. I also said
" I think I now see where you are coming from. In effect you may be right to say that the limit cannot be reached in reality because concept of ever reaching infinity is not reality. But when we put ellipses at the end of a decimal string we are invoking the concept of a decimal string that is indeed infinite contary to reality. This decimal string is most often precisely defined and in the case of .999... is precisely equal to 1."
The definition of course comes from the theorey of limits.
I think the major thought I contributed in my post was the idea that infinity is not a concept well structured by the "reality" of being reachable as the Op seems to be insisting on.

 

[Mark44]

Right here...

But the decimal .333... is also a decimal number, the ellipses "..." is just shorthand notation for the infinite string that constitutes this decimal number. Once you are talking of an infinite string of 3's there is no sense in asking of when does the sum reach 1/3 since it has by then reach 1/3. Any additional 3 in the string would be only adding a zero to the sum since n/[infinity] where n is a finite number = 0. That is a property of infinity.[/color]

You are dividing some finite number n by infinity.

 

[ramsey2879]

Right here...

You are dividing some finite number n by infinity.

To get the limit of A/n as n goes to infinity you have to know what A/[infinity] is. P.S. My high school teacher said A/[infinity] was 0.

 

[micromass]

To get the limit of A/n as n goes to infinity you have to know what A/[infinity] is. P.S. My high school teacher said A/[infinity] was 0.

I think your high school teacher was wrong then... (unless he was talking about extended reals, but I doubt it).
And to calculate limits of A/n, you don't need to know what A/[infinity] is. You can simply get it from the definition:

\forall \epsilon>0:\exists n_0\forall n\geq n_0:~|A/n|<\epsilon

The existence of the n₀ in question follows from the axiom of Archimedes. So there's no need for calculations with infinity. Indeed, the whole purpose of limits is to avoid calculations with infinites, as these are ill-defined. Of course, you can still do it with infinites in your intuition (that's how I do it), but it's not rigourous mathematics.

 

[Hurkyl]

To get the limit of A/n as n goes to infinity you have to know what A/[infinity] is. P.S. My high school teacher said A/[infinity] was 0.

No, to get the limit of A/n as n goes to infinity, you need to know the Archimedean principle.

To compute the limit of A/n as n goes to infinity instead by comparing to A/[infinity], you need a lot more information. One set of information would be

• A number system containing an element called [infinity] along with all real numbers
• Knowledge that A/[infinity] = 0 if A is finite
• Knowledge that division in this new number system gives the same results as division in the real numbers, when both numbers are real
• Knowledge that division is continuous in this new number system (at least, at (A, [infinity]))
• Knowledge that the limit of n as n goes to infinity converges to [infinity]
• Knowledge that limits computed in this new number system agree with limits computed in the real numbers when it would make sense.

(For the record, my thought processes probably would compute the limit by invoking continuity of division in the projective real numbers before any other approach)

 

[ramsey2879]

No, to get the limit of A/n as n goes to infinity, you need to know the Archimedean principle.

To compute the limit of A/n as n goes to infinity instead by comparing to A/[infinity], you need a lot more information. One set of information would be

• A number system containing an element called [infinity] along with all real numbers
• Knowledge that A/[infinity] = 0 if A is finite
• Knowledge that division in this new number system gives the same results as division in the real numbers, when both numbers are real
• Knowledge that division is continuous in this new number system (at least, at (A, [infinity]))
• Knowledge that the limit of n as n goes to infinity converges to [infinity]
• Knowledge that limits computed in this new number system agree with limits computed in the real numbers when it would make sense.

(For the record, my thought processes probably would compute the limit by invoking continuity of division in the projective real numbers before any other approach)

I supposed that calculus back in high school in the 1960's may have treated infinity differently from how the treat it today. I believe that all we had to do was simply recognized that infinity was like a much much bigger number than A to deduce that A/[infinity] was 0. PS I don't understand Micromass's math notation as I never had much math beyond High School.

 

[JaredJames]

I believe that all we had to do was simply recognized that infinity was like a much much bigger number than A to deduce that A/[infinity] was 0.

Not saying it's right or wrong, but that's also how I was taught to treat infinity in this situation.

 

[statdad]

" infinity was like a much much bigger number than A"

The problem is that when you are working with the real number system infinity is not a number, so performing calculations with it makes no mathematical sense.

