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jennijen on May 31, 2007

um, no.

zhiqiang on May 31, 2007

Yes.
x = 0.999...
10x = 9.99...
10x - x = 0.999 - 9.99...
Therefore,
9x = 9.0
x = 1.0
Therefore,
0.999... = 1.0

aaroncampbell on May 31, 2007

10 * 0.999... - 0.999... != 9 unless you limit the repeating decimal in some way (and you have to limit it in 2 different spots, so if you use 99 decimal places, you have to use 100 decimal places for the one that you multiply by 10)

zhiqiang on May 31, 2007

There are some mistakes in my previous solution. I corrected it below:

x = 0.999...
10x = 9.99...
10x - x = 9.99... - 0.999...
Therefore,
9x = 9.0
x = 1.0
Therefore,
0.999... = 1.0

aaroncampbell on May 31, 2007

Yes, you had the 9.999... and 0.999 reversed on the one step, but you still didn't address the fact that to do what you did you have to round x, and you have to round it differently in 2 different places.

zhiqiang on Jun 01, 2007

Consider the value 9.99..., the digits after the decimal point are all 9s, infinite number of 9s.
This is the same for 0.999..., where there are infinite number of 9s after the decimal point.
So
0.999... - 0.999... = 0.0
9.999... - 0.999... = 9.0

davethegr8 on Jun 01, 2007

@zhiqiang -

Sorry, no. .999... != 1.0...9

If you let x = .9999, then 10x = 9.9990, and 10x - x = 1.0009

zhiqiang on Jun 01, 2007

@davethegr8:

0.9999 is not repeating. 0.9999 != 0.999...
My solution obviously failed for non-repeating values.

Wile on Jun 01, 2007

visibly, some of us are quite tired ;o)
10*0.999.... (with 9 present n times) equals to 9.9999 (with 9 present n times, then n-1 times after the dot),
so 9.99999.... - 0.99999.... = 8.999...91, with n-1 9...
and this, even if infinite 9 series ... The limit of this diff tends to 0, but this is just asymptotic, not equal ....

wmostrey on Jun 01, 2007

2 + 2 = 5 for very large values of 2

zelda013 on Jun 01, 2007

wmostrey : could i ask you to speak to my bank ?

fuseunderground on Jun 01, 2007

You can look at it this simple way:

10 divided by 3 = 3.33 recurring
3.33 recurring times 3 = 9.99 recurring and/or 10

Lunargent on Jun 01, 2007

Sorry to break the party, but I do think this article covers most of this discussion:

en.wikipedia.org/wiki/Proof_that_0.999..._equals_1

whoisgregg on Jun 01, 2007

Thanks for that link Lunargent. :)

Liquix on Jun 01, 2007

Thats scary... Now I have to twist my mind a bit

Grusella on Jun 01, 2007

@wmostrey: 2 + 2 = 6 for very large values of 2

SpacesEnd on Jun 01, 2007

It's true that .9 repeating is proved to equal one...but you do have to consider the limitations theory...after all, it has been around for over 500 years stating: a repeating number can continue to come close to another number, but will never quite get there.

Granf on Jun 01, 2007

think about this:

if you go 9/10ths of the way to a wall, have you reached it? no. 99/100ths? no. if you continue to go 9/10ths of the way to the wall for an infinite number of times, will you ever reach it? no. Because once you DO rach the wall, you have gone 10/10ths of the way, not nine thenths. note zeeheeqwang's mistake in line two, where he asumes ALREADY that 0.9 continued equals one, before he proves it. his equation is a catch 22.

Stratification on Jun 01, 2007

No, it will get there, that's the nature of infinity. If you've read the wikipedia article above (Link here for the lazy), it covers the "never quite getting there" part.

SpacesEnd on Jun 01, 2007

To start, ANYONE can edit wikipedia, so it's not a reliable source.

SpacesEnd on Jun 01, 2007

Besides, I'm gonna go with a 500-year-old theory over some stupid web article.

Stratification on Jun 01, 2007

Which just might contain other 500 year old theories, and could in fact contain some valid arguments. Just because you're paranoid doesn't mean they're not after you.

iconmaster on Jun 01, 2007

I find that one-third proof pretty convincing.

0.333... = 1/3

Multiply both sides by 3

0.999... = 1

opello on Jun 01, 2007

It's only complicated because you're discussing two different representations of the same value.
3 * 1/3 = 1
3 * (any representation of 1/3) = 1

Liquix on Jun 01, 2007

Well, 0.999 will never be 1. There is 0.000...1 missing. The reason its missing is because we dont see the whole number. 0.333... * 3 is 1, not 0,999.

