Simple mathematical problem (?)

[Cabbage]

Do you know how to compute a derivative using infinitessimals?

Yes, I do. You do realize there are no infinitesimal real numbers, don't you?

 

[Suggestologist]

Yes, I do. You do realize there are no infinitesimal real numbers, don't you?

That's a matter of mathematical philosophy.

 

[Ziggurat]

Sorry, Suggestologist, but I'm going with Cabbage on this one. I don't think you know what you're talking about. Since you seem fixated on a certain flawed logic, maybe I can show you where you're wrong. You had:

X = .9999999......
10X = 9.99999999.......90
10X-X = 8.999999999.....91
9X = 8.99999999999.....91

But that makes some assumptions that you can somehow keep track of decimal places AFTER you've had an infinite number of 9's in between the decimal point and the "last" decimal. But you can't. Is 0.999... 9 any different than 0.999... 90? Did we tack on an extra zero one space farther down, or did we change the last 9 to a zero? That's just not a question that makes sense, because we've got an infinity of 9's already. We can't talk about a "last" digit because there IS no last digit. That's the whole bloody point. It's like asking if infinity + 1 is bigger than infinity.

 

[Cabbage]

That's a matter of mathematical philosophy.

And this is where you are 100% wrong. There are no infinitesimal real numbers, period. It is not a matter subject to debate.

First, an analogy. On one hand, for example, we have integers ("whole" numbers). On the other hand, we have rational numbers, which consist of all possible fractions (with nonzero divisors) of integers. While it is true that there is a (rational) number strictly between 0 and 1 (1/2, for an obvious example), it is not true that there is an integer strictly between 0 and 1. This is beyond dispute.

Similarly, there are infinitesimals (see, I do have some understanding of this topic (of which you have some, but apparently very limited, knowledge as well)). There are various field extensions of the reals known as non-Archimedean fields, which have infinitesimals in them (an infinitesimal being an x>0 such that nx < 1 for all positive integers n, violating the Archimedean property held by other fields, such as the reals).

The point which has apparently escaped you, however, is that the real numbers are not a non-Archimedean field! There are no infinitesimals in the real numbers!

And, as I just said, there are number systems with infinitesimals in them, such as the surreals and the hyperreals, which you yourself have mentioned in this thread. The point, however, just in case you missed it the first time, is, once again, that THE REALS THEMSELVES DO NOT HAVE INFINITESIMALS!. There are non-Archimedean fields. The reals are not one of those fields.

Surreals and hyperreals have other (meaning non-decimal) notations to represent them. Just as, not surprisingly, integers and rationals have different notations representing them (integers can be represented as rationals of course, but not conversely).

Ever try to write 1/2 as an integer? You can't do it!.

You keep trying to write a nonreal surreal number as a decimal (.999...), but You. Can't. Do. It!

.999... is a real number, and is exactly equal to one. Period.

 

[xouper]

sorgoth: You see, 1/3 is an infinitely small bit larger than 0.333...

You are wrong. This has already been thoroughly refuted in this thread. No mathematician believes that 1/3 is not exactly equal to 0.333...

Haven't you been paying any attention to what's been posted in this thread?

 

[aerosolben]

When you multiply .9999999.... by 10, is? the LAST digit -- the same position that the LAST 9 used to occupy -- now a 0 or now a 9?

THERE IS NO LAST DIGIT. The string of 9's is infinite, thus never-ending. 0.9~ * 10 = 9.9~.

I agree that no such real number exists (at that level of granularity). As I said earlier, there is no real number closer to and less than 1 than .9999999...... . The idea that numbers exist between any two real numbers does not apply to infinite (non-zero) digits of real numbers.

Numbers exist between any two distinct real numbers. All your 'exception' above indicates is that 0.9~ and 1 are not distinct.

The fact is that .999999......... inserts talk of infinity into the discussion. Whether by limits or series or recurrence of a digit or whatever. Trying to limit use of infinity to what YOU think is appropriate is like trying to have your cake and eat it too.

