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

[raven1]

i came across am argument about does .999~=1 and someone used this webpage to show they are not equal http://www.math.fau.edu/Richman/HTML/999.htm
this page seems somehow wrong to me but i haven't gone far enough in math
to disprove it , i just started to learn calclus so when it involes any in depth discussion of limits i try to be real carful what i say

(Edited by HallsofIvy so that the url could be directly used.)

 

[shmoe]

The conclusion is perfectly correct. I can't be bothered to read that page in any detail, but it looks alright at a first look.

This has come up countless times before, so there's lots of nonsense to wade through:

https://www.physicsforums.com/showthread.php?t=125474&highlight=repeating+equal+1
https://www.physicsforums.com/showthread.php?t=110434&highlight=repeating+equal+1
https://www.physicsforums.com/showthread.php?t=106212&highlight=repeating+equal+1

etc. probably more than you would care to read.

 

[Hurkyl]

shmoe: the guy's arguing that 0.\bar{9} \neq 1, not that 0.\bar{9} = 1.


Basically, he's just arguing semantics. He doesn't like how things are named. He knows full well that, in the decimals, 0.\bar{9} = 1. However, he wants to call some other number system the "decimal numbers". He also wants to allow the technical term "real number" to refer to something other than its technical definition.

 

[mathwonk]

thhhbbbbpt!

.99999... = the smallest real number not smaller than any finite decimal of form .9999...9


therefore it equals 1. case closed.

 

[HallsofIvy]

I notice the website includes
"A skeptic who accepts the series interpretation could say that 0.999... converges to 1, or that it is equal to 1 in the limit, but is not equal to one. There is an ambiguity in standard usage as to whether the expression on the right stands for the series or the limit."

That is completely wrong. It is not necessary to "accept the series interpretation". It is not an "interpretation", it is the definition of "base 10 representation" that 0.abc...= \Sigma a/10 + b/100+ c/1000+ \cdot\cdot\cdot. There is no "ambiguity" except for people who simply do not know the definition of "series". Any calculus book will tell you that \Sigma_{n=0}^\infty a_n is defined as the limit of the sequence of partial sums.

In particular, the notation 0.999... means, by definition, the series
\Sigma_{n=0}^\infty \left(0.9\right)\left(\frac{1}{10}\right)^n[/itex]<br /> That&#039;s a simple geometric series which has a simple formula: its sum (limit of the partial sums) is <br /> 0.9\frac{1}{1- \frac{1}{10}}= 0.9\frac{1}{\frac{9}{10}}= 0.9\frac{10}{9}= 1.<br /> End of discussion!

 

[shmoe]

shmoe: the guy's arguing that 0.\bar{9} \neq 1, not that 0.\bar{9} = 1.

That's what I get for not reading carefully! Somebody give me a beating.

 

[nocturnal]

It's listed on his website that he received his Phd from the University of Chicago I thought that was supposed to be a great school for math. Did this guy blow a fuse?

 

[HallsofIvy]

Oh, and I love this quote

Dedekind cuts are usually defined in the ring of rational numbers, but if we are interested in decimal numbers, we will want to work with a different ring.

Because "decimal numbers" are some kind of numbers other than "rational numbers"?

(When I am feeling really hard-nosed, I point out the "decimal numbers" is itself a mis-nomer. It should be "decimal numerals".)

 

[Hurkyl]

I thought that was supposed to be a great school for math. Did this guy blow a fuse?

The thing is -- the guy is actually talking about reasonable and interesting stuff. It's just that, for some inexplicable reason, he's decided to graft it onto the 0.9~ vs 1 "debate".

 

[chronon]

Because "decimal numbers" are some kind of numbers other than rational numbers"?

Yes, as I read it, his 'decimal numbers' are infinite strings of digits. The ring he is starting with isn't the 'decimal numbers' or the rational numbers, but the terminating decimals. What he is doing is adding some extra numbers x^- to the real number system, so that 0.9~ can be taken to represent a different object to 1.0 . Now in mathematics you can invent whatever system takes your fancy, but whether it is of any interest is a different matter. It's not entirely clear whether he's adding a whole new copy of the reals R^-, (in which case some of his numbers are no longer representable as infinite decimals, rather defeating the point) or just adding a copy of the terminating decimals. In either case the resulting object is no longer a field, or even an additive group - I'd prefer to be able to do subtraction.

