Does .999~=1? A Beginner's Perspective

[matt grime]

What makes you think there is such a point, or two, at infinity? There isn't in euclidean geometry - parallel lines do not meet. There is no coordinate (x,y) with x,y real numbers where parallel lines meet.

If you wish to introduce points at infinity then you need projective geometry.

 

[Littlepig]

What makes you think there is such a point, or two, at infinity? There isn't in euclidean geometry - parallel lines do not meet. There is no coordinate (x,y) with x,y real numbers where parallel lines meet.

If you wish to introduce points at infinity then you need projective geometry.

ok...don't know that...never heard about euclidean geometry and projective geometry...

but is that actually possible?? or is a terrible mistake saying it??

because basically, what I'm doing, is trying to separate 2 lines, but what happens is the more i try to separate then, the far way point "p' " is from p, however, it never separates, as lines are infinite, and the more degrees you rotate, the "faster" the point "p' " moves correct?? that's why i made such afirmation...in infinity, 89.9(9) is equal to 90º...so in infinite(paralelism) there is 2 intersections...which are infinity distants from "p". however, you can't say they don't exist, because otherwise you must assume that, rotating the line "b" you will make disapear point p, which actually don't apears to seems...

regards, littlepig

 

[ramsey2879]

ok...don't know that...never heard about euclidean geometry and projective geometry...

but is that actually possible?? or is a terrible mistake saying it??

because basically, what I'm doing, is trying to separate 2 lines, but what happens is the more i try to separate then, the far way point "p' " is from p, however, it never separates, as lines are infinite, and the more degrees you rotate, the "faster" the point "p' " moves correct?? that's why i made such afirmation...in infinity, 89.9(9) is equal to 90º...so in infinite(paralelism) there is 2 intersections...which are infinity distants from "p". however, you can't say they don't exist, because otherwise you must assume that, rotating the line "b" you will make disapear point p, which actually don't apears to seems...

regards, littlepig

Since infinity is just a concept and not a number, you can't say that there are points on the line at infinity. A point on a number line is always a finite distance from the zero point of the line or else it doesn't exist. Rotating the intersecting line so it is parallel and spaced from the number line doesn't make it any points disappear, it just moves the line so that all points thereon are a fixed distance from the number line. What is confusing about that?

 

[Littlepig]

Rotating the intersecting line so it is parallel and spaced from the number line doesn't make it any points disappear, it just moves the line so that all points thereon are a fixed distance from the number line. What is confusing about that?

no, my point is: there is no fixed point, the "fixed point" is the infinit, is the sucession, and the more you rotate, the far the point go, then, if you rotate till 89.9(9) degrees, the point is at inf distance from point origin but it is still there...
imagine you can't stop rotating, but you can't reach 90º...is like that...then, in the extrem, the point exists, and the degree is 89.9(9)º.
that is like dividing 1 by 3 and then multiply by 3..0.9999(9)never ends...but you know that in the end, it is 1, don't know where is the end, but you know it exists...is the same, you don't have an ending, but you know, that in the end, there's a point...

So, you don't need to ask where's the fixed distance, ask what happens to the degree when the fixed distance reaches to inf...

 

[matt grime]

Sigh. Take this part:

but you know that in the end, it is 1, don't know where is the end, but you know it exists

I'm sure the point has been repeatedly made in this thread that this is completely wrong.

You are, as is almost always the problem. Using your intuition about what happens at every stage after a finite number of decimal places to assert something about the infinitely long decimal expansion. Your intuition is wrong. Infinity is not 'a really big real number'. It is not a real number.

Parallel lines do not meet in the Euclidean plane. If I'm wrong (and I'm not), then feel free to write down the point of intersection: hint there is no such point as infinity on the Euclidean plane.

The place to use points at infinity is projective geometry, and there need not be just one point at infinity.

 

[ramsey2879]

Sigh. Take this part:



I'm sure the point has been repeatedly made in this thread that this is completely wrong.

You are, as is almost always the problem. Using your intuition about what happens at every stage after a finite number of decimal places to assert something about the infinitely long decimal expansion. Your intuition is wrong. Infinity is not 'a really big real number'. It is not a real number.

Parallel lines do not meet in the Euclidean plane. If I'm wrong (and I'm not), then feel free to write down the point of intersection: hint there is no such point as infinity on the Euclidean plane.

The place to use points at infinity is projective geometry, and there need not be just one point at infinity.

