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chris

I don't mean this as a flame, just as a point of reflection.

I wonder if Polymath could post a survey to see where those with formal mathematics training (at least an undergrad degree in math) stand vs. those without. If the mathematicians don't unanimously support Polymath then we've got bigger things to worry about.

"Nonbelievers" (I quote it because it really has nothing to do with belief): If you went to every doctor in the country and they agreed unanimously that you had Serious Disease X, would you refuse treatment because you feel fine and don't detect any symptoms? How about if every judge in the country told you that you did indeed break a law, would you still insist that you didn't, even if you felt like you didn't do anything wrong?

These situations are far, far, far less convincing than the fact that every reputable mathemetician in the country/world agrees that .999... = 1. It is fact. Actually, we've seen in the comments that there is at least one with a BS who is simply wrong in this case. However, this post is getting some serious attention and (I haven't read every post but) I don't see any professors signing up to refute Polymath's claim. The experts agree. Hmmm...

Of course if we're not talking about the realm of mathematics, then it's just a forum of opinions with no means of determining the the correct answer. So if you believe that this problem has an answer, that .9999... = 1 is either true or false, and you believe that this is a question of mathematics, wouldn't it make sense to trust a unanimous body of experts in spite of any feeling of uneasyness? What's more likely: that every mathematician in the world is wrong or that you're lacking sufficient training (it has nothing to do with intelligence) to fully understand? Remember the doctors and judges.


Mike Hoopes

What we have here, in Real Analysis, is the concept of convergence. 0.999... repeated is an infinite arithmetic series that converges on 1. Convergence is the same as equality in Real Analysis.

See http://en.wikipedia.org/wiki/Infinite_series for an example.

passerby

Teach this to students if you like, but it will only serve to convince the best students that they've been disserved by the education system.

The simple fact is, .3 repeating is an APPROXIMATION of one third, not EXACTLY one third. It adds up to 1 BECAUSE of the approximation, in exactly the same way that .9999 can be said to be APPROXIMATELY 1. Is is NOT and WILL not EVER be exactly 1, because, by DEFINITION, it is LESS than 1. If you fail to understand this, then you simply fail to understand the definition of .9 repeating. I was sick in school, and never went on to uni and higher math, but even I know you're waay wrong. And, more importantly, irresponsible to be teaching this.

p.s.: this is a random, in-passing post. I've never seen your blog before, and will probably never happen upon it again, so don't feel the need to attack me.

Michael

Me: Anyone want an apple?

Random Person X: Sure!

*Hands over apple*

Me: Sir, how many apples do you have?

Random Person X: ...One?

Me: No, you have .99999... of an apple.

Random Person X: You're weird...

schnubbi

I think the discussion here is all about the definition of 0.99|9.

People voting for 0.99|9 != 1 often have weird concepts in their mind of what 0.999... actually is, concepts having nothing to do with any entity in the world of integers, rational or irrational numbers.
The definition is mostly "intuitive" and fuzzy. If you want to say anything about 0.99..., you first have to mathematically define what it means to you!
There are definitions including limits or sequences, but most arguments for 0.99... != 1 are based on "intuitive" (mis-)concept(ion)s of 0.99xxx.

Some of them:

a) "0.999... is the number just below 1.0"

No. There is no real number "just below" another real number. between two arbitrary real numbers there are is an infinite amount of numbers in between. So the number you have in mind does not exist in the world of real numbers.

c) "0.99... is strictly irrational, thus it can't be equal to an integer"

You probably think that it is irrational because it is "somehow unlimited" just like pi or sqrt(2) is. But just because it's _written_ that way by definition, it's nevertheless rational (even an integer, namely "1"). Just like 0.333... is. 0.333... is just another symbol for 1/3, which is a rational (consisting of a fraction of two integers, which is the definition of "rational"). Using another base than 10 makes that clear. In base 9, 1/3 is simply 0.3. You can't do that with e.g. Pi in any base.

schnubbi

> The simple fact is, .3
> repeating is an
> APPROXIMATION of one third,
> not EXACTLY one third.

no, 0.3 repeating is just another _representation_ of 1/3. The decimal system doesn't allow another representation, so we need that one for calculations.
When using base9 instead of base10, you get 1/3 = 0.3.
0.3 repeating is not a number for itself such as e.g. Pi is, it's just *another* *symbol* for 1/3!

