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Old 10-28-2003, 01:29 PM   #1 (permalink)
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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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Old 10-28-2003, 01:50 PM   #2 (permalink)
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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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Old 10-28-2003, 02:05 PM   #3 (permalink)
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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...
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Old 10-28-2003, 02:07 PM   #4 (permalink)
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CSflim is exactly correct, although as an engineer I'd be more comfortable at 0.95.
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Old 10-28-2003, 02:37 PM   #5 (permalink)
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Quote:
Originally posted by redlemon
CSflim is exactly correct, although as an engineer I'd be more comfortable at 0.95.
Lol. .9999.... == .9999.... in my opinion
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Old 10-28-2003, 03:10 PM   #6 (permalink)
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I think Artsemis has it... no-one would argue that 1/3 + 2/3 is 1, so 0.33333... + 0.6666..... (=0.99999....) must be too?
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Old 10-28-2003, 03:22 PM   #7 (permalink)
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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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Old 10-28-2003, 06:33 PM   #8 (permalink)
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Old 10-28-2003, 06:48 PM   #9 (permalink)
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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
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Old 10-29-2003, 12:46 AM   #10 (permalink)
 
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Quote:
Originally posted by CSflim
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).
Quote:
Originally posted by redlemon
CSflim is exactly correct, although as an engineer I'd be more comfortable at 0.95.
Just so everyone knows, CSFilm is not exactly correct.
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!
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Old 10-29-2003, 05:52 AM   #11 (permalink)
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Let x = 0.999999....

10x = 9.9999999.........

10x - x = 9.99999........ - 0.999999....

10x -x = 9

9x = 9

x = 1
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Old 10-29-2003, 09:38 AM   #12 (permalink)
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Quote:
Originally posted by Jakejake
Let x = 0.999999....

10x = 9.9999999.........

10x - x = 9.99999........ - 0.999999....

10x -x = 9

9x = 9

x = 1
nice one...
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/(100-1)
x = 73/99
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Old 10-30-2003, 12:37 AM   #13 (permalink)
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Im so confused.

someone shoot me.

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Old 11-03-2003, 05:21 PM   #14 (permalink)
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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 Cantor-Bernstein Theorem).
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Old 11-03-2003, 07:01 PM   #15 (permalink)
 
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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?
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Old 11-04-2003, 01:30 AM   #16 (permalink)
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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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Old 11-04-2003, 01:56 AM   #17 (permalink)
 
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Shadowz, go ahead and try!
These proofs are either the trivial field or mistakes of the author!

Which one will yours be?

Last edited by KnifeMissile; 11-04-2003 at 02:03 AM..
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Old 11-04-2003, 07:14 AM   #18 (permalink)
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Quote:
Originally posted by Shadowz
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.
Go ahead. Try.

Don't forget that dividing by zero is not a legal arithmetic operation!
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Old 11-04-2003, 08:34 AM   #19 (permalink)
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Quote:
Originally posted by KnifeMissle
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?
Agreed, the word number has no general definition. I would think that most people think of integers though...

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 Cantor-Bernstein, 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 two-element 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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