What is the largest number less than 1?

[Hurkyl]

I can represent real numbers with points on a line.
I can represent (some) real numbers algebraically.
I can represent (some) real numbers with combinations of elementary functions.

(assuming the axiom of choice) There are uncountable index sets I such that each real number can be written uniquely in the form

<br /> \sum_{\iota \in I} c_{\iota} \iota<br />

Where all but a finite number of the c_{\iota}'s are zero, and the nonzero ones are rational numbers.


The point is, the countability of the digits cannot be a fundamental property of real numbers because digits are not a fundamental property of real numbers.

 

[russ_watters]

Originally posted by Organic
...and there is no mathematical law that does not allow me to use the open interval idea on this range of digits, existing in these infinite levels of scales.

Therefore [1.0999...8) is a legal notation, which is the result of

[0.99999...9)
+
[0.09999...9)
=
[1.09999...8)

As stated earlier, the law there is the mathematical definition of infinity. You have it wrong and as a result your proof is wrong.

Incidentally, this is the Math forum. In it, we discuss the accepted version of how math works. If you wish to invent a new type of math, you need to post it in the Theory Development forum (I think you already have though...).

 

[Integral]

Originally posted by Hurkyl
I can represent real numbers with points on a line.
I can represent (some) real numbers algebraically.
I can represent (some) real numbers with combinations of elementary functions.

(assuming the axiom of choice) There are uncountable index sets I such that each real number can be written uniquely in the form

<br /> \sum_{\iota \in I} c_{\iota} \iota<br />

Where all but a finite number of the c_{\iota}'s are zero, and the nonzero ones are rational numbers.


The point is, the countability of the digits cannot be a fundamental property of real numbers because digits are not a fundamental property of real numbers.

I rest my case, you have proven my point. In your own words

<br /> \sum_{\iota \in I} c_{\iota} \iota<br />
Even you use the integers to index your representaion. That is all I am saying, it is fundamental that any representation can be indexed with the integers. You do it in your general representation, I do not care what method you use, a real number has at most a countable number of terms in the sum which represents it.

 

[Organic]

Hi russ_watters,

Please show me why I cannot use the idea of the open interval on a single number, represented by base^power method, for example:

[0.999...8)

[0.999...9)

[0.000...1)

and so on.

Be aware that #) is not the last digit but an unreachable limit exactly like in [0,1)

 

[suyver]

The open interval [a,b) consists of all numbers that are equal to or larger than a and also smaller than b. This open interval has no largest number. Only a smallest upper bound: b. Look up the difference between maximum and supremum!

 

[Hurkyl]

Originally posted by Integral
I rest my case, you have proven my point. In your own words

<br /> \sum_{\iota \in I} c_{\iota} \iota<br />
Even you use the integers to index your representaion. That is all I am saying, it is fundamental that any representation can be indexed with the integers. You do it in your general representation, I do not care what method you use, a real number has at most a countable number of terms in the sum which represents it.

None of the other ways I represenetd a real number even have a sum in them...

Well, you really want a sum with an uncountable number of terms? Fine, consider this sum of hyperreal numbers:

2 = \mathrm{st} \sum_{n=0}^{H} 2^{-n}

Where H is a transfinite hyperinteger. There are an uncountable number of terms, and the value of the sum is 2 - 2^{-H}. 2^{-H} is infinitessimal, so when you take the standard part, the answer is 2.

Alternatively, I believe that one can sensibly define (for some hypersequences) a hyperreal analogue of an infinite sum, so that:

2 = \sum_{n=0}^{\infty} 2^{-n}

(as a hyperreal sum) converges to 2. There is a term for each nonnegative hyperinteger, so there are an uncountable number of terms.

 

[Organic]

Hi suyver,

So the fundamental property of the interval's idea can be used on sequence of digits based on base^power method, where the right side of it has no smallest scale (represented by ...), but the 'digit' + ')' are the notation of the sequence's infimum http://mathworld.wolfram.com/Infimum.html

Therefore [0.000...1) is a legal mathematical notation,
and also [0.999...9).

