What is the largest number less than 1?

[Shahil]

hey organic

actually my correction was on the suggestion prudenceoptimus made!

 

[Organic]

When 1.000... and 0.999... are two representations of the same number then:

1.00... = 0.999...

0.100... = 0.0999...

0.0100... = 0.00999...

0.00100... = 0.000999...

0.000100... = 0.0000999...

0.0000100... = 0.00000999...

Therefore we can write:

0.100... + 0.0100... = 0.0999... + 0.00999...

0.0100... + 0.00100... = 0.00999... + 0.000999...

But this is not true because:

0.1100... not= 0.0999... + 0.00999... = 0.10999...8

0.01100... not= 0.00999... + 0.000999... = 0.010999...8

and so on ...

The unreachable digit is 8 and not 9, therefore 1.000... cannot be represented by 0.999...

 

[jcsd]

by defintion a number represented by an infinite number of decimal places does not have a last decimal place, i.e. it does not terminate.

 

[Organic]

This is not the last digit but the limit digit or the unreachable digit of 0.999...


Therefore 1.000... is not the limit of 0.999...

 

[suyver]

You don't give up, do you? Where did you get the idea for an 'unreachable' digit?

This limit stuff is not a process that is happening. It's not as if nature (or math) is contineously writing nines after your 0.999999... For all intents and purposes, that has already happened.

Look again at jcsd's very elegant proof:

For any nunber x where:

0 < x < 1

we know that:

0 < \sqrt{x} < 1

and

x < \sqrt{x}

Lets say that there is a largest number between 0 and 1, what is it's square root? If it greater than it's square root, it is greater than 1, if it is equal to it's square root it is equal to 1 and if it's square root is greater than itself then we have generated a number that is larger than the largest number less than 1 so it can't be the largest number less than 1!

As long as you can't tell what's wrong with that, maybe you should just accept that you're wrong. Because you are.

By the way: what's the use of your double-posting?

 

[uart]

This is not the last digit but the limit digit or the unreachable digit of 0.999...

Just one question Organic. Is that a potential unreachable digit, an actual unreachable digit or just a betoid unreachable digit ?

 

[Organic]

Hi suyver,

jcsd's very elegant proof is about the non-existence of the largest number smallest than 1

where 1 is the limit of [0.999...,1.000...).

0.99999...
+
0.09999...
=
1.09999...8

and we use the interval notations not between two numbers but among range of different scales, represented by some number, and in this case the number is [1.0999...8) and the infinitely many digits of 9 cannot exist in the above addition if the limit digit 8 does not exist.

 

[suyver]

Hi Organic,

Question: why are you posting this? All the math shown previously in this thread you either didn't understand or ignore, but I am sure that by now even you must realize that nobody believes that you are correct!

Why don't you just give it up and go do something fun. Maybe read a book?

 

[Organic]

Hi suyver,

Please read this:


http://pespmc1.vub.ac.be/INFINITY.html

 

[suyver]

You're responding in a thread on MATH and you're referring to a site on PHILOSOPHY. Don't you realize how absurd that is?

In mathematics and philosopy we find two concepts of infinity: potential infinity, which is the infinity of a process which never stops, and actual infinity which is supposed to be static and completed, so that it can be thought of as an object.

Only the second kind (in this definition) is meaningful in this debate. Again, using this I can prove that
\sum_{i=1}^\infty 9\cdot10^{-i} \; = \; 1
but I guess that you won't believe that either...

 

[Organic]

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

So the "true" of the axioms is out of the scope of any mathematical research, therefore can be examined only by PHILOSOPHY.

 

[suyver]

O, now I see!
Yes, you must be completely right, Organic.


I give up. Anybody else wants a go?

 

[Organic]

Also you wrote:

but I am sure that by now even you must realize that nobody believes that you are correct!

What is the connection between belief and Math ?

 

[uart]

Thanks for finally posting a link to explain that concept that you've been talking about Organic. So now I know what the concept of "potential" vs "actual" infinity is. I must say however that the disinction is much more a philosophical one then a mathematical one.


As for your arguments of "unreachable digits", such as 1.0999...8 , I can't see how this is any different from the usual old argument that 0.9999' can't be equal to 1 becuae it "clearly" differs by 0.000...1

It does not make any sense to talk about having an infinite number of zeros followed by a one, just the same as it doesn't make any sense to talk about an infinite number of nines followed by an eight.

If you want to set it up as a proper limit then that's fine, but the result you will get is the same as everyone has already proven.

0.999...8 = Lim as n->infinity 9 * (10^(-1) + 10^(-2) + ... 10^-n) + Lim as n->infinity 8*10^(-n-1) = 1 + 0

 

[Organic]

Hi uart,

You write:

0.999...8 = Lim as n->infinity 9 * (10^(-1) + 10^(-2) + ... 10^-n) + Lim as n->infinity 8*10^(-n-1) = 1 + 0

If you write 8*10^(-n-1) then you don't understand my argument, which is based on the idea
of the open interval http://mathworld.wolfram.com/Interval.html .

Instead of using it between two different numbers, i use it on one number, represented by base 10 (we can use any other base value instead).

Through this point of view i clime that:

0.99999...
+
0.09999...
=
1.09999...8

and we use the interval notations not between two numbers but among range of different scales, represented by some number, and in this case the number is [1.0999...8) and the infinitely many digits of 9 cannot exist in the result of the above addition if the limit of digit 8 does not exist.

 

[HallsofIvy]

If you write 8*10^(-n-1) then you don't understand my argument, which is based on the idea
of the open interval http://mathworld.wolfram.com/Interval.html .

Instead of using it between two different numbers, i use it on one number, represented by base 10 (we can use any other base value instead).

Then you are using it incorrectly. There is no open interval consisting of one number.

You are still using the phrase "among range of different scales" without defining "different scales". As long as you do not define your terms no one will understand what you are saying.

 

[Organic]

HallsofIvy

The Indian-Arabic number system, based on some base > 1 and powered by
0 to -n or n, is actually a fractal with -n or n finite levels or
-aleph0 | alaph0 infinite levels, where each level has a different scale, depends on base^power.

Any infinite fraction is some unique sequence of digits along these scales, 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)

 

[NateTG]

Originally posted by Magnus
How can you represent the largest # that is less than 1?

Under convetional ordering, in the reals or the rationals, this number does not exist.

Let's say that there is a real number x with that property.
Then we have x < (1-x)/2 + x <1, which contradicts the desired property of x.

However, if you choose a different ordering on the real numbers then there can be a number x such that x is the smallest number less than one.

 

[Hurkyl]

you don't understand my argument

How can he understand it if you don't understand it?

 

[Integral]

What HallsofIvy was trying to say is that this is an important feature of the decimal representation (or base-n representation) of the real numbers, not an important feature of real numbers themselves.

The countability of digits is certainly is a fundamental feature of the real numbers. I do not care how you represent it. The sum

\sum_{i=0}^\infty d_i b^{-i}

is the general representation of the fractional part of a real number, b is the base the d_i is a selection from a set of digits. For example
d_i \in \{0,1,2,3,4,5,6,7,8,9\} if b =10
or

d_i \in \{0,1\} if b =2
The one thing that is consistent across all representations is that the summation index i is countable. Thus any representation of a Real number can have only a countable number of d_i associated with it. Am I still not clear enough?

 

[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"!