You cant multiply an unlimited number and round it. Because thats what everybody is doing

Liquix on Jun 01, 2007

Forgot to say that the whole reason the number is unlimited long is because we cant place the last 0.000...1. If we try to share that number on three we end up with another 0.000...1 with one more decimal.

iconmaster on Jun 01, 2007

But why does 0.333... * 3 = 1, instead of 0.999...?

Whatever rule you supply to show that 0.333... * 3 = 1, it will necessarily also prove that 0.999... = 1.

Liquix on Jun 01, 2007

Because in 0.33... * 3 we got 0.00...1 we cant share between the three 1/3, so it gets tossed away. Thats why 0.33... is 0.99...

opello on Jun 01, 2007

That 0.0...1 "leftover" doesn't exist, it can't. It would be at the n+1st position, where n is infinity. Since the length is uncountably infinite, it can't be used like that. Someone with a better background in real analysis might have to elaborate as I don't recall all of the details.

iconmaster on Jun 01, 2007

Liquix: you can prove to yourself that 1/3 = 0.333... just by doing the division.

Stratification on Jun 01, 2007

Another interesting way of going at it:
here

Wile on Jun 01, 2007

500 years of theory dosn't prove this theory is right. If that was the case, then we can say that Earth is flat and that the sun turns around it !!!

Anyway, the problem is that nobody can write 1/3=0.3333..., because you'll never have enought space to write it down. I think there is a confusion here, between a number (1/3) and a representation of it that is only a mind view. I mean 0.999.... is the mathematical limit of this writing. The limit equals to 1, but the written number does not. It's the same than saying F:x->1/x equals to 0 if x equals to infinity ! It's just an asymptotic approx, not an equality. And to fix some minds, Infinity does not belong to the real numbers, it's a mind view ... nothing can be equal to infinity ... that's a mathematical definition ... just something can tend to it

davethegr8 on Jun 01, 2007

@iconmaster - 1/3 is an exact number, while .333... is an approximation. While it is true that one third time three is equal to one, anytime you work with approximations you get to fudge a little. So it seems that .333... * 3 is equal to one, but it is only so because of rounding. The same way that 2/3 = .666...7

@zhiqiang -
that's the principle. put any number or nines for x, and you get the same result, so long as the number is finite. If it is infinite, well, you can't really do math on those kind of numbers, you can only approximate.

Wile on Jun 01, 2007

@davethegr8 : yep's, that's not far from what I tried to say in my previous post. But I'm not sure of my words, because I am just a frenchie, and I did maths in hi school, but in french ;o)

opello on Jun 01, 2007

@Wile: 0.333... is an "exact number" when you consider that we're using the ellipsis in a similar capacity as a vinculum. This is no limit, it is a finite value, with both finite and "repeating decimal" representations. And limits are another discussion all together.

@davethegr8: Similarly, when we say 0.333... (or 0.3 repeating, or 0.3 with the vinculum) it is the "exact" value, as we can write it in decimal form. The only rounding that is occurring in your examples is your own. 2/3 is exactly equal to 0.6 repeating.

iconmaster on Jun 01, 2007

Yeah, I guess it depends on whether or not you accept 0.333... is equivalent to 1/3. I would tend to say that 0.333 is the approximation, but 0.333... (where "..." indicates an infinite repeating of the last digit) is exactly equivalent. From what I can tell, that's the generally accepted mathematical understanding.

Stratification on Jun 01, 2007

Another (more thorough) link:
Here

Wile on Jun 01, 2007

I can't agree with that (no, don't delete my next tokens iconmaster ;oD) ...
In fact, I partially agree, One can say 0.333... equal to 1/3, but no for mathmatical calc, just for an local ellipse. That's why 0.333... * 3 = 0.999..., but you can't consider that this is a proof that 0.999...=1. This is because 0.333... is just a writing facility, not another way to write the same number. That's why you can write 1/3=0.333..., but you can make any operation onto it as if it was a real number representation in an equation.
So I admit the writing, but not the operations on it.

Wile on Jun 01, 2007

@stratification : Your page does says that all numbers can be written in expansion form, but not that you can do some operations on expansion form of numbers ... so the proof is not valid by construction ...

