why does 1/3=.333333333
#1
Posted 22 February 2005 - 10:29 AM
My grubby halo, a vapour trail in the empty air.
Across the clouds I see my shadow fly,
Out of the corner of my watering eye.
A dream unthreatened by the morning light,
Could blow this soul right through the roof of the night.
There's no sensation to compare with this,
Suspended animation, a state of bliss.
Can't keep my eyes from the circling skies,
Tongue-tied and twisted, just an earthbound misfit, I.
#2
Posted 22 February 2005 - 10:36 AM
The Wrath, on Feb 22 2005, 10:29 AM, said:
<{POST_SNAPBACK}>
1/3 doesn't really = .3333 repeating. 1/3 is an irrational number, meaning that it can't be accurately represented as a decimal. Anytime you represent 1/3 as a decimal it is only an approximation, no matter how many 3s you put on the end of it
Edit: I made a mistake, 1/3 is not an irrational number. An irrational number is a number that cannot be represented as an exact ratio of two integers. 1/3 is just a ratio that cannot be represented as a terminal decimal.
This post has been edited by Bryan: 22 February 2005 - 10:41 AM
#3
Posted 22 February 2005 - 10:46 AM
When you multiply .33333~ by 3 you get .9999~, an infinitely recurring sequence of 9s, which is equal to 1.
The argument given in the other thread to show that .999~ = 1 is informal and not particularly rigorous, but you can prove it properly using elementary calculus.
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1/3 is a rational number. Any number of the form p/q where p and q are integers is rational. The difference between rational and irrational numbers is that the decimal expansion of a rational number is periodic, ie your infinite sequence will repeat a certain pattern over and over again.
For instance, 2/7 = 0.285714285714285714, which is just '285714' repeating to infinity, and 1/6 = 1.66666~ where the 6 repeats to infinity. However an irrational number such as root(2) or pi will have a non-periodic decimal expansion, ie there is no sequence that repeats over and over again (for instance, the number represented by the decimal expansion 0.101001000100001000001etc, where the number of zeros inbetween the 1s gets larger and larger, will be irrational since there is nothing that repeats).
This post has been edited by Hal: 22 February 2005 - 10:50 AM
#4
Posted 22 February 2005 - 10:57 AM
Hal, on Feb 22 2005, 10:46 AM, said:
The argument in the other thread is based on an algebraic error, not calculus. You can never say that .999999~ = 1, it approaches 1 but never quite equals it.
#5
Posted 22 February 2005 - 10:59 AM
Bryan, on Feb 22 2005, 11:57 AM, said:
<{POST_SNAPBACK}>
Yes you can, at least in the standard construction of the real numbers where you don't have infinitesimals. The real numbers are dense, which means that in between any 2 real numbers is a third. If 0.999~ is not equal to 1, then which number is between then?
This post has been edited by Hal: 22 February 2005 - 10:59 AM
#6
Posted 22 February 2005 - 11:03 AM
0.999~ is the infinite series 0.9 + 0.09 + 0.009 + ... which is a geometric progression with initial term 0.9 and ratio (1/10). From calculus, the sum of an infinite series is (initial term)/(1 - ratio). Substituting in here we get (0.9)/(1-(0.1)) = 0.9/0.9 = 1.
#7
Posted 22 February 2005 - 11:05 AM
Hal, on Feb 22 2005, 10:59 AM, said:
<{POST_SNAPBACK}>
There is no number between them, .999~ is the largest number less than 1.
0.999~ doesn't actually exist in reality.
#8
Posted 22 February 2005 - 11:06 AM
Bryan, on Feb 22 2005, 12:05 PM, said:
<{POST_SNAPBACK}>
There is no largest number less than 1, as I said the real numbers are infinitely dense. I'm not sure what you mean by 0.99~ doesnt exist in reality.
It makes more sense (to me at least) if you think of 0.999~ as being an infinite series (0.9 + 0.09 + 0.009 + ...), which is how it's defined mathematically, rather than being a big long list of 9s
This post has been edited by Hal: 22 February 2005 - 11:08 AM
#9
Posted 23 February 2005 - 03:04 PM
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...
There is no number between them, .999~ is the largest number less than 1.
0.999~ doesn't actually exist in reality.
I'm confused by these statements.
Is .999~ a number, or not? If so, how do you define this number, what is it?
What do you think it means for a number to exist in reality?