 

[mathwonk]

the fact that .9999... = 1.0000 ... is about as deep as the fact that

f. delano roosevelt = franklin d. roosevelt

 

[CRGreathouse]

Yes, as I read it, his 'decimal numbers' are infinite strings of digits. The ring he is starting with isn't the 'decimal numbers' or the rational numbers, but the terminating decimals. What he is doing is adding some extra numbers x^- to the real number system, so that 0.9~ can be taken to represent a different object to 1.0.

Well, that's a fine thing to do, but it brings up an important question to me. If 0.9~ != 1 in his system, then either 1/3 != 0.3~ or 0.3~ x 3 != 0.9~. Can he even do arithmetic on (the equivilent of) basic rational numbers?

 

[HallsofIvy]

the fact that .9999... = 1.0000 ... is about as deep as the fact that

f. delano roosevelt = franklin d. roosevelt

The way my father used it, that "f." was an obscenity!

 

[saltydog]

the fact that .9999... = 1.0000 ... is about as deep as the fact that

f. delano roosevelt = franklin d. roosevelt

Might I disagree?

I believe the question has profound philosophical consequences:

It concerns the concept of a limit, that of the one Hall describes above. The limit exhibits a profound property of the Real numbers: they are dense. It is this simple property of the Reals which I believe is responsible for Mathematics working so well in describing Nature. Nature too appears dense: no smallest small nor largest large. This synergy between math and nature emerges (my opinion) as a survival strategy by life as it seeks to live in a massively non-linear world: when in New York, act like a New Yorker. Thus evolves a likewise massively non-linear brain that creates a non-linear geometry we call Mathematics that enables life to ponder this question.

 

[HallsofIvy]

Then I will jump in and disagree completely- this does not involve any property of the real numbers- it is entirely a matter of how we represent the real numbers in a "base 10 positional notation". If we were to use, say, base 3, then "0.9999...= 1.0" would not be true (although, I imagine that "0.2222...= 1.0" would be). If we used some representation that was not a positional notation, the question would never arise.
This is a question about representation only, not about the real numbers. Indeed, 1 (and 0.9999...) are integers so, in particular, this has nothing to do with the "density" of the real numbers. (Whatever that means. The only "densities" I know of are relative to some set. Do you mean the fact that the rational numbers are dense in the real numbers?)

 

[matt grime]

I think saltydog is attempting to describe the 'continuum' property. Which nature does not necessarily follow, or use, at all, saltydog. Lots of parts of nature behave in a quantized manner. The rest just seems to belong in philosophy, not mathematics, though I have no idea what geometry has to do with any of this, nor have I ever come across the term 'linear geometry' before.

 

[saltydog]

Then I will jump in and disagree completely- this does not involve any property of the real numbers-

Let me attempt a defense then:

The sum:

0.9\sum_{n=0}^{\infty}\frac{1}{10}^n

converges to 1 because between any two real number lies another real number (no holes). In this way the reals are "dense". My argument was not in regards to notation but to its relation to this property of the number system we create which bears a striking similarity to the apparently infinitely divisible nature of the Universe. Discussions about "0.99...=1" in my opinion reflect this beautiful connection between the geometry of real numbers and the geometry of nature.

 

[matt grime]

Let me attempt a defense then:

The sum:

0.9\sum_{n=0}^{\infty}\frac{1}{10}^n

converges to 1

red alarm light comes on: that is 1. It's partial sums converge to 1.

because between any two real number lies another real number (no holes). In this way the reals are "dense".

the rational numbers also have the property that 'between two rationals there is another rational', and they do not posses limits of sequences.

However, they are dense in the reals in the proper meaning of the word (a set is dense in itself tautologically, if the notion of denseness makes sense at all.)

