Does 0.9999 repeating 9 equal 1?

[SonOfWashington]

No number in between them...yes they are the same.

[Fergasun]

So 1 minus 0.999... is equal to zero?

[Fergasun]

I think I just read through most of this thread and find the "proofs" to be unsatisfactory. In short; I think that these infinite numbers break the way we represent numbers in a finite way.

For example; the "multiply by 10" proof. x is a simplification of a complex number; and if we are trying to say that x = .999... x is just converting it back to the simple representation our brain can understand. What if instead of saying x = .999... you said y.yyy (repeating) is equal to .999... you'll still end up with the same problem of representing an infinite number in our finite thinking.

I don't know why "math people" are throwing around proofs that contain the same flaw; use fractions (10x and 1/3) to hide the simplification... when fractions *are* the simplification.

If we don't allow this type of fraction simplification we can just "prove" that .999... is not 1 because 1 - .999... would be equal to 1/999(repeating) and 1-1 is equal to zero. Since zero is not equal to 1/999(repeating).

Someone explain to me why this isn't allowed by the math rules?

[Spec138]

I think I just read through most of this thread and find the "proofs" to be unsatisfactory. In short; I think that these infinite numbers break the way we represent numbers in a finite way.

For example; the "multiply by 10" proof. x is a simplification of a complex number; and if we are trying to say that x = .999... x is just converting it back to the simple representation our brain can understand. What if instead of saying x = .999... you said y.yyy (repeating) is equal to .999... you'll still end up with the same problem of representing an infinite number in our finite thinking.

I don't know why "math people" are throwing around proofs that contain the same flaw; use fractions (10x and 1/3) to hide the simplification... when fractions *are* the simplification.

If we don't allow this type of fraction simplification we can just "prove" that .999... is not 1 because 1 - .999... would be equal to 1/999(repeating) and 1-1 is equal to zero. Since zero is not equal to 1/999(repeating).

Someone explain to me why this isn't allowed by the math rules?

What's wrong with my proof earlier?

Ok let's try it this way then.

Our error is reduced by a factor of ten as I sad before each time we add a number to the series I posted.

For the record, my series is .999...=.9/10^0 + .9/10^1 + .9/10^2...

This would mean the error between 1 and (.9/10^0 + .9/10^1 + .9/10^2...) is equal to 1/(10^x), where x is equal to how many numbers we have added (i.e.: For .9/10^1, x = 1; for .9/10^2, x = 2 and so on.)

So our series is equal to one minus our error or:

.9/10^0 + .9/10^1 + .9/10^2... = .999... = 1 - error

As 1/(10^x) approaches infinity it equals 0.

Therefore, .999... = 1 - 0 = 1

[Fergasun]

The premise on your first line belies the simplification. If you are reducing your error that would imply there is a difference between your series and 1. No matter how many 9's are added to your series; you keep reducing the error, but you can never get that error down to zero, even if we *could* somehow wrap our minds around the number infinity.

Additionally, you'll also have to prove that 1/(10^x) is equal to zero as x approaches infinity. I think you'll run into the same problem we are having here. It is an infinitely small gap between zero and 1/(10^x) but there still exists a gap. You could divide by 1/(10^x) no matter how far infinity is, but you can never divide by zero.

Again, these are all things that will get calculators thrown at me by math majors. What's wrong with just admitting the system can't handle infinite numbers... instead of going through these proofs? Of course for all intents and purposes 0.999... is close enough to be treated like 1, but it can never be 1.

[Spec138]

The premise on your first line belies the simplification. If you are reducing your error that would imply there is a difference between your series and 1. No matter how many 9's are added to your series; you keep reducing the error, but you can never get that error down to zero, even if we *could* somehow wrap our minds around the number infinity.

Additionally, you'll also have to prove that 1/(10^x) is equal to zero as x approaches infinity. I think you'll run into the same problem we are having here. It is an infinitely small gap between zero and 1/(10^x) but there still exists a gap. You could divide by 1/(10^x) no matter how far infinity is, but you can never divide by zero.