 

[Mark44]

I supposed that calculus back in high school in the 1960's may have treated infinity differently from how the treat it today. I believe that all we had to do was simply recognized that infinity was like a much much bigger number than A to deduce that A/[infinity] was 0.

I was in high school in the 60s also, but I don't recall that we were told to treat infinity as just a big number. No, I don't believe that infinity was presented any differently back then as compared to now.

" infinity was like a much much bigger number than A"

The problem is that when you are working with the real number system infinity is not a number, so performing calculations with it makes no mathematical sense.

Right, and this was my point to ramsey2879. An expression such as n/[infinity] explicitly uses infinity in the division, which isn't valid.

 

[JaredJames]

I was in high school in the 60s also, but I don't recall that we were told to treat infinity as just a big number. No, I don't believe that infinity was presented any differently back then as compared to no.

I was in school with this stuff 5 years ago.

I was always told that if you see something over infinity it's just like having an extremely big number so you just assume 0.

Again, not arguing this either way, just pointing out the way it was taught.

 

[Raphie]

So... uh... where do "fluxions" figure into this recent discussion?

Is there a place in mathematics for what one might term "virtual" infinity or "virtual" zero?

- RF

Definition: "Virtual Infinity": The largest number one can possibly imagine at a given point in time, plus 1, at a point in time just subsequent to the point in time one initially imagined... ( e.g. Skewe's Number + 1)

 

[Hurkyl]

So... uh... where do "fluxions" figure into this recent discussion?

They don't.

Is there a place in mathematics for what one might term "virtual" infinity or "virtual" zero?

- RF

Definition: "Virtual Infinity": The largest number one can possibly imagine at a given point in time, plus 1, at a point in time just subsequent to the point in time one initially imagined... ( e.g. Skewe's Number + 1)

That's not a very good definition.

For the record, the most basic technique of calculus/analysis is the idea of "close enough" or "big enough". You don't have to invoke some mythical number larger than you can imagine; you just need to invoke a number that is big enough for the purpose at hand.

(To guard against misinterpretation, I will point out that infinity is not mythical -- at least, the mathematical notions of infinity are not. Many laypeople seem to have some mythical notion of it, though. )


Anyways, I know the basic idea you seem to be thinking. Some people like to attach a philosophical interpretation to non-standard analysis where, for example, the "standard integers" are the ones accessible to mathematicians and all other integers are simply too big or complicated for mathematicians to access directly. Wikipedia's page on internal set theory describes this viewpoint.

And for the record, in the non-standard model, while there are only finitely^* many standard integers, there does not exist a number that says how many there are. In particular, there isn't a largest one.

Also, for the record, non-standard analysis doesn't need this philosophical interpretation. It is entirely optional. In fact, I have only ever seen it mentioned when the speaker wants to model the idea of "what is accessible to mathematicians" and sometimes in the context of internal set theory.

*: by the non-standard 'measure' of such things. There are infinitely many standard integers by the standard 'measure', of course.

 

[Raphie]

That's not a very good definition.

In a strict mathematical sense, I would concur, Hurkyl. Just trying to stake out some manner of middle ground here.

Too bad mathematical "language" does not allow for that middle ground between .99999 and .99999... Where is the possibility of the (sliding scale...) partial sum contained within that black/white notational dichotomy?

See post #22, by the way, for confirmation that we agree in principle on the basic question of this thread.

 

[Hurkyl]

In a strict mathematical sense, I would concur, Hurkyl. Just trying to stake out some manner of middle ground here.

Too bad mathmatical "language" does not allow for that middle ground between .99999 and .99999... Where is the possibility of the partial sum contained within that black/white notational dichotomy?

See post #22, by the way, for confirmation that we agree in principle on the basic question of this thread.

0.999999 is in the middle ground. So is 0.9999999 if the former didn't have enough 9's for you.

There is no middle ground between 0.999... and the set of all partial sums, though. This is a rather important geometric property of the number line.

Also, there is the sequence of numbers {1 - 10ⁿ} for those people who need to feel the need to consider a sequence of increasingly good approximations to 0.999...

And, of course, there is non-standard analysis for people who want a hyperfinite number of 9's, but still have a number that is only infinitessimally different from 0.999...
(hyperfinite for the non-standard measure of finiteness)


There is mathematical language for all sorts of ideas, and if there's not, it could be invented. The only true obstacle is when someone insists on grounding their reasoning firmly in the realm of vagueness and imprecision.