You seem to have an odd idea of what infinity is. Here's a few points:

1. Infinity is a nice upper bound on the real number system. There are no numbers larger then infinity.

2. A infinite length string goes on forever. It never ends, and it has no 'last' member.

That's a matter of mathematical philosophy.

Infinitesimals are a representation of a limit approaching zero, used when zero cannot be because it is divergent. They are not numbers.

That's a matter of mathematical reality. Sure, you could make up a 'new math', but don't presume that it can dictate how things work in an established system.

EDIT to add Cabbage's explanation here is much more thorough than mine.

 

[Zep]

All copyright to Scott Adams for this, and United Features Syndicate, and EVERYONE ESLE whose ownership needs to be recognised.

 

[Earthborn]

Not quite. You see, 1/3 is an infinitely small bit larger than 0.333...

Okay, let's assume you are right. We should then be able to figure out how infinitely small bit larger it is.

We should of course remember that all the thirds of the whole are exactly the same. We can't have the first 1/3 to be 0.333_34 and the others 0.333_33. So we'll assume the infiniteth digit is the same for all of them.

Could it be a 4? Let's see:
0.333_34 + 0.333_34 + 0.333_34 = 1.000_02

No, that can't be right, we'll end up with a number larger than 1

Could it be a 2?
0.333_32 + 0.333_32 + 0.333_32 = 0.999_96

No, that is not right either, we'll end up with a number smaller than 1.

Apperently it must be between 2 and 4. Could it be a 3?

0.333_33 + 0.333_33 + 0.333_33 = 0.999_99

That looks more like it. However, it now appears as if the infiniteth digit in the sum is a 9, and it just doesn't look right. It looks like there is a infinitely small amount missing. In fact the last digit of 0.333_ isn't exactly a 3. It is 3 + 1/3. So:

0.333_33 needs to be followed by a whole lot of 3s. Apperently the infiniteth digit isn't the last one.

 

[LW]

As I said earlier, there is no real number closer to and less than 1 than .9999999......[

Now, let's suppose for a moment that your Iclaim is true (namely, that 1 != 0.99... and there is no real number between them).

Then, what is the number c = (1 + 0.99...) * 0.5?

Since c is the average of 0.99... and 1, then it has to lie between them. But we already supposed that no such number exists.

So, if what you say is true, then the real numbers are not closed under multiplication [or alternatively not under addition]. Doesn't that sound mightily strange to you?

 

[LW]

By whose definition?

Do you accept, for example, K. H. Rosen who uses that definition in the 12th chapter of Elementary Number Theory and its Applications?

[quoteThere are none? Interesting.[/quote]

Can you find a counter example?

They generate .9999999.... for the same reason that 1/1 can be made to generate .9999999...... They are what would be required to fit the role of the denominator and nominator of the ratio.

Begin with Nominator = 9 and Denominator = 10. Multiply by 10/10 without reducing the ratio. Add 9 to the nominator. Repeat.

The main difference between an integer and a real is that an integer always has a finite representation. Every single integer has a decimal representation of the form: d_0 d_1... d_n

where n is a finite integer. There is an infinite number or integers, and their sizes grow aribtarily large (that is, for any fixed number n, you can find an integer that is longer than it), but they are all finite. Your example has an infinite length, so it is not an integer.

Reals, however, have an infinite decimal expansion. Every single real has one. For example, the expansion of the integer 5 is 5.000... with an infinite number of zeroes in there.

Rosen's book has the Theorem 12.1 (page 430) that states:

Let \gamma be a real number with 0 <= \gamma < 1, and let b be a positive integer b > 1. Then \gamma can be uniquely written as

\gamma = sum[j=1,infinity] c_j / b^j

where the coefficients c_j are integers with 0 <= c_j <= b-1 for j = 1, 2, ... with the restriction that for every positive integer N there is an integer n with n > N and c_n != b-1.

Now, what can we see from here. First, the b is the base so we can replace it by decimal base 10 to get:

\gamma = sum[j=1,infinity] c_j / 10^j.

So, the real number is the sum. So, if we set that c_j = 9 for all j = 1, 2, ..., we get that

\gamma = sum[j=1,infinity] 9 / 10^j

is a real number.