 

[Hurkyl]

Organic: [1.999...8) is not accepted mathematical notation. You're going to have to explain what you mean by this, so you might as well just say the explanation rather than invent (yet again) new notation that nobody understands.

So the fundamental property of the interval's idea can be used on sequence of digits based on base^power method, where the right side of it has no smallest scale (represented by some digit), but the 'digit' + ')' are the notation of the sequence's infimum

What sequence?

For the record:

The infimum of the sequence
0.1, 0.01, 0.001, 0.0001, ...
is 0.

The supremum of the sequence
1.98, 1.998, 1.9998, 1.99998, ...
is 2. (it's infimum is 1.98)

The infimum of the sequence
0.08, 0.008, 0.0008, 0.00008, ...
is 0.

Therefore [0.000...1) is a legal mathematical notation.

Incorrect. That is notation you've invented and have not defined, so there's no way it can be considered "legal mathematical notation".

 

[Organic]

Hi Hurkyl,

First, i corrected my previous message, so please read it again.

The base^power method is infinitely many information cells upon infinitely many scales, where their periodic changes depends on base^power values.

each information cell includes n ordered digits, which represent some base value quantity, for example:

Base 2 notated by '0','1'
Base 3 notated by '0','1','2'
Base 4 notated by '0','1','2','3'

and so on.


any unique number which has aleph0 information's cells upon infinitely many scales, is asequence made of marked digits in aleph0 cells, but the important thing is not the represented digit on each cell, but the the cell's scale.

So by writing, for example [0.000...1) i say that there are infinitely many information's cells marked by 0, that interpolated forever to some cell that marked by 1.

 

[HallsofIvy]

posted by OrganicThe base^power method is infinitely many information cells upon infinitely many scales, where their periodic changes depends on base^power values.

each information cell includes n ordered digits, which represent some base value quantity, for example:

Once again, you are using words that are NOT standard mathematics and that you have NOT defined. There is no way anyone can guess what you mean.

Your assertion that "[0.000...1)" is "legal mathematical notation" is non-sense. The interval notation [a, b) always requires TWO numbers (or points) a and b- you are using only one so this is NOT interval notation and you haven't told us what it means.

By the way, Organic said, a while back,

Math is based on different consistant systems of axioms,which are propositions regarded as self-evidently true without proof.

That is definitely NOT true. I can't imagine any mathematician believing that BOTH "given a line and a point not on that line, there exist exactly one line through the given point parallel to the given line" and "given a line and a point not on that line, there exist more than one line through the given point parallel to the given line" are "self-evidently true"!

 

[Organic]

Hi HallsofIvy,

That is definitely NOT true

Please look at: http://mathworld.wolfram.com/Axiom.html


Also see an example of 2^aleph0 information's cells over different scales here:

http://www.geocities.com/complementarytheory/FPoint.pdf

Where [.000...1) is on the interpolation side of infinitely many cells, notated by '0' and approaches some cell, notated by '1'.

By using the idea of open interval on these cells we mean, that '1' can be distinguished from '0' forever, on infinitely many scales (which means: no cell is turned to zero size).


I think i have another idea based on the above.

Let us say that:

T = Math-theory

A = Its consistent axiomatic system

Therefore by writing [T,A) T depends on A but A does not depend on T, which maybe can give a new point of view on Godel's Incompleteness Theorems.


For example:

[0.99999...9)
+
[0.09999...9)
=
[1.09999...8)

The infinitely many '9' notations of the result, depends on adding 9) to 9) of the two digits of the infimum information's cells of the two added numbers.

By this example we can understand that any change in A, immediately changes T but not vise versa.

 

[Integral]

Hurykl,
After sleeping on it and getting out my copy of Royden, I realized that indeed the word "fundamental" cannot be used in reference to any representation of a Real number. While it is true that every point on the Real line has a decimal representation, this is a fact that requires proof using the fundamental theorems of the Real numbers and is therefore not fundamental of itself.

