View Full Version : What is the largest number less than 1?
How can you represent the largest # that is less than 1?
PrudensOptimus
Nov26-03, 07:32 PM
Originally posted by Magnus
How can you represent the largest # that is less than 1?
lim x
x-->1
Can you represent that in decimal form?
one_raven
Nov26-03, 07:37 PM
Wouldn't .9 (with a line over the 9, sorry, I haven't read the "How to post math functions" thread) be sufficient?
__
.9
The line over it means infinity.
So couldn't you also say .999... ?
Among the real numbers (or the rational numbers, or the irrational numbers) there is no largest number less than 1.
Among the integers, 0 is the largest number less than 1.
How can there be no largest # less than 1? that doesn't make any sense.
Why would you think there is a largest?
Here's a short proof there isn't a largest number less than 1:
Assume that there is a largest number less than 1. Let's represent this number by \inline{x}.
Now, consider the number \inline{y=(1+x)/2}
First, notice that \inline{y>x}:
y = \frac{1+x}{2} > \frac{x+x}{2} = x
Now, notice that \inline{y<1}:
y = \frac{1+x}{2} < \frac{1+1}{2} = 1
So consider carefully what we have proven:
If we assume there is a largest number less than 1, we can find a number less than 1, yet larger than the largest number less than 1.
That is a contradiction; our assumption that there is a largest number less than 1 must be false.
one_raven
Nov26-03, 08:15 PM
Hurkyl,
I understand your reasoning, and it does make complete sense...
I do have a question, however..
Is there a number greater than .999... and less than 1?
How would you represent that number?
Originally posted by one_raven
Hurkyl,
I understand your reasoning, and it does make complete sense...
I do have a question, however..
Is there a number greater than .999... and less than 1?
How would you represent that number?
The number \inline{0.99\overline{9}} is equal to one.
The limit \inline{\lim_{x \rightarrow 1} x} also equals one.
Hurkyl is correct; there is no largest real number less than one. No matter how many nines you put in a row, I can make a number with more. In the limit as the number of nines reaches infinity, the resulting quantity is one.
- Warren
I can't believe that there is no largest # less than 1. There has to be, in theory.
If you can say that .999... = 1 then how can you not in mathamatics represent the largest # that is less than 1?
the .999... = 1 rule only works because you NEVER reach the end of infinity.
I like sound laws and such, and I love math and physics... and at the same time am a hard core programmer at heart. Logic is key to me.
I could never write .999... = 1, I could say 1 = 1 and .999... = .999... but not 1 = .999...
I would say that .999... is the largest # less than 1 because as you go out there in decimal places whatever place your at you can just add a .----1 to that to achieve your value of 1... but yea, you'll never even approach infinity cause it extends forever. Damn you infinity!
[:((]
...I love math ... ...I would say that .999... is the largest # less than 1 because... You may love math, but you don't know what it is if you don't realize that math is based on rigor. If you want to hold on to your opinion about there being a largest number less than 1, then you must find a fault with Hurkyl's proof.
If you don't understand Hurkyl's proof, here's another thing to think about. Consider the mapping
y={1\over 1-x}
and for x, use the real number interval [0->1) (This means, all the real numbers from 0 to 1, but excluding 1.) This interval gets mapped to [1->infinity). You will see that there being no largest real number x less than 1 is analogous to saying there is no largest real number y. (I am assuming you agree that there is no largest real number.)
All the proofs so far have been perfectly adequate, but here's another one:
for a 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 sqaure 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!
HallsofIvy
Nov27-03, 09:00 AM
I like sound laws and such, and I love math and physics... and at the same time am a hard core programmer at heart. Logic is key to me.
I could never write .999... = 1, I could say 1 = 1 and .999... = .999... but not 1 = .999...
I would say that .999... is the largest # less than 1 because as you go out there in decimal places whatever place your at you can just add a .----1 to that to achieve your value of 1... but yea, you'll never even approach infinity cause it extends forever.
Logic is key? Your last sentence has no logic in it at all. You seem to be confusing logic with handwaving. In particular what is your DEFINITION of .999...?
theEVIL1
Nov27-03, 09:05 AM
Originally posted by Magnus
The line over it means infinity.
So couldn't you also say .999... ?
Sorry, the largest WHOLE number less than 1 is, by definition, 0.
.9999999999999999999999999999999999999999999999999 999999999999999999999999999999999999999 etc. infinitum is not a whole number
Originally posted by theEVIL1
Sorry, the largest WHOLE number less than 1 is, by definition, 0.
.9999999999999999999999999999999999999999999999999 999999999999999999999999999999999999999 etc. infinitum is not a whole number
0.9... recurring IS a whole number it is equal to 1.
theEVIL1
Nov27-03, 09:14 AM
Originally posted by jcsd
0.9... recurring IS a whole number it is equal to 1.
um.sure it is......how then can you explan that .999 infinitum will NEVER reach 1?
Must be that new math
Originally posted by theEVIL1
um.sure it is......how then can you explan that .999 infinitum will NEVER reach 1?
Must be that new math
No it's very old maths, most peope are taught the following sometime during their secondary eductaion:
x = 0.99999.....
