10282003, 01:29 PM  #1 (permalink) 
Insane
Location: West Virginia

A popular "discussion" (math)...
Simple question: Does .99999... (repeating) = 1?
This seems to be a popular discussion, at least back in the day on mIRC and some other forums I used to visit... I'm not asking if they are "very close", that's not what "=" means Post Away!
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 Artsemis ~~~~~~~~~~~~~~~~~~~~ There are two keys to being the best: 1.) Never tell everything you know 
10282003, 01:50 PM  #2 (permalink) 
Sky Piercer
Location: Ireland

Depends who you ask:
Engineer: Hell! 0.9 = 1 as far as I'm concerned! Physicist: The difference is negible. Mathematician: As you continue the expansion, the term approaches 1. The difference between them is 1/(infinity).
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10282003, 02:05 PM  #3 (permalink) 
Insane
Location: West Virginia

Well, maybe I should input my opinion, though there might be a flaw I'm overlooking...
1/3 = .3333... 2/3 = .6666... 1/3 + 2/3 = 3/3 = 1 .3333... + .6666... = .9999...
__________________
 Artsemis ~~~~~~~~~~~~~~~~~~~~ There are two keys to being the best: 1.) Never tell everything you know 
10282003, 03:10 PM  #6 (permalink) 
Rookie
Location: Oxford, UK

I think Artsemis has it... noone would argue that 1/3 + 2/3 is 1, so 0.33333... + 0.6666..... (=0.99999....) must be too?
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I can't understand why people are frightened of new ideas. I'm frightened of the old ones.  John Cage (1912  1992) 
10282003, 03:22 PM  #7 (permalink) 
Sky Piercer
Location: Ireland

A good explaination:
http://mathforum.org/dr.math/faq/faq.0.9999.html also check out the links at the bottom of the page.
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10282003, 06:48 PM  #9 (permalink) 
Insane
Location: West Virginia

Ignore the post above me  forgot to log off my old name on this computer
Anyways  thanks for that link. The explination there is very good
__________________
 Artsemis ~~~~~~~~~~~~~~~~~~~~ There are two keys to being the best: 1.) Never tell everything you know 
10292003, 12:46 AM  #10 (permalink)  
Location: Waterloo, Ontario

Quote:
Quote:
All mathematicians agree that 0.9999... = 1 and the link that CSFilm provided explains it quite well. I've said it before and I'll say it again. Infinity is not a number! 

10292003, 09:38 AM  #12 (permalink)  
Riiiiight........

Quote:
you can pretty much do this for any decimal expansion that is periodic. A corollary is that any decimal expansion that is periodic can be expressed as a rational number ( fractions for the rest of us... ) like x = 0.73737373737373...... with period n, in this case 2, (10^n)x = 73.7373737373..... (10^n)x  x = 73 x = 73/(1001) x = 73/99 

11032003, 05:21 PM  #14 (permalink) 
Crazy
Location: Switzerland

Agreeing with KnifeMissle on the fact that 0.99999...=1, by definition of the left term.
Disagreeing with him on the fact that infinity is not a number. In fact, there are many infinities, and the subject of "cardinal numbers" is concerned with the task of adding and multiplying infinities. (See that old post on the CantorBernstein Theorem).
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Didn't remember how intense love could be... Thank you B. 
11032003, 07:01 PM  #15 (permalink) 
Location: Waterloo, Ontario

Grothendieck, it depends on what you call a number. When most people think about numbers, they think of real numbers, and infinity is certainly not one of those! It's not even a complex number...
I also hesitate to call set cardinalities numbers since I don't know of any operations between them. You seem to be well educated on the subject so perahps you know of some? 
11042003, 01:30 AM  #16 (permalink) 
Crazy
Location: Vancouver, Canada

I love the imperfections of math...
And, yes, .999... = 1, according to the rules applied above... Too itred right now, but I'll show you guys how to use this logic to prove that 1 = 0.
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You know that song that goes like... 
11042003, 07:14 AM  #18 (permalink)  
Sky Piercer
Location: Ireland

Quote:
Don't forget that dividing by zero is not a legal arithmetic operation!
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11042003, 08:34 AM  #19 (permalink)  
Crazy
Location: Switzerland

Quote:
There are three fundamental operations on sets: Disjoint Union (corresponding to addition), Direct Product (corresponding to multiplication) and Power Set (corresponding to exponentiation. Here's how it goes: Given two sets A and B, say that A <= B if there is an injection from A to B (or equivalently by CantorBernstein, a surjection from B to A). We say that A > B if there is no such injection. We say that A=B if there is a bijection. Note that the use of = is somewhat misleading, if A=B in this sense, then they are not necessarily the same set, they are only equal in cardinality, put otherwise, they are the same cardinal number! The number A+B is simply the (disjoint) union of A and B. The number AB is the product of A and B, and A^B is formally defined as the set of functions from B to A. In particular, 2^B is the set of functions from B to a twoelement set, and we can identify 2^B with the set of subsets of B (check this!). Now we can represent any positive integer n by a set with n elements (e.g. 0 is the empty set), and arithmetic with positive integers is part of the above mentioned cardinal arithmetic. Now for some fun: if A is infinite, and B is any set, then A+B=max(A,B). For example, if A is infinite and B is finite, then A+B=A. We always have A<2^A by an adaption of Cantors diagonal trick. 0 times A is always 0. What you can see is that we don't get a group or ring, but we do get some operations. Here's some food for thought: Given an infinite set A, is there a set B such that A < B < 2^A (with strict inequalities?). The continuum hypothesis says that for A=(all natural numbers) there is no such set B. By the way, we can prove that 2^A = (all real numbers). For general infinite A, this is what is called the generalized continuum hypothesis (GCH). It turns out that this question is independent of the usual set theoretical axioms we use to construct mathematics.
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Didn't remember how intense love could be... Thank you B. 

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