What does it mean for a number to 'approach' another number?
You know that there is no greatest number less than one, right? If you think you have such a number x, then the (1 + x)/2 is still less than one, but greater than x.
If 0.999~ doesn't exist in reality, what do you think about any other irrational number like pi or the square root of two? Since we can only give finite decimal approximations of these numbers, do they also not exist in reality?
#10
Posted 23 February 2005 - 03:34 PM
#11
Posted 23 February 2005 - 05:10 PM
Nate T., on Feb 23 2005, 03:04 PM, said:
I'm confused by these statements.
Is .999~ a number, or not? If so, how do you define this number, what is it?
I'm in the minority here, but I still hold that .999~ is not 1. Abstractly speaking it is the largest number that is less than 1. The difference between .999~ and one is inifinitely small, but there is a difference.
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A number can't really approach a number. When I said the .999~ approaches 1, I meant that the infinite sum 0.9 + 0.09 + 0.009 + ... approaches 1. It never actually equals one because the sum never ends, but everytime you carry out another iteration of it, you get a number that is closer to 1.
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<{POST_SNAPBACK}>
Obviously pi and the sqrt(2) exist, They just can't be 100% accurately represented using decimal numbers. That has nothing to do with their nature, it has to do with the nature of the things they describe.
#12
Posted 23 February 2005 - 06:11 PM
0.999~ doesn't refer to anything in reality? You'd better be careful about statements like that. Does -1 refer to anything in reality? How about i = sqrt(-1)? If these numbers are concepts of method, then why isn't 0.999~? After all, just as negative and complex numbers have many applications, so do summing series and limits (in fact, they make all of calculus possible, which is the only thing that makes modern physics possible).
Also, I wasn't able to get a straight answer out of you about the status of 0.999~. Is it a real number? An 'abstract' number (abstract in what sense)? One minus an infinitesimal (what is an infinitesimal)? The greatest number less than one (which can't happen, btw)? A sequence of rational numbers? You seem to be using all of these definitions interchangably.
I contend that the only way to make any sense whatsoever of the expression 0.999~ is for it to represent the limit of the sequence (0.9, 0.99, 0.999, ...). Since one can show that this sequence converges to 1, it follows that 0.999~ = 1. I agree that the sequence never does reach one--that's fine. But the limit of the sequence is what 0.999~ denotes, and that is equal to 1.
#13
Posted 23 February 2005 - 07:42 PM
Nate T., on Feb 23 2005, 06:11 PM, said:
0.999~ doesn't refer to anything in reality? You'd better be careful about statements like that. Does -1 refer to anything in reality? How about i = sqrt(-1)? If these numbers are concepts of method, then why isn't 0.999~? After all, just as negative and complex numbers have many applications, so do summing series and limits (in fact, they make all of calculus possible, which is the only thing that makes modern physics possible).
No, I don't see the particular number .999~, directly referring to anything in reality. Infinity only exists as an abstract concept; therefore infinite series only exist as concepts. Let's be clear about mathematics. In ITOE, Ayn Rand aptly defines mathematics as "the science of measurement". Calculus isn't made possible by infinite series and limits, it is made possible by reality. It was developed as a manner of describing and measuring things in reality. Infinite sums and limits are used to describe calculus concepts, not the other way around.
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I'm not using the definitions interchangeably, its just that .999~ fits all these definitions.
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<{POST_SNAPBACK}>
The limit of the infinite sum 0.9 + 0.09 + 0.009 + ... does indeed equal 1. But this does not mean that the actual number .999~ is equal to 1. Just because the limit of something equals 1 does not mean that the something is 1.
0.999~ can be approximated by 1 because its pretty damn close, but its still not exactly 1. Nothing is exactly 1 but 1.
#14
Posted 23 February 2005 - 10:59 PM
I appreciate your candor in trying to defend the idea that there is no metaphysical infinity. But don't misunderstand what it means-- in mathematics it's perfectly valid to talk about infinite sets, infinite series, etc., because mathematics is a discipline involving concepts of method.
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She also mentions that mathematics is a concept of method, which includes the valid concept of infinity to help aid the ultimate goal of mathematics to measure things.
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...? This isn't how real numbers are defined. It merely steals the concept of 'complex numbers', which are defined in terms of the real numbers.