 

[BluE]

Okay, so is there a decimal place on a number line that can represent 0.9999... ?

 

[NateTG]

Originally posted by BluE
Okay, so is there a decimal place on a number line that can represent 0.9999... ?

There is no 'decimal' place on the number line. Decimal refers to representation in base 10.

The location on the number line that corresponds to the decimal 0.999999... is the same as the location that corresponds to the decimal 1, since they are the same real number. (Decimal representations of real numbers are not necessarily unique.)

 

[BluE]

Ah, okay. Sorry, and thanks. And since there is no "greatest number less than one" then that means there is no "greatest number less than x" when x is equal to any real number, right?

 

[HallsofIvy]

Yes, that is true. For x any real number, the set of all real number "less than x" is an open set and has no largest member.

 

[suyver]

Now we can wait for somebody to post that the number x - 0.0000...1 is the largest number in this set and that it is well-defined. And then we can go [zz)]. Honestly, I can't understand the near-infinite patience of some of the Senior Members here, but you have all my respect!

-Freek.

 

[uart]

Here's a question for Organic.

If x = 0.000...1 is valid number and is other than zero, then what is 10 times x equal to ?

Ok, I know that you'll say something like 10x = 0.000...10 where the "..." in the 10x expression represents one less zero than the original x.

So where does that logic get you? In the expression for x there are Infinity zeros (represented by the "...") whereas in the 10x expression there are (Infinity - 1) zeros - and they are different!.

So obviously you must believe that (Infinity - 1) is differnt than infinity. If that is so then just how do you define Infinity minus one ?

 

[russ_watters]

Originally posted by Organic
Hi russ_watters,

Please show me why...

I'm sorry, Organic, I can't help you here. I've already stated that you are arguing against the DEFINITION of infinity. As Halls said, what you posted there is not an accepted mathematical expression.

So now it is quite simply up to you to accept the definition or continue to be wrong.

There is a third choice of course - you could invent a new type of math to replace the entire existing structure. But that would take decades to do (if it could even be made to work - the definition you appear to be advocating is not self-consistent) and even then, its pretty unlikely that you'd be able to get the entire world to adopt your new math. Clearly that is what you are attempting to do - your website is full of things that don't fit with the way math actually works. But it'll be a long and uphill struggle. So it would probably be better to accept math as it is.

 

[HallsofIvy]

"Honestly, I can't understand the near-infinite patience of some of the Senior Members here, but you have all my respect!"

The problem with the inter-net is that the obvious remedy for people like this- beating about the head and shoulders with a two by four- is not applicable.

 

[Organic]

Hi russ_watters,

Please show me what have you found in my work, which is not self-consistent.

Thank you.


Organic

 

[russ_watters]

Originally posted by Organic
Hi russ_watters,

Please show me what have you found in my work, which is not self-consistent.

Thank you.


Organic

1.000...1 is not self consistent. We've been over this before though - there can't be an infinite amount of zeros before the 1 by the definition of infinity.

 

[jcsd]

Organic is the number 0.999...9 rational or irrational?

 

[BigRedDot]

Let us assume, for the sake of contradiction, that 0.999... and 1 are distinct real numbers. Then, since the rational numbers are dense in the reals, it follows that there must exist a rational number q such that 0.999... < q < 1.

Organic, would you please demostrate to us explicity a rational number which lies between 0.999... and 1?



I should also mention that formal developments of the real numbers do not rely on decimal expansions, per se. Indeed, Strichartz points out why such a development is often avoided: "However, it has two techincal drawbacks. The first is that the decimal expansion is not unique: 0.999... and 1.000 are the same number." One favorite method to define the real numbers is in terms of equivalence classes of Cauchy sequences. That is, real numbers are idetified with sequences of rational numbers that do not converge to rational numbers (yet satisfy the Cauchy criterion), but more than that, more than one sequence is identified with the same real number, since many sequences can belong to the same equivalence class. This is fine however, since members of equivalence classes are, well, equivalent.