Stratification on Jun 01, 2007

@Wile: I believe it says that all real numbers are by nature in expansion form and writing them another way is simply shorthand to make them manageable. It's an important distinction.

opello on Jun 01, 2007

Your premise that 1/3 and 0.333... are different because the symbols that make them up are different is incorrect. As stratification's link says, you can write any real number with an infinite sequence of numbers following the decimal point. Whether they're zeros (in the case of 1 as 1.000...) or threes (in the case of 1/3 as 0.333...).

Stratification on Jun 01, 2007

So, I've posted several pages with a number of proofs that they do equal. I'm going to say the case is closed unless someone can find something similar saying they don't equal. Something besides reasoning that they find some flaw somewhere that doesn't look right to them. (Hopefully that isn't overly antagonistic.)

Honey on Jun 01, 2007

It's. I learned that on school a while ago.

SpamDog on Jun 01, 2007

does .9999... = 1.0?? I say "Close enough for government work" I mean unless you are using this for sending rockets into space or splitting the atom.. I would say it is close enough. then again.. I usually do all my math as an estimataion anyway.. who needs to be precise

katylava on Jun 01, 2007

actually, i think it equals 1.0 in rocket science.

marcelofigueroa on Jun 01, 2007

@SpacesEnd:

1) You can edit Wikipedia, are you reliable?

2) Mathematic arguments are independent from wheter they are presented on a website or on a 500-years-old manuscript. A proof is a proof.

@Stratification: I agree with your last post.

There's no discussion about this subject: the statement is indeed true; there are many formal proofs available at the time.

-- My two cents --

The following is no proof at all, but it's a nice way to at see the equalitie in action.
In school we're taught to deal with periodic decimal numbers as follows:

342.12784784...784... = 342 + (12784-12)/99900

1.- Separate the integers from the decimal numbers (342.12784... = 342 + 0.12784...).

2.- Take the decimal numbers until one repetition of the periodic (repetitive) bit and write it as an integer (0.12784 => 12784).

3.- Substract the non-repetitive decimal part writed as an integer (12784-12).

4.- Build a number composed by: one nine for each digit counted in one of the repetitive bits, and one zero for each digit counted in the non-repetitive decimal part (0.12784 => 99900).

5.- Finally divide the obtained in (3) by the number obtained in (4) and add the integers separated in (1).

So!, we get:

342.12784784...784... =

342 + (12784-12)/99900 = 342+12772/99900

Uff!

Well applying the same to 0.9...9... we get:

0.9...9... = 0+(9-0)/9 = 9/9 = 1 :o)

Greetings from Chile!

SpacesEnd on Jun 01, 2007

Um...ANYONE can edit wikapidea. There's an EDIT button on the corner of EVERY page!!

SpacesEnd on Jun 01, 2007

Besides, the limitations theory has not ONLY been around for 500 years, it's been around for 500 years AND no one has proven it wrong!

SpacesEnd on Jun 01, 2007

The world is obviously not flat.

SpacesEnd on Jun 01, 2007

Additionally, .3 repeating is not equal to 1/3. It's just really really close.

Stratification on Jun 01, 2007

Limit theory is covered in one of those links for a proof that it does equal. Wikipedia hasn't been an issue here for a large number of posts if you're wondering.

SpacesEnd on Jun 01, 2007

Put it this way, you are at point 'A' and you're trying to get to point 'B.' Every time that you attempt to reach your point, you go 9/10 of the way there. You keep going 9/10 of the way there over and over and over again, but you'll never get to point 'B,' because once you've got there, you haven't gone 9/10ths of the way.

Starry on Jun 01, 2007

SpacesEnd said: Additionally, .3 repeating is not equal to 1/3. It's just really really close.

I agree with that. I heard somewhere its impossible to get to 1/3 exactaly in a base 10 number system.

In my opinion, the important question here is can it still apply in a base 7 system ;)

SpacesEnd on Jun 01, 2007

Stratifacation: The limitations Theory proves that .9 repeating DOESN'T equal 1. Trust me. Look it up. Although I take it back. Wikipiedia is not a bad site.

Steax on Jun 02, 2007

Err.. can a pattern be a valid argument here?

2/9 = 0.22...
7/9 = 0.77...
8/9 = 0.88...

if the pattern is to continue, then 0.99... should be equal to 9/9 = 1, right?

[still reading wikipedia article]

wallphone on Jun 03, 2007

Steax: very interesting.