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The reason I brought up that expression was because it helped to prove that there is no greatest number less than one. The argument goes as follows: Suppose there is a largest value less than one. Call it 'x'. Then (x + 1)/2 is strictly greater than x, but strictly less than one. Therefore, by the definition of x, it follows that (1 + x)/2 <= x. But then x < (x + 1)/2 <= x. This means that x is something other than itself. But A is A, so contradictions do not exist; one of our premises is wrong. The wrong premise is the beginning one, that x is the largest number strictly less than one-- no such number can exist.
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If you open ItOE to the section "Definitions", Rand defines a definition to be "a statement that identifies the nature of the units subsumed under a concept." Moreover, it is the statement which condenses and explains all the other knowledge of the concept in question under the context of one's current knowledge. If you try to do this for the various 'definitions' for 0.999~ you present above, you'll find little more than an anti-concept, since most of those definitions don't even describe the same objects and are contradictory.
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What else is the 'actual number' .999~ besides the limit of the sequence (0.9, 0.99, 0.999, ...)? I agree that just because the limit of a sequence is one does not mean the sequence is one. Sequences aren't numbers, and I never claimed that they were.
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How close is it? Blank out.
#15
Posted 24 February 2005 - 03:24 AM
A is 0.9, 0.99, 0.999, 0.9999, etc.
B is 1.1, 1.01, 1.001, etc.
C is 0.9, 1.1, 0.99, 1.01, 0.999, 1.001, etc.
We define limA, limB, limC to be the limits of the sequences.
If limA differs from 1 then so does limB (and that limA does not equal limB). If this is true then you have to say that limC is undefined since it consists of two infinite subsequences converging to two different values.
If however you say that limC = 1, then you have to say any infinite subsequence of C is 1 and, in particular, that limA = limB = 1
Do you really want to go so far as to say limC is undefined?
This post has been edited by punk: 24 February 2005 - 03:24 AM
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The queen is their slave
#16
Posted 24 February 2005 - 09:28 AM
Bryan, on Feb 23 2005, 08:42 PM, said:
What do you mean by calculus is "made possible by reality"? I suppose this is trivially true in the sense that if nothing existed then noone would have been able to come up with calculus, but calculus can be (and usually is) defined in a purely formal manner that makes no mention whatsoever of physical or geometric objects, and it still works perfectly well. Calculus is useful because of certain aspects of reality, yes.
This post has been edited by Hal: 24 February 2005 - 09:31 AM
#17
Posted 24 February 2005 - 09:51 AM
Nate T., on Feb 23 2005, 10:59 PM, said:
I appreciate your candor in trying to defend the idea that there is no
metaphysical infinity. But don't misunderstand what it means-- in
mathematics it's perfectly valid to talk about infinite sets, infinite
series, etc., because mathematics is a discipline involving concepts
of method.
I understand that it is valid to discuss infinity; my point of this
whole thread is that you can't use the concept of infinity to set a
number equal to a number that it is not equal to.
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the infinite series 0.999~ less real or less valid than the concept of
1?
related to things in reality.
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concept of 'complex numbers', which are defined in terms of the real
numbers.
Do you care to give a definition? Mine was ok, just kind of sloppy.
Here are some other's from
google:
" any number that is not imaginary Example:"1.23156..., 5, 8/6, e,
square root (3)"
" A number with an integer and a fractional part. The primitive types
double and float are used to represent real numbers. "
" A real number is one-dimensional and can be placed somewhere on the
number line. The set of real numbers includes all rational and all
irrational numbers. "
" Any finite or infinite decimal. Any rational or irrational number. "
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"infinitely small amount away", which is what an infinitesimal is.
Care to elaborate on what an "infinitely small amount" is?
infinitely small. The only limit on the smallness is that it can never be zero; it always has to be something.
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that there is no greatest number less than one. The argument
goes as follows: Suppose there is a largest value less than one.
Call it 'x'. Then (x + 1)/2 is strictly greater than x, but strictly
less than one. Therefore, by the definition of x, it follows that (1
+ x)/2 <= x. But then x < (x + 1)/2 <= x. This means that x is
something other than itself. But A is A, so contradictions do not
exist; one of our premises is wrong. The wrong premise is the
beginning one, that x is the largest number strictly less than one--
no such number can exist.
