0.999... = 1, dammit!

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0.999... = 1, dammit!

#1  Postby Vorticity » Feb 25, 2010 6:01 pm

Since this is a brand new forum to which I myself am new, I thought it fitting that I should start by contributing this inevitable and indispensable thread to the mathematics section. Each and every forum of this sort is guaranteed to have this as an acrimonious 30+ page thread eventually, so we might as well do it now. So let's have at it...

I assert that the repeating decimal 0.9999999... is exactly, rigorously, and indisputably equal to 1.

I submit that anyone in disagreement with this is, in fact, a nematode.

Thoughts?
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Re: 0.999... = 1, dammit!

#2  Postby RPizzle » Feb 25, 2010 6:07 pm

Vorticity wrote:Since this is a brand new forum to which I myself am new, I thought it fitting that I should start by contributing this inevitable and indispensable thread to the mathematics section. Each and every forum of this sort is guaranteed to have this as an acrimonious 30+ page thread eventually, so we might as well do it now. So let's have at it...

I assert that the repeating decimal 0.9999999... is exactly, rigorously, and indisputably equal to 1.

I submit that anyone in disagreement with this is, in fact, a nematode.

Thoughts?


Isn't it just infinitesimally close to 1, but not actually 1.
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Re: 0.999... = 1, dammit!

#3  Postby Shaker » Feb 25, 2010 6:18 pm

Isn't it just infinitesimally close to 1, but not actually 1.

That's what I'd have thought, but no mathematician I - I have to take my shoes and socks off to do sums with numbers bigger than 10 in. I can't see how 0.9999999999 ... can be equal to 1, only an infinitesimally closer and closer approach to it.
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Re: 0.999... = 1, dammit!

#4  Postby Vorticity » Feb 25, 2010 6:20 pm

RPizzle wrote:Isn't it just infinitesimally close to 1, but not actually 1.


Nope. In the set of numbers known as the Reals, to which (regardless of whether or not they are distinct) both the numbers 0.9~ and 1 belong, there is no such thing as two numbers which are "infinitesimally close" to one another.

In other words, if a and b are both real numbers, then either a = b exactly, or the difference between them is some finite, non-zero, real number. Since we can name no finite, non-zero, real number corresponding to the difference between 0.9~ and 1, the only alternative is that they are equal.

(By the way, 0.9~ is shorthand for 0.9999(repeating forever).)
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Re: 0.999... = 1, dammit!

#5  Postby RPizzle » Feb 25, 2010 6:22 pm

0.9*(1+1/10+1/100+1/1000...) If you were to graph these individually you'd end up with an asymptote so not equal to 1 no matter how many iterations you go.
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Re: 0.999... = 1, dammit!

#6  Postby Shaker » Feb 25, 2010 6:26 pm

In the set of numbers known as the Reals, to which (regardless of whether or not they are distinct) both the numbers 0.9~ and 1 belong, there is no such thing as two numbers which are "infinitesimally close" to one another.

In other words, if a and b are both real numbers, then either a = b exactly, or the difference between them is some finite, non-zero, real number. Since we can name no finite, non-zero, real number corresponding to the difference between 0.9~ and 1, the only alternative is that they are equal.

Yes, yes, that's exactly what I meant to say :lol:

But then RPizzle comes along and says the opposite ... there's only one way to find out ...
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Re: 0.999... = 1, dammit!

#7  Postby Spearthrower » Feb 25, 2010 6:27 pm

Vorticity wrote:Since this is a brand new forum to which I myself am new, I thought it fitting that I should start by contributing this inevitable and indispensable thread to the mathematics section. Each and every forum of this sort is guaranteed to have this as an acrimonious 30+ page thread eventually, so we might as well do it now. So let's have at it...

I assert that the repeating decimal 0.9999999... is exactly, rigorously, and indisputably equal to 1.

I submit that anyone in disagreement with this is, in fact, a nematode.

Thoughts?



I am the one who cannot grasp it! :thumbup:

Please feel free to batter your head against my blank mathematical wall in frustration.

Ready?