As a non-numeric reasoning for this, I observed that .999... is less than one by an _immesurably_ small amount.

Since its so small, it can't be measured, when you measure .999... you get 1.

Steax on Jun 03, 2007

Well, 0.999... is equal to one minus ten to the power of infinity. Yes, an infinitely small number. So for all intents and purposes, an infinitely small number can never amount to anything, no matter how many times it's multiplied (except infinity), thus can be regarded as zero.

With that in mind, 0.999... = 1 - (0) = 1.

I'm just trying not to go overboard here.

Liquix on Jun 03, 2007

c = 0.999...
10c = 9.999... - 0.00...9
10c - c = 9.999... - 0.00...9 - 0.999
9c = 9.999...
c = 0.999...

Remember that that 10c is an infinity minus one digit

Liquix on Jun 03, 2007

And to steax:
0.999... = 1 - (0.00...1)

SpacesEnd on Jun 03, 2007

2/9 is not equal to 0.22...
just as 7/9 is not equal to 0.77... ect...

The whole root of the argument bases around the idea that if a number gets so infinitesimally close to another number, it gets there. But the truth is, it doesn't. Just as the limitations theory says.

SpacesEnd on Jun 03, 2007

The main argument that I find interesting in this case, is that: If .9 repeating doesn't equal 1.0, then what's in between? I don't think .99999... is equal to 1.0, but to be honest, I can't answer this question. Is there anyone that can?

Steax on Jun 03, 2007

The number 0.00...1 is impossible. It would have to be infinitely small. The question is, do you want to ignore it (which makes perfect sense), or count it in (for some reason).

There's really no purpose to include 0.000...1. No matter how many times you multiply it, it will never become a big enough number.

Liquix on Jun 03, 2007

The reason we don't use infinite numbers is because we are too dumb to get the whole image.

If if you start parting atoms in half we will never get to 1, because there is alway a small difference between the two numbers.

Even though not even god can measure it, the difference is still there

SpacesEnd on Jun 03, 2007

What it really seems to come down to (or at least from my *humanly dumb* understanding) is that .9999... is NOT equal to 1.0, but it's so friggin' close that for all practical reasons/purposes, it's effectively one.

But theoretically, it's still not ;)

SpacesEnd on Jun 03, 2007

Please say something if I'm wrong, or if someone disagrees.

iconmaster on Jun 03, 2007

zhiqiang posted this proof, but it bears repeating:

Say that c = 0.999... (infinitely repeating 9s).

10c = 9.999... (move the decimal point one space to the left)

10c - c = 9.999... - 0.999 = 9 (we're just subtracting the trailing decimals)

9c = 9

c = 1 (divide by 9)

Thus, c = 1 and c = 0.999..., so 1 = 0.999...

There are too many ways to demonstrate this for it not to be taken seriously. It's not that 0.999... is really, really close to 1; 0.999... is 1, without any equivocation.

tamashii on Jun 03, 2007

there is some serious nerd up in this thread…

marcelofigueroa on Jun 03, 2007

Hi!,

The number between 1.0 and 0.999... doesn't exist. Admiting that it does leads to a contradiction.

Liquix on Jun 03, 2007

Lets say this:
You are driving with your spaceship trough space. You got the 9/10 of the speed of light and started to travel as soon as the universe got created. Lets say you travel for about 10 years.

Each year you will gradually fall behind. When you have traveled 9 light years, the light has traveled one more. Thats the same Principe here. You can't part up 0.999... to measurable sizes, but there is still a small difference from 0.999... and 1.

Wile on Jun 04, 2007

I do agree with Liquix

Liquix on Jun 04, 2007

Another example:
The universe is always expanding. If you got this spaceship again. This time you travel at the same speed as the light. After the 10 first meters the light has traveled, you are one meter behind. We count the radius of the universe as 1. Then you would be 0.9 away from the center of the universe. After another 10 meters you are still one meter behind the light, but only 0.99 meters away from the center of the universe.
Another 10 meters: 0.999 and so on.

Even though how far you go, you will never catch up to that last meter. If you drive your spaceship for ever you will be 0.999... meters away from the center of the universe, not 1.

One thing you have to note though: The universe's expanding ratio is increasing all the time, but unlike the universe, numbers increase with the same ratio. Our system increases by 10 for instance

SpacesEnd on Jun 04, 2007

Iconmaster: It's a good trick you have there, but I think Granf already proved you wrong.