Again, these are all things that will get calculators thrown at me by math majors. What's wrong with just admitting the system can't handle infinite numbers... instead of going through these proofs? Of course for all intents and purposes 0.999... is close enough to be treated like 1, but it can never be 1.

No, there is only error when using a finite amount of 9s. That is why there must be an infinite amount of 9s.

The limit as x-> infinity of 1/x is 0.

[Fergasun]

Looks like I came to this thread and everyone has already had the discussion that my thinking will lead is into with PokerPacker on pages 5-8 (I've got 40 posts per page, doesn't everyone else max it out?). After reading more closely I have to say I agree with him.

This debate seems like the old "what's is..." debate. What does "equal" mean. For mathematicians the number .999... behaves just like 1, so it is equal to 1.

[Thiebear]

I thought we proved with the Patriot missile system.

Sometimes close isn't good enough. Sure, for a lot of issues rounding it off is fine.

But when you miss the incoming scud due to 100 hours of usage and the .01 being off, we need to get the smart people to come up with a new way.

[Mark The Homer]

Furguson, you must have missed this post (which ironically happens to be post #333...)

For the naysayers:

http://wiki.answers.com/Q/What_is_one_third_as_a_decimal

http://mathforum.org/library/drmath/view/55845.html

http://math.wikia.com/wiki/Proof:The_Decimal_0.999..._is_Equivalent_to_1

http://uk.answers.yahoo.com/question/index?qid=20090225122725AApnPuX

http://www.ehow.com/how_2239655_change-fraction-decimal.html

Found this on usenet:

http://groups.google.com/group/rec.puzzles/browse_thread/thread/7e94aa5aea335146/b7e0f1d98afeb4dc?hl=en&ie=UTF-8&q=decimal+one+third

We're laboring under the assumption that when you write "0.333..."

you mean "decimal point followed by repeated 3s forever." A formal

_

notation for this is 0.3 (horizontal bar over the 3).

If this is not what you mean, if instead you mean "decimal point

followed by a lot of 3s, for example maybe a million or two," then

let's stop right now, because such a number is not equal to 1/3.

If you really mean "3s forever" then it is obvious that you cannot

write all the digits of this number. If you were to attempt to

write down all the digits, you would fail: you would either use up

all the paper and ink in the world or die first.

All the division algorithm can do for you is to generate the digits

of 0.333... one at a time. If after each step you still have a non-zero

remainder, well, you haven't finished the division yet!

>Besides since 1/3 * 3 = 1... shouldn't 0.333... * 3 be 1 as well then?

Of course it is, because 0.999... = 1. But that begs the question

of what an infinitely long decimal means in the first place, so I

don't particularly like this as a "proof" that 0.999... = 1.

>Is 1/3 = 0.333... or is it just a convention that is used because it is

>otherwise impossible to feed 1/3 into a computer (or for another reason,

>point is, is it a convention)?

It certainly has nothing to do with computers, since any real computer

is completely incapable of storing an infinite number of decimal digits.

I suppose you might regard the notation as a "convention." I prefer

to call it an idealization; it is mathematics, not mere arithmetic.

The long division algorithm is arithmetic and will never actually

generate more than a finite number of digits.

In mathematics, however, we can define infinite sets and manipulate

them according to well-defined, consistent rules, so we can handle

infinite decimals provided that we make appropriate definitions of

what we want to do with them. It turns out that there is a very

useful set of definitions that identifies each infinite decimal with a

real number. This makes it easy to prove that there are more real

numbers than rational numbers (by the Cantor diagonalization proof).

According to these definitions, we would write 0.333... = 1/3 and

0.999... = 1. A disadvantage of these definitions is that there are a

few real numbers (anything of the form n*10^(-m) for n, m integers)

that end up with two representations, e.g. 0.999... = 1.000...

and 0.124999... = 0.125000... (the latter example, IIRC, is from

Tom Apostol's _Calculus_), but this is a relatively minor problem.

>third question: should I ask this in sci.math?

I would recommend reading the sci.math FAQ first, which will give

another explanation of why 0.999... = 1.

David A. Karr

I googled for the sci math faq...