But it can't be 1 because it's got a 0 at the beginning! :eatu:
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Re: 0.999... = 1, dammit!

#8  Postby Vorticity » Feb 25, 2010 6:33 pm

Shaker wrote:
Isn't it just infinitesimally close to 1, but not actually 1.

That's what I'd have thought, but no mathematician I - I have to take my shoes and socks off to do sums with numbers bigger than 10 in. I can't see how 0.9999999999 ... can be equal to 1, only an infinitesimally closer and closer approach to it.


A common confusion in discussions of this sort. You're thinking of the number 0.9~ as a process, as a thing evolving in time that is "getting somewhere", almost as if it is a quantity that is unreeling and evolving right before your eyes. In fact this is not the case. To put it in an unrigorous way: If an infinite-decimal number seems as if it is "approaching" something, then the value of that number is the same as the thing it is approaching. This is the way that infinite-decimal numbers are defined in the real number system.

In order to explain what I mean in more detail, I'll need to get all numbery on you. Think of some number with an infinite, non-repeating, decimal expansion. Take pi for example:

pi = 3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510(and so on)

Now consider the following sequence of numbers:

x_0 = 3
x_1 = 3.1
x_2 = 3.14
x_3 = 3.141
x_4 = 3.1415
and so on...

Each number in the sequence is known as a "partial expansion" of pi, with x_n corresponding to 3 followed by n digits of pi after the decimal point. Note that the actual number pi is defined to be exactly equal to the limit of this sequence. By "limit" I mean, roughly, the number this sequence is "approaching". (This notion can be made more rigorous.) This is how the system of real number is constructed.

Now consider the sequence

x_0 = 0
x_1 = 0.9
x_2 = 0.99
etc...

These are the partial expansions of the number 0.9~. What number is this sequence "approaching"? You yourself say that it is approaching 1. Therefore, by definition, it is equal to 1. Thus 0.9~ = 1.
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Re: 0.999... = 1, dammit!

#9  Postby Spearthrower » Feb 25, 2010 6:38 pm

Vorticity wrote:I submit that anyone in disagreement with this is, in fact, a nematode.


Image

Has anyone got a vacant digestive tract for me to settle down in?
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Re: 0.999... = 1, dammit!

#10  Postby Shaker » Feb 25, 2010 6:40 pm

Thanks guys.

I think I'll be leaving this particular board alone from now on :eh:
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Re: 0.999... = 1, dammit!

#11  Postby RPizzle » Feb 25, 2010 6:45 pm

I'm no math major, so I can't really put up much of a counter-argument. I am likely a nematode.
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Re: 0.999... = 1, dammit!

#12  Postby Spearthrower » Feb 25, 2010 6:56 pm

RPizzle wrote:I'm no math major, so I can't really put up much of a counter-argument. I am likely a nematode.


Once I find a digestive tract, you can take the anus end!
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Re: 0.999... = 1, dammit!

#13  Postby RPizzle » Feb 25, 2010 6:57 pm

Spearthrower wrote:
RPizzle wrote:I'm no math major, so I can't really put up much of a counter-argument. I am likely a nematode.


Once I find a digestive tract, you can take the anus end!


I'm going to write a very shrill and strident post about you saying that! :mob:
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Re: 0.999... = 1, dammit!

#14  Postby Vorticity » Feb 25, 2010 6:57 pm

RPizzle wrote:0.9*(1+1/10+1/100+1/1000...) If you were to graph these individually you'd end up with an asymptote so not equal to 1 no matter how many iterations you go.


The key to this is what you mean by the word "these". It seems clear to me that the objects referred to by this word are the following numbers:

x_1 = 0.9*(1) = 0.9
x_2 = 0.9*(1 + 1/10) = 0.99
x_3 = 0.9*(1 + 1/10 + 1/100) = 0.999

with, in general,

x_n = 1 - 1/10^n

It is certainly true that if you graph x_n as a function of n, you get a curve of points that approaches 1 asymptotically as n becomes large. It is also true that it never reaches 1 for any finite n. This, however, in no way implies that 0.9~ and 1 are distinct. Here are two explanations of this:

1) When we think of the number 0.9~, we do not consider it as being represented by any number x_n of the above sequence for finite n. In fact, the axioms of the real numbers define it as the limiting value of the sequence of numbers x_n. This limiting value can be rigorously demonstrated to be 1.