Liquix: I agree with you as well

iconmaster on Jun 04, 2007

It's not a trick, it's simple algebra. :)

Regarding the "never quite get there" angle, this is actually an ancient problem (one of Zeno's paradoxes, the dichotomy) for any finite progression. Say you throw a rock at a tree. To get from you to the tree, the rock will first have to cross half of the distance; then it will have to cross half of the half-distance; then half of that, and half of that, and so on. It will never reach the tree, because it will always be halfway between its last point and the tree.

However, clearly the rock does reach the tree, so something is wrong with the proof.

I.e., the rock doesn't travel 0.999...9% of the way there -- it travels the whole way. ;)

More thorough explanation of the dichotomy paradox.

Lunargent on Jun 04, 2007

*amazed that there are still people discussing this, when there really is very little to discuss (in the real numbers, at least)*

rudydanielle on Jun 04, 2007

are you kidding me?

eury on Jun 04, 2007

Chef: Reading the paradox part, what if the rock stops short of 99.9999...9%, would the rock considered as being arrived?

I am not math genius. but reading the wikipedia article, I sense that there is a general acceptance that .9999...9 is equal to 1.

I said this because we know in chemistry, the atomic value of an element is never rounded to the next of anything. When people rounded, it is never entered the equation, only saved on top of their heads for quick conversations.

iconmaster on Jun 04, 2007

Oops, I did mean 99.999...9% there. 99.999...9% isn't all the way there, because the "...9" means there is a final, terminating 9.

99.999...% (no terminating 9) would be all the way there though.

The notation is confusing, isn't it?

Stratification on Jun 04, 2007

@SpacesEnd: Regarding using a limit argument to say they don't equal. Did you read this article? Go down to the section labeled "Limit Argument". I get the feeling you're not reading these, just arguing off the top of your head.

iconmaster on Jun 04, 2007

Now... can someone explain why 0.888... doesn't equal 1? Or even 0.9?

marcelofigueroa on Jun 04, 2007

@Lunargent: I agree.

@Iconmaster: ????

SpacesEnd on Jun 07, 2007

Stratification: I read it. I don't agree. When I said that the Limitations theory proved that it doesn't, This was because I had thought that you said that the limitations theory was proof that .999... DOES equal 1.0. Sorry for the mistake!

marcoftheweb on Jun 07, 2007

Interesting and amusing discussion… But how come no one has thoroughly brought up the question of "precision" when regarding two numeric values? When we're speaking mathematics, we're talking about the language of science. All of the sciences I am aware of depend upon things like precision and margin of error when talking about any calculation.

So asking the question "Does 0.9 repeating equal 1.0?" is much too ambiguous. We need to know what domain of values we're working with. Most of the reasonable proofs people are referring to in this thread do actually refer to a proper domain. Then, you can actually start to prove something.

I'm with iconmaster. I say 0.8 equals 1.0 as far as I'm concerned (as long as our problem domain rounds to the nearest integer). See, we can prove whatever we want. ;-)

wallphone on Jun 08, 2007

According to Google:
http://www.google.com/search?q=0+%2B+.9999999999

Microsoft:
http://search.live.com/results.aspx?q=0+.9999999

Ask:
http://www.ask.com/web?qsrc=167&q=.99999999999+%2B+0

christopher on Jun 08, 2007

Oh no - its SURDS!!!

bdelisio on Jun 10, 2007

First let me say that I applaud the efforts of the middle schoolers to wrap their minds around complex mathematical concepts. Well done! And thank you for sparking our minds a bit, too.
The problem with the argument that 0.999... equals 1 is that you are trying to compare apples and oranges, and getting funny results.
The premise of this argument, that 0.999... is a rational number is incorrect. Is is not! A rational number is a number which can be expressed as a ratio of two integers. Since, 1/1 = 1 != 0.999..., 0.999... is not a rational number. To argue otherwise would be Reductio ad absurdum.
Is 0.999... an irrational number? The decimal expansion of an irrational number never repeats or terminates, therefore, the answer is no.
Is 0.999... even a real number? Since real numbers may be described informally as numbers that can be given by an infinite decimal representation then, indeed, 0.999... would be a real number.
As Liquix correctly points out above, as soon as you plug-in rational numbers, the logic of the argument that 0.999... is equal to one falls apart. There is alway going to something left over: 1 x10expN, where N=the number of decimal places.
Therefore the "proof" that 0.999... equals 1, offered by mathematicians and taught in textbooks, is limited only to the one special case, where N=infinity. And only because of the ambigueous nature of infinity.
While this problem may be interesting to number theorists it has no practical application in the real world. It is the exception, not the rule.
Note: Much of the above was lifted from Wikipedia, the free encyclopedia that can be edited by anyone. The logic is mine :-)

iconmaster on Jun 10, 2007

But rational numbers can be equivalent to their non-rational counterparts. Again, the example of 1/3 = 0.3333... (which the middle schooler can demonstrate as well as anyone).