In a certain sense, your position is tenable. People have certainly investigated non-standard analysis, and in particular Abraham Robinson founded a logically satisfactory basis for the real numbers using "bonafide" infintesimals. However, this kind of non-standard analysis is actually harder to justify (it requries us to assume more) and gains us nothing in what we can prove. I'm also pretty sure you didn't have any of this in mind however.

Now,I happen to believe that mathematics is by and large, if not entirely, a human creation, whose sole justification is pragmatic sanction. So the whole question is somewhat meaningleess. You are free to believe whatever you want about things you call numbers. The only important questions are: is your system useful and is it not obviously inconsistent. I don't know about your real number system, but the real number system where 0.999... does equal 1.000... (the one that is almost universally used and that all the analysis textbooks describe) has answered both those questions in the affirmative long, long ago.

 

[Lonewolf]

I can't believe this thread is still going...

 

[Hurkyl]

Even in the hyperreals, there is no largest number less than 1.

 

[R12]

Is "bumping" (posting on threads that haven't been posted on in a while) allowed here? Didn't find it in the rules.

Okay, so in theory, .999... = 1. This could mean that .888... = 1 as well. After all, it is only .111... different.

But wait, .111... supposedly equals 1, so .888... would be 1 (.999..., to clarify.) minus 1 (.111...) But wouldn't that be zero?

Also, he just asked how to represent the greatest number less than 1, not if .999... was a real number. So it seams as if

lim x
x-->1

was the answer, as PrudensOptimus said.

Forgive me if I'm wrong. Trying to wrap my head around this is hard, considering I am only 16.[/color]

 

[gb7nash]

Okay, so in theory, .999... = 1. This could mean that .888... = 1 as well.

Not quite. There's a multitude of reasons, but one reason .888... is not equal to 1 is that there exists a number between .888... and 1 that is not equal to .888... or 1.

This is a property of the real numbers that you'll learn in real analysis:

If two real numbers are distinct, there are an infinite number of real numbers between them. The contrapositive* of this statement is that if there are not infinite number of real numbers between two different numbers, then they are, in fact, the same number.

For .999... and 1, there's no number that's between them, so they're the same. There's a ton of other reasons too why this is true but this is one of them.

 

[Phrak]

Is "bumping" (posting on threads that haven't been posted on in a while) allowed here? Didn't find it in the rules.

Not that I know of, and it's a very good thread.

Okay, so in theory, .999... = 1. This could mean that .888... = 1 as well.

No. Within the system of numbers within the discussion of the year 2003, where .999...=1, then 0.888...=0.889.

 

[HallsofIvy]

Is "bumping" (posting on threads that haven't been posted on in a while) allowed here? Didn't find it in the rules.

Okay, so in theory, .999... = 1. This could mean that .888... = 1 as well. After all, it is only .111... different.

So? 0.1111... is not 0. 0.8888... is, by definition of the "decimal representation of the real numbers" the infinite sum
\frac{8}{10}+ \frac{8}{100}+ \frac{8}{1000}+ \cdot\cdot\cdot
which is the "geometric series"
\sum_{n=1}^\infty \frac{8}{10^n}= \sum_{n=1}^\infty 8(0.1^n}

There is a simple formula for the sum of a geometric series:
\sum_{n=0}^\infty ar^n= \frac{a}{1- r}

Here, the sum starts at n= 1 rather than n= 0 but that is easy to fix- with n= 0 8(.1^n)= 8(.1^0)= 8 so we only have to subtract off the missing first term, "8".
\sum_{n=1}^\infty 8(0.1^n)= \frac{8}{1- 0.1}- 8= \frac{8}{.9}- 8
= \frac{80}{9}- 8= \frac{80}{9}- \frac{72}{9}= \frac{8}{9}
which is NOT equal to 1.