2) Let's assume for the moment that 0.9~ is less than 1. Consider the number d = 1 - 0.9~. Under this assumption, we should have d strictly greater than zero. Furthermore, since 0.9~ and 1 are both real numbers, d must also be a real number. What real number can d be?

Define the sequence of real numbers

d_n = 1 - x_n

with x_n defined as above. Since 0.9~ is greater than x_n for every finite n, this implies that that d is less than d_n for every finite n. Any value you choose as a candidate value for d then yields more problems than it solves. For example, suppose we say that d = 0.0000001 (six zeroes). We can then choose n = 7, to get

d_7 = 1 - x_7 = 1 - 0.9999999 (seven 9's) = 0.0000001 (six zeroes)

But we know that d < d_7! Thus d must be less than our guess of 0.0000001. The thing is, 0.0000001 is nothing special. You can make the same argument for any positive candidate value of d, and prove that d must be less than that value. The only possible value of d that doesn't lead to a contradiction is zero itself. Therefore d = 0, and since d = 1 - 0.9~, 1 = 0.9~ exactly.
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Re: 0.999... = 1, dammit!

#15  Postby Spearthrower » Feb 25, 2010 7:00 pm

Can you explain without numbers?
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Re: 0.999... = 1, dammit!

#16  Postby Vorticity » Feb 25, 2010 7:01 pm

Spearthrower wrote:But it can't be 1 because it's got a 0 at the beginning! :eatu:


Inherent in this statement is the assumption that a given real number can have only one decimal expansion, and hence that 0.9~ and 1 must be distinct since their first digits are distinct. In fact, this is false. Every real number with a finite decimal expansion (such as 1) has at least two decimal expansions. Examples:

3.1 is the same as 3.0999999~
4 is the same as 3.999999~
1 is the same as 0.999999~
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Re: 0.999... = 1, dammit!

#17  Postby Vorticity » Feb 25, 2010 7:02 pm

Spearthrower wrote:Once I find a digestive tract, you can take the anus end!


Or a Chick tract.
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Re: 0.999... = 1, dammit!

#18  Postby Vorticity » Feb 25, 2010 7:03 pm

Spearthrower wrote:Can you explain without numbers?


Hmmm.

No, I guess not. I mean, not without numbers in it anywhere at all. I'd have to use at least the numbers 1 and 9.
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Re: 0.999... = 1, dammit!

#19  Postby Spearthrower » Feb 25, 2010 7:11 pm

Vorticity wrote:
Spearthrower wrote:But it can't be 1 because it's got a 0 at the beginning! :eatu:


Inherent in this statement is the assumption that a given real number can have only one decimal expansion, and hence that 0.9~ and 1 must be distinct since their first digits are distinct. In fact, this is false. Every real number with a finite decimal expansion (such as 1) has at least two decimal expansions. Examples:

3.1 is the same as 3.0999999~
4 is the same as 3.999999~
1 is the same as 0.999999~



See, I get what you are saying logically, but it seems self defeating for the purpose of mathematics to me.

If 0.999 is 1, then there's no delineation marked, surely it's just as easily 0 as 1?
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Re: 0.999... = 1, dammit!

#20  Postby RPizzle » Feb 25, 2010 7:15 pm

Vorticity wrote:
Spearthrower wrote:Can you explain without numbers?


Hmmm.

No, I guess not. I mean, not without numbers in it anywhere at all. I'd have to use at least the numbers 1 and 9.


A physicist vs. a pre-med in a maths battle. Well, I'm boned. Too bad there isn't a white flag smiley or I'd be throwing it up here. I'm going to concede defeat because I don't have nearly as strong of a grasp of mathematics as you do. I'm going to turn in for the, well its actually the afternoon, but if you'll take a bit of time later, I may have a few questions for you if you were willing to help me understand this concept better. Cheers! :cheers:
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