What is the largest number less than 1?

[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