Here's the proof within - two proofs actually, one using limits... references included...

http://faqs.cs.uu.nl/na-dir/sci-math-faq/0.999999.html

In conclusion, the answer to the question in the thread title in an unequivocal YES.

/thread

The sci math.faq proves that you and posterpacker, and all the other fairly young guys who grew up with math calculators who can't agree with this fact, are incorrect.

[Zguy28]

If .9999 repeating was equal to 1, it would be 1, not .99999 repeating.

[Fergasun]

Mark,

Did you even read my posts? Did you even read the FAQ? The *question* posed in this thread was "does 0.999... equal 1". Clearly the answer is *no*. Did you even read the FAQ?

In modern mathematics, the string of symbols 0.9999... is understood to be a shorthand for ``the infinite sum 9/10 + 9/100 + 9/1000 + ...''. This in turn is shorthand for ``the limit of the sequence of real numbers 9/10, 9/10 + 9/100, 9/10 + 9/100 + 9/1000, ...''. Using the well-known epsilon-delta definition of the limit (you can find it in any of the given references on analysis), one can easily show that this limit is 1. The statement that 0.9999... = 1 is simply an abbreviation of this fact.

Again; dealing with a limit... and even the sci.math FAQ has to readily, freely and comfortably admit they are dealing with a limit. So what this means is that number .999... "for all intents and purposes behaves like 1". However; it is not 1. It will always, into infinity, where we will have rulers capable of measuring infinitely small measurements be a number, that... is... less.... than... one. No matter how hard anyone tries to convince me, I know that even though that number behaves like 1; it is not 1! If it were 1 no one would have to couch the answer as a limit or whatever or go through a couple steps of hand-waving that hand-wave the whole point of the discussion.

All these proofs are doing is saying "these numbers which don't fit into our system behave and can be substituted freely with other numbers that do fit into our system"... but... to make it work; a substitution still must be made. That's it... it's a simplification... I'm not arguing the simplification is wrong, I'm not saying that 0.999... doesn't exhibit mathematical behavior of 1... I'm saying, it does not equal 1! And like I said, we can go through the definition of "equal", and I'm sure not planning on conceding the point, and I'm sure others on this thread and PokerPacker are not going to concede the point.

[Thinking Skins]

If .9999 repeating was equal to 1, it would be 1, not .99999 repeating.

its called having more than one way to represent the same number.

like 1/4 = 0.25 = 0.250 = 0.2500 = 0.25000 ...

[Fergasun]

The simple fact that we are having this discussion and smart people exist and can argue both sides prove to me that the numbers are not equal. Wow, it's not a proof that is probably mathematically sound, no doubt. Obviously no one here has proved anything, other than we can argue about "what equal is".

This is not the same was saying 1/4 = 0.250. The terminating 0 means something, it means nothing of something... whereas the repeating 9 has meaning... it means 9 of something. If there is 9 of something there is still something! Now I'm even more convinced that I'm correct!

Just admit that it's a simplification! These repeating, infinite numbers break the math system of thinking. All the proofs are doing is figuring out a way to patch them in.

[Fergasun]

To a mathematician if you had a twin brother who behaved like you, walked and talked like, you, ate everything you did, burned calories at the exact same rate, had the same job, and blah blah blah and eventually died the same time as you; you wouldn't exist. It would be 2x your twin brother (or he would be 2x you). However; in truth you *and* your twin would both exist. You would exist as unique but indistinguishable entities. There would be no way for any observer to tell you guys apart, but you still would be two unique people!

And this is why I hated proofs and other math BS in college.

[Fergasun]

But... at the same time... you and your twin would exist as separate; and would not exist in the same time-space continuum. Just as .999... and 1. They have identical behavior, however if you looked at the number-line, .999... will always be on the less-than side of the 1.

Hand me a Guinness because that is brilliant!

[Mark The Homer]

Ferg, of course I read the faq.

When I first came into this thread, I was on your side. But I saw the light. I think most people who come into this thread initially think the answer is no. But after considering the proofs and thinking about it, they change their position.

What really does it for me is the .333... = one-third thing. They equal. They are the same. They don't just behave the same, they are the same. To me, that is so clear. And I realized this long ago, probably when I was in high school.