 

[Char. Limit]

0.999999 is in the middle ground. So is 0.9999999 if the former didn't have enough 9's for you.

There is no middle ground between 0.999... and the set of all partial sums, though. This is a rather important geometric property of the number line.

Also, there is the sequence of numbers {1 - 10ⁿ} for those people who need to feel the need to consider a sequence of increasingly good approximations to 0.999...

And, of course, there is non-standard analysis for people who want a hyperfinite number of 9's, but still have a number that is only infinitessimally different from 0.999...
(hyperfinite for the non-standard measure of finiteness)


There is mathematical language for all sorts of ideas, and if there's not, it could be invented. The only true obstacle is when someone insists on grounding their reasoning firmly in the realm of vagueness and imprecision.

I'm going to guess that that sequence is supposed to read {1 - 10^-n}.

 

[TylerH]

actually, there is a decimal expansion between .999... and 1. it's .9999... (the extra 9 and . signifying that it is still expanding and that there will always be a 9 between the two)

The number of 9s and the number of .s is arbitrary. They just mean to continue in the "obvious way." (Hence the ambiguity Fredrick keeps talking about.) The "obvious way," in this case, being a repeating decimal. In other words, ".999... = .9999... ." Although, it is commonly accepted that three is the correct number of .s for an "ellipsis," which is what ... is. The concept is also flawed. You are assuming ∞ + 1 > ∞ when, in fact, ∞ + c = ∞, where c is a constant. But ∞ isn't a real number, so one shouldn't use it as I just did. I was just illustrating my point. In other words, one more 9 than an infinite number of 9s is the same number of 9s as an infinite number of 9s.

how can the two meet if .999... is expanding onward forever? it would have to stop expanding to reach 1.

They meet only because .999... repeats forever. Numbers, or "expressions," don't "expand." They have a set value. You can define lots of ways to get that set value, the easiest is the infinite sum everyone keeps using. They don't just come up with it arbitrarily, it's origin comes from the very definition of the real numbers. See Wikipedia: http://en.wikipedia.org/wiki/Decimal_representation, in this case, we would define a=.9 for all i. Then we take the infinite sum, to get the same proof we've already seen 100 times, just with a little background behind why it's valid, this time. Does that help?

 

[disregardthat]

And, of course, there is non-standard analysis for people who want a hyperfinite number of 9's, but still have a number that is only infinitessimally different from 0.999...
(hyperfinite for the non-standard measure of finiteness)

Is 0.9999 = 1 in the non-standard reals? Is the non-standard reals even complete? I should think so by the transfer-principle, but it is clear that 0.99999... cannot converge since every term is less that 1-e for some infinitesimal e. If so, how does 0.999 even make sense in the non-standard reals? Maybe the transfer-principle doesn't apply to completeness.

EDIT: what does an hyperfinite number of 9's mean?

 

[Raphie]

There is no middle ground between 0.999... and the set of all partial sums, though. This is a rather important geometric property of the number line.

In a philosophical vein... That I recognize the validity of this statement (i.e "no middle ground") is what has had me thinking much of late about Durkheim's distinction between the sacred and the profane.

 

[Hurkyl]

If so, how does 0.999 even make sense in the non-standard reals? Maybe the transfer-principle doesn't apply to completeness.

When you transfer, you have to transfer everything; you can't pick and choose.

In the standard model, a decimal numeral has its places indexed by integers. When you transfer that notion to the non-standard model, the corresponding notion of a "hyperdecimal numeral" has its places indexed by hyperintegers.

The partial sums of the non-standard infinite summation

\sum_{n=1}^{+\infty} 9 \cdot 10^{-n}​

that would define the hyperreal value of the hyperdecimal numeral 0.999...
(don't forget n ranges over hyperintegers!) are well-defined (since everything appearing is internal), and it's easy to check that they are an increasing (internal) bounded sequence, that they satisfy the transfer of the Cauchy criterion for convergence of a sequence, and so forth.

Of course, it's easier to compute this sum by just recognizing that it's the transfer of a standard sum that converges to 1.

EDIT: what does an hyperfinite number of 9's mean?

