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Artsemis 10-28-2003 01:29 PM

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!

CSflim 10-28-2003 01:50 PM

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).

Artsemis 10-28-2003 02:05 PM

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...

Redlemon 10-28-2003 02:07 PM

CSflim is exactly correct, although as an engineer I'd be more comfortable at 0.95.

bennyb 10-28-2003 02:37 PM

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

cliche 10-28-2003 03:10 PM

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?

CSflim 10-28-2003 03:22 PM

A good explaination:
http://mathforum.org/dr.math/faq/faq.0.9999.html

also check out the links at the bottom of the page.

jgorrell 10-28-2003 06:33 PM


Artsemis 10-28-2003 06:48 PM

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 :)

KnifeMissile 10-29-2003 12:46 AM

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!

Jakejake 10-29-2003 05:52 AM

Let x = 0.999999....

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

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

10x -x = 9

9x = 9

x = 1

dimbulb 10-29-2003 09:38 AM

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

Doesn't Matter 10-30-2003 12:37 AM

Im so confused.

someone shoot me.

kthx

Grothendieck 11-03-2003 05:21 PM

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).

KnifeMissile 11-03-2003 07:01 PM

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?

TheShadow 11-04-2003 01:30 AM

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. :D

KnifeMissile 11-04-2003 01:56 AM

Shadowz, go ahead and try!
These proofs are either the trivial field or mistakes of the author!

Which one will yours be?

CSflim 11-04-2003 07:14 AM

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. :D

Go ahead. Try.

Don't forget that dividing by zero is not a legal arithmetic operation!

Grothendieck 11-04-2003 08:34 AM

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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