=>
10x = 9.9999.... =>
10x -x = 9x = 9 =>
x = 1
I can't believe that there is no largest # less than 1. There has to be, in theory.
Why?
If you can say that .999... = 1 then how can you not in mathamatics represent the largest # that is less than 1?
I don't see the connection.
the .999... = 1 rule only works because you NEVER reach the end of infinity.
No, the rule works because it is a logical consequence of the definitions.
I could never write .999... = 1, I could say 1 = 1 and .999... = .999... but not 1 = .999...
Can you write \inline{1/2=2/4}? \inline{1} and \inline{0.\bar{9}} are two different representations of the same number, just like \inline{1/2} and \inline{2/4}.
I would say that .999... is the largest # less than 1 because as you go out there in decimal places whatever place your at you can just add a .----1 to that to achieve your value of 1... but yea, you'll never even approach infinity cause it extends forever. Damn you infinity!
This is where your problem lies; you are imagining \inline{0.\bar{9}} as some sort of process instead of as a number.
While it is certainly true you can get the value \inline{0.\bar{9}} through a process (such as taking the limit of \inline{0.9, 0.99, 0.999, \ldots}), \inline{0.\bar{9}} is a number. It does not change, it does not approach anything; it is simply a number.
HallsofIvy
Nov27-03, 05:52 PM
......how then can you explan that .999 infinitum will NEVER reach 1?
Must be that new math
No one needs to explain it- it's not true.
.999 "infinitum", by which I presume you mean the infinite sequence of 9s, is, by definition, the infinite series .9+ .09+ ...+ 9(.1)n+... which can be proven to be exactly equal to 1 (it's a very easy geometric series- you should have learned how to sum those in secondary school).
Hi Magnus,
Let us look at the opposite side of this problem.
x = 0
Can you find the smallest number, which is bigger then x?
Excellent point.
I do believe you guys.
The way I see .999... is basically.. a number that extends forever, it starts in the tens, goes to hundereds, thousands, etc. etc.. each time becoming closer and closer to 1. I see it becoming infinitely close to 1 as itself extends infinitely. Its just so hard to picture a # that starts off not as 1 become one just because it has no end.
IE: if you line it up.
1.000
0.999...
1 != 0
. = .
0 != 9
0 != 9
0 != 9
Thats my hangup.
I understand the big picture that infinity has no bounds. It's just mind boggling really.
Kina like, what's outside of the universe?
Integral
Dec1-03, 02:19 PM
Consider this.
First of all we must work with Real numbers, this is a matter of how the Real number system is defined. So what is a Real number? One important feature of Real numbers is the identity of each digit with an integer. To restate, there is a one to one coorespondenc beteen the digits of a Real Number and the integers. This where the construction which has "an infinite number of zeros followed by a 1" fails the test of a Real number, what integer cooresponds to that one?
In that sense the smallest Real number cannot be written specifically but we can write:
10^{-N} , \in \bold {N}
and claim that in general this is the form taken by the smallest Real number. Of course we cannot actually represtent the smallest Real number as there is no largest Integer.
Ok, here is the whole point of this post.
consder this
.1 + .999... = 1 + .0999...
.01 + .999... = 1 + .00999...
Can you see that if I have added a small number to .999... to get one plus a number consisting of a finite number of zeros followed by an infinte "tail" of 9s.
Now we can do this in general to get
10^{-N} + .999... = 1 + . (N-1 zeros)..999...
I can now write
10^{-N} + .999... > 1 \forall N \in \bold N
So no matter how small of a Real number I add to .999... I get 1 + a bit more.
There is only one number for which it is true, 1. Thus .999... =1
HallsofIvy
Dec1-03, 09:14 PM
One important feature of Real numbers is the identity of each digit with an integer.
No, that is not an important feature of the real numbers. That is a feature of the symbols used in one specific way of representing the real numbers. One could represent the real numbers in Roman numerals and they would still have the same properties. The properties of the real numbers are independent of how they are represented.
russ_watters
Dec1-03, 10:14 PM
Originally posted by Magnus
I can't believe that there is no largest # less than 1. There has to be, in theory. Here's another way to think of it without the mathematical proof (sorry guys). Can you think of a number greater than .9 and less than 1? Sure: .99. How about greater than that and less than one? Sure: .999. How about....
As you can see, you can keep doing that forever. Thats how infinity works. Its about the same as asking if there is any number greater than infinity: Nope.
Integral
Dec2-03, 12:11 AM
Originally posted by HallsofIvy
No, that is not an important feature of the real numbers. That is a feature of the symbols used in one specific way of representing the real numbers. One could represent the real numbers in Roman numerals and they would still have the same properties. The properties of the real numbers are independent of how they are represented.
It certianly is a key feature, I do not care what number system you use there is only a Countable number of digits in a Real number. That is the key feature, it is indeed indepentent of the representation but the countablility of the digits is essential and that is what I am refering to.
Edit;
I thing I am seeing the point of confusion. Perhaps you thought I was speaking of the actual digits, ie 0,1,2,3,4,5,6,7,8,9. That was not my meaning at all.