But doing exactly the same thing with 0.9999... rather than 0.88888... gives
\sum_{n=1}^\infty 9(0.1)^n= \frac{9}{.9}- 9= \frac{90}{9}- 9= 10- 9= 1

But wait, .111... supposedly equals 1, so .888... would be 1 (.999..., to clarify.) minus 1 (.111...) But wouldn't that be zero?

No, you have simply misunderstood everything that was said here.

Also, he just asked how to represent the greatest number less than 1, not if .999... was a real number. So it seams as if

lim x
x-->1

was the answer, as PrudensOptimus said.

Forgive me if I'm wrong. Trying to wrap my head around this is hard, considering I am only 16.[/color]

As anyone who has taken basic Calculus or precalculus knows, \lim_{x\to 1} x= 1 so it is NOT "less than 1".

And, of course, 0.99999... is a real number- any number written in decimal notation like that is a real number.

 

[Dragonfall]

The largest number less than 1 is \frac{9,007,199,254,740,991}{9,007,199,254,740,992}. This is a fact.

 

[TylerH]

1 &gt; \frac{9,007,199,254,740,992}{9,007,199,254,740,993 } &gt; \frac{9,007,199,254,740,991}{9,007,199,254,740,992 }

 

[HallsofIvy]

I suspect that DragonFall meant that as a joke!

 

[TylerH]

I would hope so! :P I was just going along. I should have added a sarcastic "This is a fact." at the end to make it clearer, I guess.

 

[Dragonfall]

No computer scientists here? Tough crowd, tough crowd.

Seriously though, if I'm not mistaken, the number above should be the largest number less than one expressible in double precision. 1-2^{-53}

Or one minus "machine epsilon".

 

[HallsofIvy]

I was under the impression that "machine epsilon" could vary from one processor to another.

 

[TylerH]

Dragonfall's going by the IEEE standard for a double precision floating point. So, even though there are other machine epsilons, the one he used is the standard.

 

[Stocko]

A simple proof that 0.999... = 1 (for ... read "recurring". I'm new here and haven't figured out how to do symbols yet).

1/9 = 0.111...

Therefore 9 x 1/9 = 0.999...

But 9 x 1/9 = 1

So 0.999... = 1.

My small, modest and very late contribution to this remarkably long lived thread.

 

[camilus]

This question is rather simple, you just have to be more precise.

Analyze "What is the largest number less than 1?" in terms of set theory, you can't go wrong. Greatest number less than 1 in:

1. N? none because 1 is the smallest element in the set N.
2. Z? 0
3. R? none because the set of Reals is uncountable.

 

[n.karthick]

This question is rather simple, you just have to be more precise.

Analyze "What is the largest number less than 1?" in terms of set theory, you can't go wrong. Greatest number less than 1 in:

1. N? none because 1 is the smallest element in the set N.
2. Z? 0
3. R? none because the set of Reals is uncountable.

What about in the set of rational numbers Q which is countable

 

[Nebuchadnezza]

Sorry for not reading through all of the pages...

But can't you simply prove that 0.999... is equal to 1 in this matter?

0.999... can be expressed as a geometric series. As such

\frac{9}{{10}} + \frac{9}{{{{10}^2}}} + \frac{9}{{{{10}^3}}} + ... + \frac{9}{{{{10}^n}}}

An infinite geometric series is an infinite series whose successive terms have a common ratio.

The formula for the sum of a infinite geometric series is as follows: \frac{{{a_1}}}{{1 - k}} where k, is the ratio between the terms and a_1 is the first number in the sequence. Plugging in everything we have. We now get.

= 0.999...