Let’s do this with numbers:
y = (x+1)/2, where x is .99: y = .995
Now substitute .995 in for x:
y = (.995+1)/2: y = .9975
Keep repeating over and over, replacing x each time with the result for y the time before. What is the “final” answer after and infinite amount of iterations? .999~
This is why .999~ is the largest number that is less than 1.
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sequence (0.9, 0.99, 0.999, ...)?
It is the result of the infinite sum of the sequence (.9, .99, .999, .9999, …).
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<{POST_SNAPBACK}>
It is an infinitely small amount less.
Edit: Sorry about the text wrapping of this post, it turns out the going from the message board to gmail to word back to the message board messes it up
This post has been edited by Bryan: 24 February 2005 - 09:56 AM
#18
Posted 24 February 2005 - 10:19 AM
Hal, on Feb 24 2005, 09:28 AM, said:
Declaring Miss Rand's definition to be bad depends on your definition of measurement. If you define measurement as the identification of relationships, mathematics certainly qualifies as the science of measurement.
As far as complex analysis goes, it can be directly used to measure the magnitude and phases of voltages and currents in electrical systems.
I have no experience in number theory but I googled it and it said "A branch of mathematics that investigates the relationships and properties of numbers." That being said, you could describe number theory as the measurement of numbers themselves.
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<{POST_SNAPBACK}>
I was trying to make a distinction between calculus being used as a mathematical tool to describe reality as opposed to the mathematical tools that are used to describe calculus (i.e. limits and infinite sums).
#19
Posted 24 February 2005 - 10:35 AM
For f(x) = (x+1)/2 as x gets arbitrarily close to 1 the limit = 1. The key word is arbitrarily which is usually taken to mean infintesimally close. But the infintesimal or the arbitrary doesn't exist in reality. And that's why .999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
99999999999999999999999999999999999999999999999999999999999999999999999~ = 1. A is A
In matters of style, swim with the current; in matters of principle, stand like a rock.--Thomas Jefferson
#20
Posted 24 February 2005 - 10:41 AM
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whole thread is that you can't use the concept of infinity to set a
number equal to a number that it is not equal to.
Suppose that 0.999~ < 1. Then e = 1 - 0.999~ is a positive number. Hence, there exists an integer N such that 1/10^N < e. Now it should be clear that 0.99...9 (N + 1 9's) < 0.999~. But this implies that e = 1 - 0.999~ < 1 - 0.99...9 (N + 1 9's) < 1/10^N <= e. Once again, we get e is not itself, which is a contradiction.
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related to things in reality.
Okay, that's a good step, and you're right-- the concept 1 is far less abstract than 0.999~.
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The way the real numbers are constructed from the rational are as equivalence classes of Cauchy sequences of rational numbers, equipped with termwise addition and multiplication. It's a result in Analysis to show that this structure is an algebraic field and has the familiar order properties of the real line, as well as being complete.
Essentially, this means that every real number can always be identified with a sequence of rational numbers (a, b, c, ...). In our case, each of the sequences are equivalent to the real number 1:
(1, 1, 1, ...)
(0.9, 0.99, 0.999, ...)
(1.1, 1.01, 1.001, ...)
(0.9, 1.1, 0.99, 1.01, ...)
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infinitely small.
You're right, to this extent: if you accept axiomatically the existence of an infinitely large number, you can build a system in which infinitesimal numbers exist, and in such a system the series 0.9 +0.09 + ... will converge to a number less than one by an infintiesimal. You can even say which infintiesimal.
But if you'd like to stay in the real number system, where there are no 'infintiely large numbers', just sequences of numbers increasing without bound, you cannot have a smallest positive number. If you think you have such a number, then half of it is always both positive and strictly smaller, which is a contradiction.
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y = (x+1)/2, where x is .99: y = .995
Now substitute .995 in for x:
y = (.995+1)/2: y = .9975
Keep repeating over and over, replacing x each time with the result for y the time before. What is the “final” answer after and infinite amount of iterations? .999~
Okay, well let's work with this. Let's define f(x) = (1 + x)/2, as you've done above, and consider the sequence of numbers
f(x)
f(f(x))
f(f(f(x)))
...
You claim that the "final answer" a = f(f(f(...))) is equal to 0.999~, right? Okay, well clearly by the definition of a, we have f(a) = a. Therefore, (1 + a)/2 = a, whence 1 + a = 2a, so a = 1. QED.
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Such a number doesn't exist. I gave you the proof of that already. If you can find an incorrect step in that proof, please show me.

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