The "with the restriction" part of the theorem states that every real number has a unique expansion that doesn't end in an infinite number of digits b-1. In the decimal case it means that every real number has a unique representation that doesn't end in an infininite number of 9s.

However, this doesn't mean that a real number may not have such a representation. The theorem says that every real number has exactly one expansion that doesn't end in infinite nines, but it doesn't forbid it to have two representations: one that doesn't end in nines and one that does end.

Moreover, since 0.999... is a real number and according to the Theorem 12.1, each real number has an expansion that doesn't end in an infinite number of 9s, then it has to have also another expansion. That expansion is 1.000... .


[Edited to add stuff about integers]:

Rosen's book has also a theorem about representations of integers. It is Theorem 2.1 on page 40. It goes like this:

Let b be a positive integer with b > 1. Then every positive integer n can be written uniquely in the form:

n = a_k * b^k + a_{k-1} * b^{k-1} + --- + a_1 * b + a_0

where k is a nonnegative integer, a_j is an integer with 0 <= a_j <= b-1 for j = 0, 1, ..., k and the initial coefficient a_k != 0.

The difference here to the case of real numbers is that k is an integer. That is, an integer may not have an infinite representation as infinity is not an integer.

 

[whitefork]

I submit this this thread is the one true spawn of the thread that will not die.

 

[Suggestologist]

Sorry, Suggestologist, but I'm going with Cabbage on this one. I don't think you know what you're talking about. Since you seem fixated on a certain flawed logic, maybe I can show you where you're wrong. You had:

X = .9999999......
10X = 9.99999999.......90
10X-X = 8.999999999.....91
9X = 8.99999999999.....91

But that makes some assumptions that you can somehow keep track of decimal places AFTER you've had an infinite number of 9's in between the decimal point and the "last" decimal. But you can't. Is 0.999... 9 any different than 0.999... 90? Did we tack on an extra zero one space farther down, or did we change the last 9 to a zero? That's just not a question that makes sense, because we've got an infinity of 9's already. We can't talk about a "last" digit because there IS no last digit. That's the whole bloody point. It's like asking if infinity + 1 is bigger than infinity.

Exactly. I was asked to show how "Infinity + 1" was relevant to the discussion earlier in this thread and this is where it is relevant. .99999..... has omega nines following the decimal point. If multiplying by 10 produces 9.99999...... such that 9.99999..... - .9..... = 9; then you have to assume that omega+1 9's were actually following the decimal point in .99999.....; and that omega+1 = omega which is not correct in the transfinite system. Now someone will say, who cares about the transfinites, we're talking about the reals. And my reply is that we are talking about the properties of a real number, and so the transfinites apply.

 

[Cabbage]

Exactly. I was asked to show how "Infinity + 1" was relevant to the discussion earlier in this thread and this is where it is relevant. .99999..... has omega nines following the decimal point. If multiplying by 10 produces 9.99999...... such that 9.99999..... - .9..... = 9; then you have to assume that omega+1 9's were actually following the decimal point in .99999.....; and that omega+1 = omega which is not correct in the transfinite system. Now someone will say, who cares about the transfinites, we're talking about the reals. And my reply is that we are talking about the properties of a real number, and so the transfinites apply.

There is no (omega+1)th digit in a real number.

What's the (omega+1)th digit of pi? What if you multiply it by 10? Divide it by 10?

 

[aerosolben]

then you have to assume that omega+1 9's were actually following the decimal point in .99999.....

No.

.9 * 10 + .09 * 10 + .009 * 10 + ... = 9.99...

You would have to worry about an extra digit if 9.99... had a terminal digit, but it does not have a terminal digit. Both 0.9~ and 9.9~ have an infinite string of nines following them, as does (0.9 x 10^(50 bajillion)).

Infinity+1 is meaningless, as there are no real numbers larger than infinity by definition.

You are repeatedly attempting to use infinity as a finite number, which it most certainly is not. Therein lies the disagreement.

 

[69dodge]

Originally posted by Suggestologist
Wrong. That notation is correct and is used by mathematicians.