 

[saltydog]

I think saltydog is attempting to describe the 'continuum' property. Which nature does not necessarily follow, or use, at all, saltydog. Lots of parts of nature behave in a quantized manner. The rest just seems to belong in philosophy, not mathematics, though I have no idea what geometry has to do with any of this, nor have I ever come across the term 'linear geometry' before.

Very well Matt. I am struck by the similarities between the properties of non-linear systems and the geometry of math itself. Not non-linear geometry but the very geometry of mathematics itself: nested, fractal, and ergodic (the last property explaning why we can get to the same result from so many ways). But Philosophy it should be then.

 

[Hurkyl]

However, they are dense in the reals in the proper meaning of the word (a set is dense in itself tautologically, if the notion of denseness makes sense at all.)

What saltydog stated was the definition of a "dense ordering" -- an order is dense iff for any two elements, you can find a third between them.

 

[matt grime]

What saltydog stated was the definition of a "dense ordering" -- an order is dense iff for any two elements, you can find a third between them.

who knows what was meant. i focussed on the 'no holes' part.

 

[HallsofIvy]

Let me attempt a defense then:

The sum:

0.9\sum_{n=0}^{\infty}\frac{1}{10}^n

converges to 1 because between any two real number lies another real number (no holes). In this way the reals are "dense". My argument was not in regards to notation but to its relation to this property of the number system we create which bears a striking similarity to the apparently infinitely divisible nature of the Universe. Discussions about "0.99...=1" in my opinion reflect this beautiful connection between the geometry of real numbers and the geometry of nature.

That simply isn't true. The partial sums of
0.9\sum_{n=0}^{\infty}\left(\frac{1}{10}\right)^n
converge to 1, and the sum is equal to 1, in the field of rational numbers- which is not complete and has "holes". In fact, it is easy to show that any geometric series in which "a" and "r" are both rational converges, in the field of rational numbers, to a rational number. There is no need to bring real or irrational numbers into it.

And the "apparently infinite divisible nature of the Universe" is just that- "apparent". Have you never heard of atoms? The universe is not "infinitely divisible".

 

[HallsofIvy]

I think saltydog is attempting to describe the 'continuum' property. Which nature does not necessarily follow, or use, at all, saltydog. Lots of parts of nature behave in a quantized manner. The rest just seems to belong in philosophy, not mathematics, though I have no idea what geometry has to do with any of this, nor have I ever come across the term 'linear geometry' before.

Blast you, matt! I wondered why this same topic showed up in "philosophy"- it's your fault! This isn't philosophy, it just bad mathematics- and mysticism.

 

[saltydog]

That simply isn't true. The partial sums of
0.9\sum_{n=0}^{\infty}\left(\frac{1}{10}\right)^n
converge to 1, and the sum is equal to 1, in the field of rational numbers- which is not complete and has "holes". In fact, it is easy to show that any geometric series in which "a" and "r" are both rational converges, in the field of rational numbers, to a rational number. There is no need to bring real or irrational numbers into it.

Very well Hall. I obviously don't have it then. Thanks.

And the "apparently infinite divisible nature of the Universe" is just that- "apparent". Have you never heard of atoms? The universe is not "infinitely divisible".

And atoms are made of quarks and those of perhaps strings. But I do not in the least believe that is the end of it nor are super-clusters the end at the other extreme. Rather I suspect we encounter singularities which change the rules: "Infinitely divisible" is then a reflection of our limitations with understanding Nature.

 

[Dragonfall]

Representations of some real numbers by decimals is not unique, just like the fact that representations of some real numbers by fractions is not unique: 1/2, 2/4, etc. Surely that's not hard to grasp.