> by DEFINITION, it is LESS
> than 1.

What is your definition of 0.99... then? Show it to me, and I will show you how it differs from any definition of 0.99... in math.

> I was sick in school, and
> never went on to uni and
> higher math, but even I know
> you're waay wrong.

So you have an "intuitive" _feeling_ what 0.99... means to you. That has nothing to do with a mathematical definition. When you make a statement in science like "A != B", "Apples = Oranges", "0.99... != 1" you first have to strictly _define_ what A, B, Apples and Oranges _are_. Without that, the statements are just pointless.

The world of numbers (integers, rational.., irrationals...) consists of a set of clearly defined terms and some axioms. With the rules of logic, you proof theorems, based on the axioms.
One of these theorems is 0.99... = 1.

If you disagree, you probably have another (intuive) conception of rational numbers, the symbols representing them, or 0.99... in particular. That's fine, there isn't one true conceptualization of numbers resembling the "real world". But, the existing mathematical models have shown to be useful, don't contradict themselves, and solve a lot of real-world problems. Any alternative must copete with that.

However, it doesn't make sense
for us to discuss statements about apples and oranges if we don't agree on what apples and oranges actually are. Then we basically speak different languages where the statements have different meanings and we will never agree on them.

> And, more importantly,
> irresponsible to be teaching
> this.

I'd rather say people not knowing much about a particular subject, but nevertheless judging people who do, act irresponsibly.

Beans

First off, way too many comments on here to read all of them. If I repeat something, sorry, but I gave up on reading like 378 posts ago.

Seems obvious to me that you can't do mathematical operations with a repeating decimal. You can only perform the operation once you round it off at some point. (and you wouldn't round 0.66666 to 0.6666 in truncating the decimal, it would be 0.6667) A brief justification for the previous statement about rounding:

5800 + 6400 = 12200
5858 + 6464 = 12322
5858.58 + 6464.64 = 12323.22

When you round off to do the mathematics, you put zeros at the right end of the number to replace all continuing digits.

The original post unwittingly gave the material to disprove his own argument:
He correctly said that .9 repeating is essentially the sum of
9/(10^n) from n=1 to infinity. That is the only effective mathematical representation of .9 repeating. Without having to type out the equations, I will just say that since 9<10, you will never "recover" the 1/10^n that is lost in each addition.
In essence,
1 = (the sum (n=1 to n=x) of (9/(10^n)))+(10/(10^(x+1))) for any value x.

Try it. Truncate the sum at a whole number n rather than infinity and the equation will work.

Now consider what this means:
In my representation of the number 1, I have included an element "(10/(10^(x+1)))" which separates our .9 repeating from 1. Any fraction in which the denominator is represented by a positive value greater than 1 powered to infinity is by definition the smallest possible positive value. "What if I put 1 in the numerator instead of 10, wouldn't that be smaller?" No. Infinity is infinite, the relative value of all other entities is insignificant except to be non-zero when rationed against infinity.

The end result is that we have demonstrated that when x = infinity, that:
1 - the smallest possible positive value = the sum (n=1 to n=infinity) of (9/(10^n)) = .9 repeating.

Beans

Please note in my argument that n/infinity where n>0 is infinitely small and incomparable to all other values of n except to say that it is NON-ZERO and positive.

Lzygenius

Ok, if .9 repeating is 1 then a simple subtraction would prove it.

1-.9 repeating = 10^-(infinity)

pEnG

0.9999999999999... = 1
If you disagree, you're a fucking retard.

Mikeer

All 3 of your proofs are false.
1: Here you toy with number of digits after the dot, which do not equal. x has infinite digits after dot, 10x has infinite-1 digits. thus 9x = 8.(9)1
2: Here you toy with rounding up. You take 0.6666... There are infinite digits in 0.(6), and you do agree with that, right? There is an infinite+1 digit, which equals 6. Thus the infinite digit is 7 due to rounding up. In 0.(3) the infinte digit remains 3 because the infinite+1 digit is 3. Now we sum the numbers : 0.(3)3 + 0.(6)7 = 1. 3/3 = 1. 1 = 1.
In all the other examples the infinte+1 digit equals in one of numbers [5-9] and [0-4] in the other, thus the infinte digit in one number is rounded up and in the other one is not. Whatever numbers you take you get a 1.
3: Here you toy with > and >= expressions. Are you sure you're a math teacher? Because even 6th grade kids know difference between these expressions. On top of that according to the definition of an infinite geometric series the number, that none of components of sum exceeds in the sum. And what is number, taht none of the components exceed? No, it's not 1, it's 0.9. That proves that 0.(9) DOES NOT equal 1.