Matt, I said finite, not fixed. My point is that it is error to refer to a point at infinity in Euclidean space, for a point to exist in Euclidean space, it must be a finite distance from the orgin. The definititon of parallel lines is that every point on one line is a fixed distance from the other line. I was referring to the face that a line can always be positioned so that every point is a fixed distance from another line. This has nothing to do with points at infinity.

1 = .99999... and 89.999... = 90 but there is no point at infinity so what is the problem?

 

[matt grime]

Matt, I said finite, not fixed.

I neither read your post in detail, nor replied to it. So why address me about something I haven't commented on?

 

[ramsey2879]

no, my point is: there is no fixed point, the "fixed point" is the infinit, is the sucession, and the more you rotate, the far the point go, then, if you rotate till 89.9(9) degrees, the point is at inf distance from point origin but it is still there...
imagine you can't stop rotating, but you can't reach 90º...is like that...then, in the extrem, the point exists, and the degree is 89.9(9)º.
that is like dividing 1 by 3 and then multiply by 3..0.9999(9)never ends...but you know that in the end, it is 1, don't know where is the end, but you know it exists...is the same, you don't have an ending, but you know, that in the end, there's a point...

So, you don't need to ask where's the fixed distance, ask what happens to the degree when the fixed distance reaches to inf...

I don't understand what you are getting at. Infinity means that there is no end. How can there be a point at the end when there is no ending?

 

[ramsey2879]

I neither read your post in detail, nor replied to it. So why address me about something I haven't commented on?

Yeah, I was wrong and apologise

 

[Gib Z]

I haven't read any of the posts, because we've had this topic far too many times. All i know is that, in the article I quote "These can only approximate 1/3, for example, so we don't have an exact expression for 1/3." and "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" God Damn these people, x=0!

Or "Moreover, we have no interpretation for the number -3.14159265." Looks like negative pi to me...

From that, I already know that I know more math than this guy does.

This topic has been discussed far too many times, and it is pointless. They are equal, live with it.

 

[CRGreathouse]

From that, I already know that I know more math than this guy does.

I think I found a better nonsense quote:

The fact that you can't compute the decimal expansion of a sum from the decimal expansions of its addends is a well known phenomenon that was noticed by Turing. In a fully constructive treatment of the real numbers, this is often stated by saying (informally) that not every positive real number has a decimal expansion. More precisely, there is no constructive proof that every positive real number has a decimal expansion (or at least we don't know of one).

 

[Hurkyl]

I haven't read any of the posts, because we've had this topic far too many times. All i know is that, in the article I quote "These can only approximate 1/3, for example, so we don't have an exact expression for 1/3." and "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" God Damn these people, x=0!

Or "Moreover, we have no interpretation for the number -3.14159265." Looks like negative pi to me...

From that, I already know that I know more math than this guy does.

This topic has been discussed far too many times, and it is pointless. They are equal, live with it.

I think I found a better nonsense quote:

The fact that you can't compute the decimal expansion of a sum from the decimal expansions of its addends is a well known phenomenon that was noticed by Turing. In a fully constructive treatment of the real numbers, this is often stated by saying (informally) that not every positive real number has a decimal expansion. More precisely, there is no constructive proof that every positive real number has a decimal expansion (or at least we don't know of one).

You guys aren't being fair. IIRC, his math is quite sound. His only problem is an equivocation fallacy -- he is invoking alternative treatments of the notion of "real number", when everybody else is using the term in its standard meaning.

 

[CRGreathouse]

You guys aren't being fair. IIRC, his math is quite sound. His only problem is an equivocation fallacy -- he is invoking alternative treatments of the notion of "real number", when everybody else is using the term in its standard meaning.

Actually, I rather agree that most of what he says appears sound, and much of the remainder could be salvaged with some rigor. Certainly I have no problem with people exploring alternate systems.

But saying that Turing proved that not all reals have decimal expansions... that's just crazy talk.

 

[Hurkyl]

But saying that Turing proved that not all reals have decimal expansions... that's just crazy talk.

I wouldn't be so sure -- remember that, there, the author was talking about constructive analysis, which doesn't have many of the nice properties we're used to having. There, a real number is usually defined in a way similar to this:

A real number is a (computable) function f that takes an integer n and returns a fraction r, satisfying the property that:

|f(n) - f(m)| < 2^-n + 2^-m

(To connect with the "usual" model of the reals, f(n) is a Cauchy sequence)

We could define a "decimal number" in a similar fashion -- it takes an integer n and returns a decimal digit... and satisfies the property that there exists a bound M such that n > M implies f(n) = 0. (To connect with the "usual" model, f(n) would be the n-th place in a decimal number)

And I believe that, in fact, there does not exist a (computable) function that takes a real number as input and returns a decimal number as output.