Which mathematicians? Where? I'd like to see it for myself.

If 0.00000.....0001 is meaningless under decimal notation then so is 0.99999999.....

I disagree. I will try to explain why.

A decimal expansion consists of a sequence of digits. Each digit position is labelled with an integer: there is a first position after the decimal point, a second position, a millionth position, etc. Since there are an infinite number of integers to label the positions with, there are an infinite number of digit positions. Since each of those integers is finite, each digit position has as its label a finite number.

Now obviously we can't actually write down an infinite number of digits. However, if the digits follow some sort of pattern, we might still be able to say what each digit is, using only a finite-size description. The decimal expansion of a rational number, for example, follows a particularly simple sort of pattern: after a while, a finite-length sequence of digits repeats forever. So, to describe the decimal expansion of a rational number, we can simply write down the part before the repetition starts, as well as one copy of the part that repeats, and say which is which.

Describing 0.9999. . . is easy: each digit after the decimal point is a 9.

If we try to describe what you wrote as 0.000. . .0001, we run into trouble. (Remember, a description of a decimal expansion must specify which digit appears at each integer-labeled position.) Well, ok, let's start. What's the first digit? 0. What's the second digit? 0. What's the millionth digit? 0. If n is any integer at all, what's the nth digit? 0. So where's the 1? You wrote a 1, so apparently there is supposed to be a 1 somewhere. But where is it? What is the label of the position where it appears?

All of the digits of 0.9999999.... are 9 after the decimal point. What is the last real digit of the number 0.999999999.....?

There is no last digit. Each digit is followed by another.

Why do you suppose there must be a last digit?

 

[slimshady2357]

the limit is 1 then.

Kee-Rist! This is still going on?

Lux,

0.999... is a NUMBER which is in a certain NOTATION.

In decimal NOTATION a NUMBER is (please read this carefully) DEFINED to be equal to the limit of it's SERIES expansion.

Therefore, 0.999... is NOTATION for the NUMBER that is equal to the limit of the SERIES:

0.9 + 0.99 + 0.999 + 0.9999 ....

Which, you agree is equal to 1.

And therefore 0.999.... (which is merely a notation for a specific number) is exactly equal to 1.

What part of this do you not agree with? You seem to be wanting to re-define what the decimal NOTATION actaully means!

Perhaps what you actually want to say is that 0.9 + 0.99 + 0.999 + 0.9999 .... is not equal to 1?

Adam

Edit: My apologies, I see now that Lux has finally seen all this

Sorry all. Responded to a post on the 8th page and didn't realize this thread had made it to the 10th already!

Adam

 

[LW]

If we try to describe what you wrote as 0.000. . .0001, we run into trouble.

BTW, has anyone pointed up this far that in general mathematicians use a notation of the form:

a_0 a_1, a_2, ..., a_n

where there's something on both sides of ellipsis to denote a finite case. According to this interpretation, 0.00..01 would denote a finite string. Though, usually you have the elements indexed somehow so that you can later speak of the length of the string.

 

[69dodge]

Originally posted by Suggestologist
Exactly. I was asked to show how "Infinity + 1" was relevant to the discussion earlier in this thread and this is where it is relevant. .99999..... has omega nines following the decimal point. If multiplying by 10 produces 9.99999...... such that 9.99999..... - .9..... = 9; then you have to assume that omega+1 9's were actually following the decimal point in .99999.....; and that omega+1 = omega which is not correct in the transfinite system.

Not exactly. You have to assume that the number of 9s is 1 + omega, and 1 + omega does equal omega.

Both 0.999. . ., and 9.999. . . have omega 9s after the decimal point; the latter has an extra 9 at the beginning. Hence, "1 + omega". "Omega + 1" means you stick an extra item at the end, not at the beginning.

 

[slimshady2357]

That's a matter of mathematical philosophy.

No, it's a matter of definition

Adam

 

[whitefork]

You know, 30+ years later I still have nightmares about being back in high school.

But sometimes I'd like to be a math teacher there. With a big red pencil.