You can throw whatever "proofs" you have at me, but I ill always be able to prove you're wrong.

Mike Demenok.

Monimonika

Mike Demenok, for your "disproof" #1, let's do this little math equation:

10 x 1/3 = 10/3
10/3 = 3 + 1/3 (mixed fraction)

1/3 = 0.33|3, right? So:

3 + 0.33|3 = 3.33|3

Now, can you tell me how the "0.33|3" in this equation:

10 x 0.33|3 = 3.33|3

is different from the "0.33|3" in the previous equation?

For "disproof" #2, which 6 are you rounding up? The one at the "end" of infinity? Where's that?

The flaw in YOUR arguments is that you are assuming that there is an "end" to the repeating decimals at which you can stick a "final" decimal. But if you do that, the decimals are no longer repeating infinitely, but are instead FINITE. Infinite does not mean "really, Really, REALLY, big/long", it means that there is no end point, it goes on FOREVER.

A number like 0.00...(infinite 0s)...1 cannot exist because infinity is not a "fixed" (aka finite) size. If I ask you where that 1 is supposed to be, you can't just answer "At the end of the infinite 0s" because THERE IS NO END to the 0s.

Subtracting 1 from infinity is meaningless because it is INFINITE. There is NO END to compare the lengths of two infinities.

If you can't imagine infinity without placing it within arbritrary boundaries, then just admit your mental inability and leave the teaching of math to people who can understand the concept.

I haven't read all the posts in this thread, so I'll get back to #3 once I know what you're referring to.

Monimonika

Also, the rounding up that you did from 0.(6) to 0.(6)7 doesn't work for another reason. Contemplate this equation:

2/3 + 2/3 = 4/3 = 1 + 1/3

You state that 1/3 = 0.(3) even with rounding, right? So:

1 + 0.(3) = 1.(3)

Now, let's replace the 2/3s with their rounded-up values (just like in your disproof #2):

0.(6)7 + 0.(6)7 = 1.(3)4

Oops! Where did that 4 at the end come from? And to say that one of the 2/3s should be presented as simply 0.(6)6 would go against consistency, wouldn't it?

I hope you are man enough to see how this demolishes your disproof #2 (aka arbritrary rounding-up of numbers).

We get the answer of 0.(9) = 1 because we don't go around rounding inconvenient numbers up. Unlike you, who has to round 0.(6) to 0.(6)7 just to get the answer to psuedo-equal 1.

Mikeer

I figured as much that there would be someone saying that my "disproofs" are false, but hey, they ain't much more false than the original "proofs". What can you say about the 3rd "disproof"?

Monimonika

[I figured as much that there would be someone saying that my "disproofs" are false, but hey, they ain't much more false than the original "proofs". What can you say about the 3rd "disproof"?]

I see a definite LACK of any reference (much less refutations) to any of the points I brought up. Considering that you also typed,

[You can throw whatever "proofs" you have at me, but I ill always be able to prove you're wrong.]

but didn't even attempt to prove wrong even a tiny shred of what I posted, I'll have to consider this a forfeit on your part (that means you lose and I win).

Then you have the nerve to say that Polymath's proofs are still false (despite the very reasons for the falsity you came up with were shown to be wrong). The "false" label is all yours, not Polymath's.

As for "disproof" #3, I wanted to engage it when I was able to figure out what you were trying to say. Assuming that "disproof" #3 is in reference to Polymath's first post on this subject (the previous thread), here's what confuses me:

[Here you toy with > and >= expressions.]

I have no clue as to how this relates to what Polymath wrote. More specifically, where was ">=" (which I understand to mean "greater than or equal to") even implied in Polymath's explanation of the infinite geometric series? Please, tell me.

[On top of that according to the definition of an infinite geometric series the number, that none of components of sum exceeds in the sum. And what is number, taht none of the components exceed? No, it's not 1, it's 0.9.]

Read the definition again. It's the "partial sums", not the "components" that don't exceed the sum. Here's the difference between the two terms:

Components (in your definition): 0.9, 0.09, 0.009, etc.