 

[Changbai LI]

0.99…≠ 1
1. 10÷3 = 3+ 1 ÷3，so 10÷3 = 3 …1 is not correct.
The values on each side of an equal sign means both values are strictly equal.
9÷3 = 3. it is right. It can be checked by direct computations (by times 3).
10÷3 =（9+1）÷3 =3 + 1÷3 is right now. It can be checked by direct computations (by times 3).
So 10÷3 = 3…1 is not correct. It can not checked by direct computation.
1÷3 = 0.3… is not correct either.
The right way is ：
1÷3=（0.9 +0.1）÷3 = 0.3+0.1÷3（≈ 0.3）. (1.1)
=（0.99 +0.01）÷3 = 0.33 +0.01÷3（≈ 0.33）. (1.2)
…
=（1-1/10^n）÷3 +（1/10^n）÷3 （≈ 0. 3…3） (1.3)
=（1-10/10^n+1）÷3+（10/10^n+1 ）÷3=（1-1/10^(n+1）÷3)+（1/10^(n+1 ）÷3 ) (1.4)
=…
In the division, because there is always a remainder of one, there will also always be a fraction of 3.
So 1÷3 = 0.3… is not correct.
Times 3, then 1≠0.9….
End.

 

[Hurkyl]

In the division, because there is always a remainder of one, there will also always be a fraction of 3.

There is only a remainder of 1 if you decide to compute finitely many digits.

You are right; 1/3 is equal to 0.3 with a remainder of 1. That is,

1/3 = 0.3 + 0.1 / 3

But don't forget that 0.1 / 3 = 0.0333...

 

[Office_Shredder]

0.99…≠ 1
1. 10÷3 = 3+ 1 ÷3，so 10÷3 = 3 …1 is not correct.

Do you mean to say that 1/3 isn't simply 3 repeating, but 3 repeating with a 1 at the end?

Do you realize how many threes are in between the first 3 and the 1? In fact, the number you propose here isn't even possible as an element of the real numbers

 

[HallsofIvy]

0.99…≠ 1
1. 10÷3 = 3+ 1 ÷3，so 10÷3 = 3 …1 is not correct.
The values on each side of an equal sign means both values are strictly equal.
9÷3 = 3. it is right. It can be checked by direct computations (by times 3).
10÷3 =（9+1）÷3 =3 + 1÷3 is right now. It can be checked by direct computations (by times 3).
So 10÷3 = 3…1 is not correct. It can not checked by direct computation.
1÷3 = 0.3… is not correct either.
The right way is ：
1÷3=（0.9 +0.1）÷3 = 0.3+0.1÷3（≈ 0.3）. (1.1)
=（0.99 +0.01）÷3 = 0.33 +0.01÷3（≈ 0.33）. (1.2)
…
=（1-1/10^n）÷3 +（1/10^n）÷3 （≈ 0. 3…3） (1.3)
=（1-10/10^n+1）÷3+（10/10^n+1 ）÷3=（1-1/10^(n+1）÷3)+（1/10^(n+1 ）÷3 ) (1.4)
=…
In the division, because there is always a remainder of one, there will also always be a fraction of 3.
So 1÷3 = 0.3… is not correct.
Times 3, then 1≠0.9….
End.

Simple non-sense. The fact that every term in a sequence has a property (has a remainder when divided by 3) doesn't mean that the limit has that property. That's your logical error.

 

[Littlepig]

hi there, i have 17 years old, so don't put me with complicated math..
some days ago i have a discution with friends of me exactly about this subject.
after some arguments, i acepted that 0.9(9)=1. i put myself thinking about it and i have a question about it:

we have 2 lines(don't sure the traduction in inglish, but is a infinite number of points that are alined all in "front" of the other): A and B, they are perpendicular, their intercection is the point "p" and we start rotating B like the example above:

http://img297.imageshack.us/img297/5649/screenhunter008se9.png"

my question is: as 1=0.9(9)[or 0.9...], we can assume that, in the infinite, the point "p' "(a projection of "p") is at a infinite distance from point "p" right??
and can we assume that in infinite, the degree "b" is 90º? why not as 89.9(9)º=90º. right??

then, 2 paralen lines have, at least, 1 interception point...

now, if 89.9(9)º=90º then in this http://img413.imageshack.us/my.php?image=screenhunter011tv9.png"

we can assume that degree "c" is too 89.9(9) right?? then there is another point(p'') that exist too, right? and we assume 2 parallel lines have 2 intercection points...

if not, where is the mistake??

thank you in advance,
Regards, Littlepig

 

[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.