 

[mathwonk]

GibZ do not forget the cardinal rule of the teaching profession, there is a naive student born every minute and we get paid again for teaching the same old claptrap to each new generation.

i.e. just because it bores the teacher, does not mean it seems old to the class. actually this is a license to work forever without learning a new trick.

 

[Gib Z]

I don't seem to understand you mathwonk...whose the teacher and whose the student? I'm 14 years old, I'm sure the person who wrote the article is older than me. He should know by now he can't name an alternate system the same as another one, and be surprised that we're confused and insulting him about it.

 

[CRGreathouse]

I wouldn't be so sure -- remember that, there, the author was talking about constructive analysis, which doesn't have many of the nice properties we're used to having. There, a real number is usually defined in a way similar to this:

A real number is a (computable) function f that takes an integer n and returns a fraction r, satisfying the property that:

|f(n) - f(m)| < 2^-n + 2^-m

(To connect with the "usual" model of the reals, f(n) is a Cauchy sequence)

We could define a "decimal number" in a similar fashion -- it takes an integer n and returns a decimal digit... and satisfies the property that there exists a bound M such that n > M implies f(n) = 0. (To connect with the "usual" model, f(n) would be the n-th place in a decimal number)

And I believe that, in fact, there does not exist a (computable) function that takes a real number as input and returns a decimal number as output.

So you're saying that:
1. There's no computable function that converts arbitrary binary expansions to decimal expansions
2. Turing proved this
3. The article, when writing that Turing proved that no computable function returns the decimal expansion of a real, meant (1) and (2)

Alright, perhaps. Let me think about this.

Given a binary expansion* adding up to N, where the last term's (binary) exponent is n, the number is in [N, N + 2^n] if repeating 1s are allowed. For no computable function to exist to convert this to decimal form, for almost all natural numbers k there must be some real numbers x_k and y_k such that x_k and y_k differ in the kth decimal place, and such that there exists no computable f(k) such that the first f(k) places suffice to distinguish x_k and y_k. Right?

I'll need more time to think about this.

* In which every number is either 0 or 1, else the number could change arbitrarily with each additional bit.

 

[Hurkyl]

I'm trying to be less committal than that! I'm pretty sure I remember a theorem that there are more constructive reals than constructive decimals. I don't know if Turing proved it, but seeing his name in this context is unsurprising.

I have vague recollections that if we compute their values in the "standard" reals, that {values of computabie decimals} is actually proper subset of {values of computable reals}... but I don't remember for sure. But I'm confident about what I did say in my previous post.


One important correction to your post:

1. There's no computable function that converts arbitrary binary expansions to decimal expansions

That's not what I said -- a computable real isn't a "binary expansion": it's a Cauchy sequence whose m and n-th terms differ by less than 2^-m + 2^-n. For example,

1/2, 5/4, 7/8, 17/16, 31/32, ...

is a perfectly good computable real number. I stated "fraction", but I think you get something equivalent if you substitute "binary decimal"... but it's important to note that point is that the leading digits of the m-th term and the (m+1)-th term don't have to agree, such as in the sequence I listed above.


If you want to take smaller steps into this stuff, I'm pretty sure that equality is not a computable relation. (For both the computable reals and the computable decimals)

 

[Hurkyl]

Oh, there's a short proof that equality is not decidable for the computable real numbers. I'll demonstrate with a simpler problem: telling if a real number is equal to 0.


Suppose I had a computable function Z(x) that computes whether the real number x is equal to zero.


I can use Z to solve the halting problem as follows: suppose I have some program P, and I want to know if it halts or not. I can construct a computable real number D as follows:

D(n) = 0 if P doesn't halt within n steps
D(n) = 2^-k if P halts after k steps, with k <= n

Clearly D is computable -- to compute D(n), simply run P for n steps, and see if and when it stops.

Clearly, Z(D) = true iff P doesn't halt. So, Z can be used to solve the halting problem.

Because the halting problem is incomputable, Z must be an incomputable relation.

 

[CRGreathouse]

Hmm, well, you're quite right. Still, none of this was justified by the link... all of this comes from you, including cleaning up the definition.

Thanks. I concede the point.