I don't need anybody to show me a proof because I can see it in my head. I see it. And I don't understand why you and pokerpacker don't see it, because it's as clear as day. That's why I think calculators might be to blame, because they don't see it either.

.333... equals exactly one-third. It's not close and there is no gap. It's a third. They are one and the same.

From there, the answer to the question in the thread title is obvious.

[Thinking Skins]

Mark,

Again; dealing with a limit... and even the sci.math FAQ has to readily, freely and comfortably admit they are dealing with a limit. So what this means is that number .999... "for all intents and purposes behaves like 1". However; it is not 1. It will always, into infinity, where we will have rulers capable of measuring infinitely small measurements be a number, that... is... less.... than... one. No matter how hard anyone tries to convince me, I know that even though that number behaves like 1; it is not 1! If it were 1 no one would have to couch the answer as a limit or whatever or go through a couple steps of hand-waving that hand-wave the whole point of the discussion.

All these proofs are doing is saying "these numbers which don't fit into our system behave and can be substituted freely with other numbers that do fit into our system"... but... to make it work; a substitution still must be made. That's it... it's a simplification... I'm not arguing the simplification is wrong, I'm not saying that 0.999... doesn't exhibit mathematical behavior of 1... I'm saying, it does not equal 1! And like I said, we can go through the definition of "equal", and I'm sure not planning on conceding the point, and I'm sure others on this thread and PokerPacker are not going to concede the point.

Tell me something. Do you think \pi is a real number? Or sqrt(2)?

sqrt(2)*sqrt(2) = 2. Thats how we define sqrt(2).

But if we look at the decimal representation of sqrt(2), we get

sqrt(2) = .41421 35623 73095 04880 16887 24209 69807 85696 71875 37694 80731 76679 73799...

I wrote '...' because the number goes on forever. However, unlike the .999... example, nothing is being repeated. Its called an irrational number. However we can still say that sqrt(2) EQUALS this number because we can construct a sequence that converges to a unique real number.

You seem to not deny that the sequence {.9, .99, .999, .9999, .99999, ...} converges to 1.

But the DEFINITION OF CONVERGENCE OF A SERIES is that

if the sequence of partial sums of a series converges to some number S, then that SERIES converges to S.

THE WAY THIS IS WRITTEN is that

S = a_1 + a_2 + a_3 + ... + a_n + ...

or in this example, since the sequence of partial sums {.9, .99, .999, .9999, ...} converges to 1, the mathematical representation

1 = .9 + .09 + .009 + .0009 + .00009 + ...

or

1 = .999999...

is just saying the same thing. Its the definition of equals for a series. One side of the equality is the infinite sum thats known to converge to some number. And the other side is the number that it converges to.

Trust me, if we were not able to write down equalities for infinite numbers then we would not be able to work with most real numbers because most real numbers are irrational (in that when we write them in a decimal notation, there is an infinite number digits and nothing's ever repeating). But we can work with these irrational numbers because for each one we can define a sequence that converges to that number.

Lets go back to the sqrt(2) example again. This is an irrational number. A few of the digits are listed above. If we could only work with it as an infinite set of digits, then we couldn't ever use it. So instead we say that sqrt(2) * sqrt(2) = 2 and prove that sqrt(2) is the least upper bound for the sequence of partial sums for the infinite series that goes along with this number

0.41421 35623 73095 04880 16887 24209 69807 85696 71875 37694 80731 76679 73799...

In that way, we're able to say that

sqrt(2) = .41421 35623 73095 04880 16887 24209 69807 85696 71875 37694 80731 76679 73799...

[Thinking Skins]

But... at the same time... you and your twin would exist as separate; and would not exist in the same time-space continuum. Just as .999... and 1. They have identical behavior, however if you looked at the number-line, .999... will always be on the less-than side of the 1.

Hand me a Guinness because that is brilliant!

.9999.... is an infinite series. 1 is the least upper bound for this series, so since the series is bounded and monotone (in an earlier proof I showed this), the series converges to its least upper bound.

The definition of an infinite series converging to a number S means that we can write S instead of the infinite series!!!!