The fractional part of every Real number can be represented as
\sum_{n=0}^\infty d_n 10^{-n}
Were the d_n \in \{0,1,2,3,4,5,6,7,8,9\}
The correspondence with the integers I am speaking of is the index n. This is of course a base 10 Real number if you choose to represent the number in a different base the number rasied to a power will change as will the set of basic digits.
I am not sure that Roman numerals dealt well with fractions! Seems it was the Arabic numerals and the place value system which allowed this development.
Hi russ_watters,
There are at least two kinds of infinities: actual infitiny, potential infinity.
As mauch as i know, Math language uses only potential infinity.
More about these kinds of infinity you can find here:
http://www.geocities.com/complementarytheory/RiemannsLimits.pdf
Organic
correction on the limit example:
wouldn't it be:
lim x
x --> 1 (x approaches 1 from the negative side!!)
Hi Shahil,
If you are talking about my pdf on actual and potential infinities, then (-oo,0) is the mirror image of (0,oo), so we need only one of them for the example.
One important feature of Real numbers is the identity of each digit with an integer. To restate, there is a one to one coorespondenc beteen the digits of a Real Number and the integers.
This cannot be an important feature of the real numbers because you cannot even write the statement (or even the spirit of the statement) "there is a one to one coorespondenc beteen the digits of a Real Number and the integers" in the basic theory of real numbers.
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.
There are at least two kinds of infinities: actual infitiny, potential infinity.
As mauch as i know, Math language uses only potential infinity.
It would be great if anyone knew what you meant by actual or potential infinity.
hey organic
actually my correction was on the suggestion prudenceoptimus made!
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...
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.
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...
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?
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 ?
:-P
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.
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 realise that nobody believes that you are correct!
Why don't you just give it up and go do something fun. Maybe read a book (http://www.amazon.com/exec/obidos/ASIN/0764553259/qid=1070378082/sr=2-3/ref=sr_2_3/104-0224823-8620762)?
Hi suyver,
Please read this:
http://pespmc1.vub.ac.be/INFINITY.html
You're responding in a thread on MATH and you're referring to a site on PHILOSOPHY. Don't you realise how absurd that is? [6)]
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...
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.
O, now I see!
Yes, you must be completely right, Organic.
I give up. Anybody else wants a go?
Also you wrote:
but I am sure that by now even you must realise that nobody believes that you are correct!
What is the connection between belief and Math ?
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' cant 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
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
Dec2-03, 11:26 AM
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.
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)
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.
you don't understand my argument
How can he understand it if you don't understand it?
Integral
Dec2-03, 05:59 PM
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 di 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 di associated with it. Am I still not clear enough?
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
\sum_{\iota \in I} c_{\iota} \iota
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
Dec2-03, 11:00 PM
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
Dec3-03, 02:57 AM
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
\sum_{\iota \in I} c_{\iota} \iota
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
\sum_{\iota \in I} c_{\iota} \iota
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.
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)
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!
Originally posted by Integral
I rest my case, you have proven my point. In your own words
\sum_{\iota \in I} c_{\iota} \iota
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.
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).
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".
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
Dec3-03, 06:48 AM
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"!
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
Dec3-03, 04:40 PM
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.
Okay, so is there a decimal place on a number line that can represent 0.9999... ?
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.)
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
Dec3-03, 08:35 PM
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.
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. [6)] [g)] 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.
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
Dec4-03, 11:12 AM
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
Dec5-03, 07:01 AM
"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.
Hi russ_watters,
Please show me what have you found in my work, which is not self-consistent.
Thank you.
Organic
russ_watters
Dec5-03, 11:25 PM
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.
Organic is the number 0.999....9 rational or irrational?
BigRedDot
Dec7-03, 02:30 AM
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
Dec7-03, 07:31 AM
I can't believe this thread is still going...
Even in the hyperreals, there is no largest number less than 1.
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.
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.
[color=#660066]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
Jan4-11, 06:01 AM
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.
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
Jan4-11, 04:57 PM
The largest number less than 1 is \frac{9,007,199,254,740,991}{9,007,199,254,740,992 }. This is a fact.
1 > \frac{9,007,199,254,740,992}{9,007,199,254,740,993 } > \frac{9,007,199,254,740,991}{9,007,199,254,740,992 }
HallsofIvy
Jan4-11, 07:04 PM
I suspect that DragonFall meant that as a joke!
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
Jan5-11, 03:33 PM
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
Jan6-11, 04:55 PM
I was under the impression that "machine epsilon" could vary from one processor to another.
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.
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
Apr20-11, 12:25 AM
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 cant 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
Apr20-11, 04:04 AM
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 cant 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
Apr20-11, 04:56 AM
Sorry for not reading through all of the pages...
But cant 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
Apr20-11, 09:34 AM
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
Apr20-11, 01:21 PM
\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
Apr20-11, 08:38 PM
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
Apr21-11, 01:01 AM
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
Apr21-11, 06:53 PM
Thanks for clearing that up. I got it now.
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.
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
vBulletin® v3.8.7, Copyright ©2000-2012, vBulletin Solutions, Inc.