Partial sums: 0.9, 0.99 (from 0.9 + 0.09), 0.999 (from o.9 + 0.09 + 0.009), etc.

Huge difference, isn't it? Again, go and look up the definition from a legitimate source other than the one between your ears.

Another way that you can look at this is whether there is a number greater than 0.(9) but less than 1. And no, adding another number at the "end" of the 9s in 0.(9) won't work because, as I mentioned before, there is no end.

Let me make it even clearer for you. Take the average number between the two numbers. As in, for the numbers 4 and 2, just add them together (4 + 2 = 6), divide by 2 (6/2 = 3), and you have the middle number 3.

Easy, right? Now try this with 1 and 0.(9). Go on, try it.

If you follow the basic (and I mean really basic) rules of addition and long division, you'll end up with the answer of 0.(9). If you somehow come up with a different answer, please show me your work, for I will point out where you erred.

And next time, please try to address what I wrote rather than take the coward's way out, okay?

Mike

http://www.geocities.com/clan_phetus/redcircle.JPG

Mike

http://www.geocities.com/clan_phetus/redcircle.JPG
It's a good link that provides some insight into the debate of this topic.

Donald Mills

I cant say that i read all of the posts cause there are quite a few of them. and i have to say that there is no flaw in your algebra. BTW I a student in higher mathematics currently. the only thing i have to add is that the more and more i look at it and work with numbers in my life the more i see that the decimal system in general is flawed. the decimal system is base 10 and its impossible to represent some fractions in base 10 such as 1/3 without talking about infinity and all that jazz. about the time that i realized the difference between irrational numbers and rational approximations of real numbers i realized that true mathemiticians work in fractions, and nothing else. because 1/3 = 1/3 and thats perfectly accurate and theres no controversy. hell im for throwing the decimal system out. BTW i used to think .999999999... = 1 is bull too, but ive come to decide that its trying to represent all fractions in base 10 that is incorrect.

zabong

Great thread. Haven't laughed that hard for some weeks.

Seems that Polymath has found the ultimate flypaper for the mathematically challenged. May Cantor have mercy on their souls...

Having 10 fingers to start with is a serious design flaw when trying to understand real numbers. We should be amoebae with large brains, then our intuition would not be nailed on using digits from 0 to 9.

Craig

I've read through almost all of the above comments, but in case I missed this argument already posted, I apologize.
The truth is we should be asking the question of whether .9 repeating is rational or irrational. If it is rational, then it can be written as a fraction (9/9). If it is irrational, then it cannot be written as a fraction. However, that would contradict the understanding that irrational numbers in decimal form never terminate nor repeat.

I conclude that .9 repeating does equal one.

If you want to contradict this post, please tell me a) if you think .9 repeating is rational or irrational.
b) if you say rational - what fraction (where the numberator and denominator are integers) represents it?
c) if it is irrational - what makes it the only irrational decimal number to repeat?

ffe

along side all of the math
I really have always believed that just because i am labled stupid dose not mean that nothing in this u-verse can travle faster than light how dose sound as you can hear it react in space

mmm...mmm...mmm...malek

.9 repeated can't equal 1 because 1 exists and .9 repeated does not. .3 repeated can't equal 1/3 because 1/3 exists and .3 repeated does not.

Ernie

I think the issue of .9-repeating is of the same variety: as the decimal is extended infinitely, .9 ultimately converges with 1 -- but only as the result of how it would sensibly appear, not how it purely is.

(I also concur)

Please see: http://www.thecyberprofessor.com for more clarification.

Ernie

Dave

I can't believe people are actually contesting this. Doesn't everyone learn that 0.999... = 1 when they're like 12 years old?

No sane mathematician will tell you otherwise.

sulliwan

Ok, this might be a bit late to comment, but this was the first page that came up when I searched for a discussion about this.

Your fallacy in saying that 0.(9) = 1 lies in treating infinity as a singular concept.
For example, take the set of all positive integers, the sum of all of which is infinity. Let's call this A. Next, take the set of all integers, sum up the absolute value of them all and you also get infinity. Let's call this B. However, this is easily proven to be larger than A, in fact, twice as large.

So, if
0.(9) = 1 - 1/A

then the average between 0.(9) and 1 is 1 - 1/B.

Since it has been shown there exists a number that is 0.(9), 0.(9) != 1.

QED.

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