It's not like rational and non-rational numbers inhabit different universes. They're both subject to the usual axioms.

And if "there's always something left over," how can 1/3 = 0.333...? Again, it's the same problem for both equations. If

1/3 = 0.333...

then

1 = 0.999...

bdelisio on Jun 10, 2007

Again, a rational number is a number which can be expressed as a ratio of two integers.
0.333... can be expressed as 1/3
0.999... cannot be expressed as a ratio of two integers.

Dave on Jun 10, 2007

Ok, frankly I'm amazed that this has lasted so long. It can be (and has been many MANY times already in this thread) proven that 0.9 recurring = 1
(please take time to read this if you're sceptical so that this can be laid to rest)

taking the times-by-10 argument and explaining it in a (hopefully) fool-proof way:

First - since infinity is a stand-in-word for the biggest number possible, it can be defined as:
infinity = infinity + 1
Try to solve that and you'll hit a divide by 0, proving that it isn't a real number, but that definition will help later.

0.9999... is 0.9 with infinity 9's
times that by ten and you get:
0.9999... * 10 = 9.999...
We now have infinity - 1 9's after the decimal point (one has been moved into the units), and going with SpacesEnd's argument, that means that if you do 9.999... - 0.9999, you will be left with a little bit at the end (effectively -0.000...01)

BUT!

we defined infinity = infinity + 1. Putting this into the 9.999... with infinity - 1 9's after the decimal point, we can say that it actually has infinity 9's after the decimal point.

Now that we know that, we can split 9.999... into 9 + 0.9999... and put it back into the equation to get:

0.9999... * 10 = 9 + 0.9999...

Now it's obvious to see that we can subtract 0.9999... from both sides to give

0.9999... * 9 = 9

Dividing both sides by 9 gives the final result of:

0.9999... = 1

Is that thorough enough? :P
If anyone have any arguments, explain them SPECIFICALLY. Tell me exactly which step is wrong and I will explain it further, or maybe even find that the stage is wrong after all and that a major proof of maths is in fact wrong :)

Wow that's a long post

bdelisio on Jun 10, 2007

The problem with your reasoning is that conventional mathematical operations do not behave as expected on non-terminating decimals.
Operations have to be confirmed using fractional equivalents.

You are trying to prove 0.999... is a rational number by using operations that simply do not work. 0.999... , by definition, is not a rational number.

Dave on Jun 10, 2007

Sorry but I asked for any replies to be specific. If you want me to take your argument seriously, please break it down into exactly what is wrong (explaining why the operators break down for example). Then I will listen.

Sorry to be stubborn, but IMO since this is a mathematical topic it should require mathematical reasoning.

bdelisio on Jun 10, 2007

The only thing you proved, Dave, is that conventional mathematical operations such as addition, subtraction, division, multiplication do not behave as expected on non-terminating decimals.
You are trying to prove 0.999..., a non-terminating decimal, is a rational number by using operations that simply do not work on a non-terminating decimals.

By definition, 0.999... is not a rational number.

Dave on Jun 10, 2007

Well I won't accept an argument of "By definition, 0.999... is not a rational number."

You may well be right and I'm willing to accept that, but ONLY IF you can explain why. Any explanation / web link / anything will do, but show me some kind of reasoning that isn't along the lines of "It's like this because it is"

Currently I can see absolutely no problem with the logic I posted, even accepting what you're saying, since the number 0.9999... can be split into the sum of 0.9, 0.09, 0.009, 0.0009, etc and the calculations will work when thought about like that.

btw, Wow - there are some pretty big debates about mathematics on this forum, lol

joshuasbones on Jun 10, 2007

If .9999... does not equal 1 then every calc class I've taken is wrong.

In the interest of creating even more discussion...
Pi is exactly 3!!

Dave on Jun 10, 2007

*stunned silence*

joshuasbones, I recommend you run before the mob of scientists hunts you down, lol

Candy on Jun 10, 2007

The world may never know....

bdelisio on Jun 10, 2007

I think joshuasbones is on to something...
You need to round off pi so that it equals 3,
you also need to round off 0.999... to equal 1!