 

[mr_homm]

While I fully agree with the correct subset of the arguments given above (you know who you are), it seems to me that there is a genuine ambiguity in the notation 0.999... Caveat: Of course the real number system is defined axiomatically, so questions about the notation resolve ultimately not into questions about about the real number system per se, but about whether the notation is actually referring to the real number system or to some alternate system. This seems to be the difficulty with the original reference that started this thread.

The ambiguity is in the "..." -- since the decimal number system was invented long before Cantorian ordinals, the "..." is the "naive infinity" which is usually assumed to by omega, the smallest transfinite ordinal. If you state that assumption explicitly, then there is nothing to discuss, since you have then defined the notation precisely enough that the you can apply Cauchy's theory of limits to it and prove that 0.999... = 1. End of story.

However, if you allow the decimal to continue to a higher ordinal number of decimal places, you have a notation that seems not to correspond to the standard reals. Of course, this does not by itself create a number system for the notation to point to. (If that were true, then Anshelm's ontological proof of the existence of God would actually be valid and we should all be convinced by it. I think not.)

There are of course various extensions of the reals, particularly the hyperreals and surreals, which seem (at first glance anyway) to be candidates for number fields that would be appropriately expressed by such a larger-ordinal notation. As far as I am aware, this is an open question (someone correct me if I am wrong). I do know that Conway and Berlekamp showed that one particular notation (the "Hackenbush game") for the surreals had a subset of expressions that had an obvious correspondence with the real numbers expressed in the binary system.

I'm not sure that this contribution to the thread advances the discussion in any way; I'm just trying to add something that is perhaps of interest to a discussion on a topic that is extremely tedious to mathematicians who have to deal with cranks on occasion.

--Stuart Anderson

 

[Hurkyl]

There are of course various extensions of the reals, particularly the hyperreals and surreals, which seem (at first glance anyway) to be candidates for number fields that would be appropriately expressed by such a larger-ordinal notation. As far as I am aware, this is an open question (someone correct me if I am wrong).

It's very easy for hyperreals -- hyperdecimal numbers are indexed by hyperintegers, just as ordinary decimal numbers are indexed by ordinary integers.

The hypernatural numbers are not (externally) well-ordered, but I don't see why the index set should be expected to be an ordinal.

 

[DeadWolfe]

I wonder how this topic is really number theory...

 

[mr_homm]

It's very easy for hyperreals -- hyperdecimal numbers are indexed by hyperintegers, just as ordinary decimal numbers are indexed by ordinary integers.

Thanks! I had forgotten this (or never new, perhaps), as I have been more interested in the surreals than the hyperreals lately.

The hypernatural numbers are not (externally) well-ordered, but I don't see why the index set should be expected to be an ordinal.

No special reason why, I just mentioned larger ordinals as an example of a possible extension. BTW, could you give me a reference on the hypernaturals indexing the hyperreals? My acquaintance with this topic comes from Nelson's Radically Elementary Probability Theory, which went over non-standard analysis rather briefly as background for the application to probability.

Thanks!

--Stuart Anderson

 

[Hurkyl]

BTW, could you give me a reference on the hypernaturals indexing the hyperreals?

It's a direct application of the transfer principle.

In the standard model, the decimal expansion of a real number s is nothing more than a function f:Z->{0, ..., 9} satisfying

s = \sum_{n \in \mathbb{Z}} f(n) 10^n
\lim_{n \rightarrow +\infty} f(n) = 0

and we have a theorem that every real number has a decimal expansion.


Applying the transfer principle tells us that in the nonstandard model, the hyperdecimal expansion of a hyperreal number s is nothing more than an (internal) function f:*Z->{0, ..., 9} satisfying

s = \sum_{n \in {}^\star \mathbb{Z}} f(n) 10^n
\lim_{n \rightarrow +\infty} f(n) = 0

and we have a theorem that every hyperreal number has a hyperdecimal expansion.

 

[mr_homm]

@Hurkyl

Got it! Thanks. I must be having a bad day, because it's obvious now that you mention it. Of course, the axioms are the same as for standard analysis when the new predicate "standard" is not used, so of course all the usual results follow whenever no distinction between standard and nonstandard values is made. This point was made clearly by Nelson, so I shouldn't have needed your prod. But apparently I did need it, so thanks!

--Stuart Anderson

 

[Changbai LI]

0.9... is not =1. You can see
0.9.../=1(English)
My emai is
Changbai LI

 

[cristo]

This has been discussed ad nauseum on the forums. If you wish to see why 0.9..=1, please use the search facility and read through the several threads.