It means that there is a hyperinteger H bigger than zero, and the n-th digit of the hyperdecimal numeral in question is:

• 9, if H <= n < 0
• 0 otherwise

(the 0-th place is the one's place, the 1-th place is the ten's place, the (-1)-th place is the tenth's place, etc)

 

[disregardthat]

Thanks. Does this mean that the corresponding thing to a countable sequence in the standard reals is a sequence indexed by hyper-integers?

9, if H <= n < 0
0 otherwise

And does this have a corresponding non-standard real number (I should guess so as it is a cauchy-sequence indexed by hyperintegers)? Does all non-standard real numbers have a corresponding digit representation?

Still, shouldn't every subset of the non-standard reals bounded below have a largest lower bound by the transfer principle? If we consider the subset of real numbers larger than 0 as a subset of the non-standard reals, this would have no largest lower bound in the non-standard reals. Am I not using the transfer-principle correctly here?

EDIT: Maybe it is because the subset of real numbers cannot properly defined by the transferred axioms of the reals to the nonstandard reals in order for the transfer principle to work? EDIT again: This got me wondering about the "standard part" function. If the reals is not a "properly defined" subset of the non-standards, the standard part couldn't be "properly defined" either.

 

[Hurkyl]

Thanks. Does this mean that the corresponding thing to a countable sequence in the standard reals is a sequence indexed by hyper-integers?

Yep. Well, two technicalities.

The first one is harmless -- in standard analysis, there are lots of countable ordinals (or even more general "order types") that could be used to index seqences. I'm assuming you didn't mean to include these more general sorts of things.

The second is more important -- the sequence has to be "internal".

And does this have a corresponding non-standard real number (I should guess so as it is a cauchy-sequence indexed by hyperintegers)? Does all non-standard real numbers have a corresponding digit representation?

Yep! It's the transfer of the standard theorem:

Every real number is equal to the infinite sum that computes the value of some decimal number​

which becomes

Every hyperreal number is equal to the infinite sum that computes the value of some hyperdecimal number​

Still, shouldn't every subset of the non-standard reals bounded below have a largest lower bound by the transfer principle?

Every internal set.

Am I not using the transfer-principle correctly here?

The internal / external distinction is probably the most important one to understand to avoid making mistakes.

(I think your edit is touching upon the ideas I write below down to the horizontal line)

In the non-standard model, we have the set of hyperintegers, the set of hyperreals, and so forth. And to these we can apply the tools set theory, calculus, analysis, number theory, algebra, or whatever. The transfer principle says the standard and non-standard models have exactly the same theorems.

The power of non-standard analysis comes because we can also view the hyperreals as a standard set. Even better, we can view the standard real numbers as a subalgebra of the non-standard ones!

But that's where the danger comes too -- viewed this way, the standard model has a lot more things in it than the non-standard model does. Doing set theory in the standard model let's us construct a lot of sets that the non-standard model doesn't have. (similarly for sequences, functions, et cetera)

(of course, this all transfers. The non-standard model has hyperhyperreals and a "hypernonstandard model" built on top of them...)
___________________________________________________

Generally speaking, there is a quick way to tell if an object is internal (and thus usable in the non-standard model) or not -- if it makes any reference whatsoever to the standard model other than transferring something, it's probably external.

So the standard part function is, in fact, external. The set of positive standard real numbers (viewed as a subset of the hyperreals) is also an external set. Because it is not internal, it doesn't contradict the LUB property that every internal nonempty bounded subset of the hyperreals has a greatest lower bound.

 

[disregardthat]

Thank you, these are really good explanations. I should get hold of a book on model theory, the transfer principle seems really powerful.

 

[Hurkyl]

Have you tried Keisler's calculus book? I've skimmed through part of it and found it useful. (but then, I knew some non-standard analysis already before doing so)

 

[disregardthat]

Have you tried Keisler's calculus book? I've skimmed through part of it and found it useful. (but then, I knew some non-standard analysis already before doing so)

I haven't, but this is excellent. Thanks for the reference, this will come in handy. Do you perhaps know of a good reference which treats the transfer principle as well?

 

[SewerRat]

Here's a proof that .999...=1.

.999... can be written as the infinite sum as follows:

<SNIP>

---

Hello Char - I found that little proof fascinating. Might that concept also extend to natural numbers like Pi, e, etc. ? Although they are not recurring as such, they are nevertheless convergent, getting ever closer to an asymptote value without ever reaching it (intuitively). If a proof can be established, it would yield actual values for said numbers, albeit with a huge number of decimal places but a finite number nonetheless.