So, as a scientist, it makes as much sense to me that 0.999... equals one as does one saying pi equals 3. The “proofs” are the same.

@joshuasbones: Do you believe that just because you are taught something, even repeatedly, it means that it must be right? People in the middle ages were taught the earth is flat until someone came along and challenged conventional thinking.

@Dave: read the wikipedia article mentioned above, section 6: Different answers from alternative number systems as a start. I’m sure you’re aware this argument has been raging on the internet for years and that there is plenty of “proof” for both sides.

Dave on Jun 10, 2007

"I’m sure you’re aware this argument has been raging on the internet for years"

nope - up 'till now I've heard only that 0.9 recurring = 1

It seems that that article you reference is on the side of 0.9 recurring *does* equal 1, unless you adopt a different convention. That's fair enough and I can agree with that, although I would say that with standard mathematics the convention is for 0.999.. = 1 (I believe the article says that too)

So I think that for most uses 0.9 recurring = 1, but in special cases with different number systems it isn't. Sort of like the tortoise and the arrow palindrome it seems.

Still, I don't think there's anything to gain from arguing either side so I'm calling it quits - 0.999... both is and isn't 1 as far as I care :)

SpacesEnd on Jun 10, 2007

Interesting.....I'm amazed that nobody can put their finger on a correct answer...

Dave on Jun 10, 2007

Well, it seems that there are 2 correct answers, each correct in it's own way, but one more useful than the other.

sort of like how we define e^j(theta) (or e^i(theta) for those who learned longer ago) as being sin(theta)+jcos(theta) (as far as I can remember...) That's an arbitrary statement, it would be just as easy to say it equals something else, but it wouldn't help with as much other stuff. So saying 0.9 recurring is 1 is the same sort of thing. It makes life easier.

Maths is full of things that can be done differently and still be theoretically correct - products of matrixs, imaginary numbers, and apparently this too, lol

bdelisio on Jun 10, 2007

The article I referenced above states “Although the real numbers form an extremely useful number system, the decision to interpret the phrase "0.999…" as naming a real number is ultimately a convention, and Timothy Gowers argues in Mathematics: A Very Short Introduction that the resulting identity 0.999… = 1 is a convention as well.

That is to say that there is insufficient evidence for, and sufficient evidence to the contrary that the identity 0.999… = 1 is merely a convention and not a mathematical fact.

In the same article it states: "With the rise of the Internet, debates about 0.999… have escaped the classroom and are commonplace on newsgroups and message boards, including many that nominally have little to do with mathematics."

iconmaster on Jun 11, 2007

So is 1/3 = 0.333... a convention or a mathematical fact?

Note that by stating, "cannot be expressed as a ratio of two integers," you are assuming what you're trying to prove -- i.e., that 1 does not equal 0.999...

If you accept that 1 does equal 0.999..., then it's obvious that it also equals 1/1 -- which is the ratio of two integers. 0.999... is just one of many expressions of a rational quantity.

opello on Jun 11, 2007

Heh, gotta love fallacious arguments.

I did enjoy this one by wallphone:
Since its so small, it can't be measured, when you measure .999... you get 1.

It's not quantum physics.

The standard limit arguments were good too, by traversing any fractional portion of a distance to something, you'll never get there.
For anyone invoking limits, I'd like to see a series that gives you 0.999... repeating, that cannot be solved as equal to 1.

bdelisio on Jun 11, 2007

@opello: a limit is not the same as an equality

opello on Jun 11, 2007

After reading more of the posts above, it seems that there became a disconnect where people thought 0.999... repeating wasn't a real number (as a member of the set R).

Since real numbers form an ordered field, with the important operations for this discussion: addition and subtraction.

Addition and subtraction are fully closed in the set of real numbers, so you can subtract 0.999... repeating from 1, and you must get another real number. The real number system doesn't have nonzero infinitesimals. So the difference must be 0.

In abbreviated form, it's basically the argument that is on Wikipedia, but "no nonzero infinitesimals" is a fact of real numbers, as is closure over addition and subtraction. Regardless of anyone's "wiki editing abilities." :)

opello on Jun 11, 2007

@bdelisio: However there are techniques to solve both finite and some infinite arithmetic series. That's what I meant.

opello on Jun 11, 2007

Another way to think about it, after talking with some other friends:

Given that subtraction is closed over real numbers,
you can subtract 0.999... repeating from 1,
and must get another real number.