The limit S is called the sum of the series

S = a_1 + a_2 + ... + a_n + ...

Thats the DEFINITION of the sum of an infinite series!

So we can write 1 instead of .999999... BECAUSE the series .99999... converges to 1.

The only shorthand is in writing .99999... instead of writing

sum_{n = 0 to infinity} .9*10^(-n).

But if I were to write 1 = sum_{n = 0 to infinity} .9*10^(-n), then there would be no shorthand and its a factual statement because its mathematically provable from the axioms of mathematics.

[Thinking Skins]

Ferg, tell me, do you have a problem with posts 346 or 344?

[Zguy28]

The simple fact that we are having this discussion and smart people exist and can argue both sides prove to me that the numbers are not equal. Wow, it's not a proof that is probably mathematically sound, no doubt. Obviously no one here has proved anything, other than we can argue about "what equal is".

Separate but equal?

[December90]

One question

Anything multiplied by "10" will logically end with what digit (Irrespective of the decimal placement - Although if it is to the right of the decimal place it may be left off as not needed)

Don't worry this is related to this thread:)

[LaxBuddy21]

Ferg, of course I read the faq.

When I first came into this thread, I was on your side. But I saw the light. I think most people who come into this thread initially think the answer is no. But after considering the proofs and thinking about it, they change their position.

What really does it for me is the .333... = one-third thing. They equal. They are the same. They don't just behave the same, they are the same. To me, that is so clear. And I realized this long ago, probably when I was in high school.

I don't need anybody to show me a proof because I can see it in my head. I see it. And I don't understand why you and pokerpacker don't see it, because it's as clear as day. That's why I think calculators might be to blame, because they don't see it either.

.333... equals exactly one-third. It's not close and there is no gap. It's a third. They are one and the same.

From there, the answer to the question in the thread title is obvious.

See Mark, I think your problem is your acceptance that .333.... is really one third but its not. It has the same problem .999.... has in that it asymptotically approaches 1/3 but it never reaches it. With each 3 you add to the end you systematically reduce the error but it never disappears completely.

Here is another way to think about it that might help. Say you are standing at point A and you want to go to point B. You go half way to point B and then half way again and again and again. You will reach a point that you will be so close to point B that it will be negligible and in your mind you will be there but if you are truly going half way every time you move, you will NEVER reach point B. Just like if you keep adding 9's to .999.... you will get to a point that you think the difference is negligible but you will never truly reach 1.

[Mark The Homer]

One question

Anything multiplied by "10" will logically end with what digit (Irrespective of the decimal placement - Although if it is to the right of the decimal place it may be left off as not needed)

Don't worry this is related to this thread:)

You're trying to say that .999... x 10 has a zero at the end?

Sorry. No.

[Thinking Skins]

One question

Anything multiplied by "10" will logically end with what digit (Irrespective of the decimal placement - Although if it is to the right of the decimal place it may be left off as not needed)

Don't worry this is related to this thread:)

sorry, but 10*\pi does not end. Its still an irrational number.

[Thinking Skins]

See Mark, I think your problem is your acceptance that .333.... is really one third but its not. It has the same problem .999.... has in that it asymptotically approaches 1/3 but it never reaches it. With each 3 you add to the end you systematically reduce the error but it never disappears completely.

That means that it converges to 1/3. And the definition of an infinite series converging to a sum means that we can say that the series equals to the sum.

Its the definition of convergence/

Here is another way to think about it that might help. Say you are standing at point A and you want to go to point B. You go half way to point B and then half way again and again and again. You will reach a point that you will be so close to point B that it will be negligible and in your mind you will be there but if you are truly going half way every time you move, you will NEVER reach point B. Just like if you keep adding 9's to .999.... you will get to a point that you think the difference is negligible but you will never truly reach 1.

Yeah, but the point B is an upper bound to where you will travel. Moreso, the point B is a least upper bound (in that there is no point B' before B thats also an upper bound). And since you're travelling in one direction (monotone) and since the total distance you're going to travel is bounded (by the starting point and the point , then we know that if take infinitely many of those halfsteps, you will converge to the point B.

Its an important result on the convergence of monotone and bounded sequences.