= \frac{9}{{10}} + \frac{9}{{{{10}^2}}} + \frac{9}{{{{10}^3}}}+...+\frac{9}{{{{10}^n}}}

k = \frac{{{a_n}}}{{{a_{n - 1}}}} = \frac{{\frac{9}{{{{10}^n}}}}}{{\frac{9}{{{{10}^{n - 1}}}}}} = \frac{9}{{{{10}^n}}}:\frac{9}{{{{10}^{n - 1}}}} = \frac{9}{{{{10}^n}}} \cdot \frac{{{{10}^{n - 1}}}}{9} = \frac{{{{10}^{n - 1}}}}{{{{10}^n}}} = {10^{n - 1}} \cdot {10^{ - n}} = {10^{\left( {n - 1} \right) - n}} = {10^{ - 1}} = \frac{1}{{10}}

{a_1} = \frac{9}{{10}}

S = \frac{{{a_1}}}{{1 - k}} = \frac{{\frac{9}{{10}}}}{{1 - \frac{1}{{10}}}} = \frac{9}{{10}}:\frac{9}{{10}} = \frac{9}{{10}} \cdot \frac{{10}}{9} = 1

 

[camilus]

What about in the set of rational numbers Q which is countable

\mathbb{Q} is a densely ordered set, which means that for any x and y such that x < y, the exists a z in \mathbb{Q} where x < z < y. So there in no greatest number less than 1 because whatever number x you give me when y=1, I can always find a z.

 

[olivermsun]

\mathbb{Q} is a densely ordered set, which means that for any x and y such that x < y, the exists a z in \mathbb{Q} where x < z < y. So there in no greatest number less than 1 because whatever number x you give me when y=1, I can always find a z.

Just to connect this to the discussion above, also notice that every number in the series 0.9, 0.99, 0.999, 0.9999, ... is a rational number, so there are your arbitrarily close rationals right there.

 

[Thinker8921]

Could someone please clear a doubt I have with infinity being taken as a value.
I would like to start from the beginning to be clear. The sum of a geometric series is given by:
S=(rB-A)/r-1
Where S=sum of the series, r= ratio between terms, B=last term, A=first term.
Now B=A(r^N-1). So,
S=[{rA(r^N-1)}-A]/r-1
i.e, S=[{(r^N)-1}A]/r-1
Now in the case of the infinite geometric series 0.99… ,
N= infinity A= 0.9, r= 0.1
Then there is as the 1st term 0.9, 2nd term is 0.09, 3rd is 0.009 etc. Then the sum of this series will be 0.99...
Also in the numerator is 1-(r^N). This is instead of (r^N)-1. And in the denominator it is 1-r instead of r-1. This is because r is less than 1.
So finally there is:
S=[{1-(r^q)}A]/1-r
Where q represents infinity.
Now since there is infinity as the power of r, this is seen to become:
S=A/1-r
But the problem is it correct to take infinity as a value and approximate like this in the case of 0.99… ?
I thought infinity never had a definite value, that it was instead just an ever-rising concept. Could this 'problem' make the 0.99… equal 1?
How can 0=1/q, so q*0=1 be explained then? (q= infinity)
*I don't know how to write the equations with symbols yet. So sorry for the messy work.

 

[olivermsun]

The infinite sum (if it converges) is defined as the limit of the partial sums
\sum_{n=1}^\infty{a_n} = \lim_{N\rightarrow\infty} \sum_{n=1}^N{a_n}.

Thus in your expression you take the limit where the exponent goes to infinity and call that the value of the infinite series. This is different from stating that the exponent is actually infinity.

 

[Thinker8921]

Thanks for clearing that up. I got it now.

 

[dalcde]

If you use hyperreal numbers, it might make sense that the largest number is (1-epsilon), where epsilon is an infinitesimal. However, according to my understanding, there isn't a one unified infinitesimal. An infinitesimal can be smaller than another in some weird sense.

 

[Hurkyl]

If you use hyperreal numbers, it might make sense that the largest number is (1-epsilon), where epsilon is an infinitesimal. However, according to my understanding, there isn't a one unified infinitesimal. An infinitesimal can be smaller than another in some weird sense.

Hyperreals satisfy the same theorems (interpreted internally) that the reals do -- there is no largest hyperreal smaller than 1.

e.g. (1 - epsilon) < (1 - epsilon / 2) < 1