The difference must not be infinitesimal, unless it is zero. Then by the 0 identity (if a - b = 0, a = b), the numbers must be equal.

The next thing to consider is that between any two, non-equal, real numbers, there exists an infinite number of real numbers. I would challenge anyone that disagrees with 0.999... repeating being equal to 1 to name any number between 0.999... repeating and 1.

bdelisio on Jun 11, 2007

Define "0.999... repeating"

Is it 0.999?
1-0.999=0.0001
Is it 0.99999?
1-0.99999=0.000001
Is it 0.9999999?
1-0.9999999=0.00000001
Is it 0.999...9?
1-0.999...9=0.000...1

Any rational number of the form, 9x10exp-N (where -1 =< N < infinity) pluged into this equation for 0.999... give a real remainder.

The problem is that 0.999... is not defined.

If your argument were true for 0.999... then would it not be true for the above examples?

The arguments and counter-arguments presented in this blog, including Opello'o and many others, can be found in the wikipedia article.

opello on Jun 11, 2007

"0.999... repeating" is the number zero, followed by a decimal point, followed by an infinite sequence of nines.

The only time that "0.999... repeating" is not equal to 1, is when you change number systems. Since we're all used to the set of real numbers it's safe to assume that's the domain of the original question. Again, the only valid counter example is when you take into account non-real number spaces.

However, even in your last example, the result of your subtraction cannot exist in the real number space. Which you conceded above that 0.999... repeating was a member.

I think you want to put a limit on the number of digits there are past the decimal. Which you're doing by saying it's 'N' while in fact that exponential representation of an infinitely repeating decimal is not appropriate. You don't write 1/3's decimal expansion as 3.3e^(-1).

It's not really a problem of the 0.999... repeating being defined, but that you won't accept that it's got an infinite number of decimal places.

fuseunderground on Jun 11, 2007

0.99 recurring and 1 are simply two ways of representing (writing down) the same value.

bdelisio on Jun 11, 2007

Hence, the reason 0.999... = 1 is an agreed upon convention, rather than mathematical fact.

opello on Jun 11, 2007

That's tantamount to saying that 1/3 is not the same as 0.333... repeating, because the symbols that make up the the representations are different. However, 1/3 is exactly equal, mathematically, to 0.333... repeating. That is a zero, followed by a decimal, followed by an infinite sequence of threes.

soprano on Jun 11, 2007

If we accept that the sum of an infinite geometric series is a/(1-r), then we have .999... = .9 + .1 * .9 + .1^2 * .9 + ... = .9/(1-.1) = .9/.9 = 1.

bdelisio on Jun 12, 2007

@opello: you are trying to present the Proof by Fractions.

@soprano: (Hey! You're dead!) you are trying to present the Infinite Series and Sequences.

Both of the arguments have been made before.

joshuasbones on Jun 13, 2007

@bdelisio: (from way back) Yes, because Prof. Drelles is way way way smarter than me and if he says something is true than it must be.

Also, e is exactly 3!

SpacesEnd on Jun 15, 2007

opello:
.333.... is NOT equal to 1/3. It just keeps getting closer.

memodude on Jun 15, 2007

@SpacesEnd:

from http://en.wikipedia.org/wiki/Recurring_decimal:

1. x = 0.333333…
2. 10x = 3.333333… (multiplying each side of the above line by 10)
3. 9x = 3 (subtracting the 1st line from the 2nd)
4. x = 3/9 = 1/3 (simplifying)
5. 0.333333... = 1/3

therefore:

1. x = 1/3 (given)
2. x * 3 = 1 (multiply by 3 using #1)
3. 0.333333... = 1/3 (given from previous proof)
4. x = 0.333333... (substitute #3 into #1)
5. 0.333333... * 3 = 1 (substitute #4 into #2)
6. 0.999999... = 1 (simplify)

So either 0.999999... doesn't exist or it's equal to 1.

SpasticNate on Jun 15, 2007

My head is going to implode! 1 head - 1 head = no head. NOONE HAS A HEAD! AHHH!

cheerios on Jul 19, 2007

No it doesn't. Because they're two different numbers.

Stratification on Jul 19, 2007

@cheerios: You just had to open this up again, didn't you.

SpacesEnd on Jul 20, 2007

Yes they're two different #s, but #s an value are two different things.