 Through proofs, yes, you have "proven" that .9 repeating equals 1 and also through certain definitions.

But in the realm of logic and another definition you are wrong. .9 repeating is not an integer by the definition of an integer, and 1 most certainly is an integer. Mathematically, algebraicly...whatever, they have the same value, but that doesn't mean they are the same number.

I'm getting more out of "hard" mathematics and more into the paradoxical realm. Have you ever heard of Zeno's paradoxes? I think that's the most relevant counter-argument to this topic. Your "infinity" argument works against you in this respect. While you can never come up with a value that you can represent mathematically on paper to add to .999... to equal one or to come up with an average of the two, that doesn't mean that it doesn't conceptually exist. "Infinity" is just as intangible as whatever that missing value is.

But really in the end, this all just mathematical semantics. By proof, they are equal to each other but otherwise they are not the same number. I've been through many math courses and I know they are equal in that context. I've just always been the one that takes math outside of its strict confines and puts it into topics such as philosophy.

Any debate involving infinity wouldn't be complete without Zeno's paradoxes :) Try this: the five key on my keyboard is broken. When I count, I have to use the S key instead. So, given that representation, the number S12 is two to the power of nine, and the value of pi to fifteen decimal places is 3.141S926S3S89793. It's a representation, right? Yet it breaks the rule that decimal numbers don't use letters, doesn't it? Just like .999999... breaks the rule that integers don't use decimal points. Same difference. Let A and B be functions such that: 1 = A(x) + B(x)

Let B = 1 * (10 ^ -x).

Therefore limit of A(x) as x approaches infinity is 1.
However, be definition, we know that A(x) never equals 1.

A(x) can be defined as:
A(0) = 0.
A(x) = 9 * (10 ^ -x) + A(x-1), where x > 0.

Therefore A(x) = .9(repeating) if and only if x = infinity.
However, by definition we know at 1 = A(x) + B(x).
A(x) = 1 if and only if B(x) = 0.

B(infinity) = 1 * (10 ^ -infinity) != 0.

Therefore A(infinity) != 1.
Therefore 1/9 + 8/9 != A(infinity)

The only real problem with this method is show that the assertion that: 1 = A(x) + B(x) is valid for the functions I specified. like i said in the other entry

1/3 does not equal .3333...
1/3 is an exact number while .3333... is an attempt to represent a non terminating sequence

1/3 + 2/3 = 1 0.9999999999 + 0 =1

ten or more 9's after decimal is equal to 1 as per google
check it out!!!!!!!! i decided to put my proof to your problem in general terms

if and only if 1/x is a terminating sequence of decimal places can its decimal value be used in a mathmatical equation. if 1/x is a non terminating sequence of decimal places, the estimated value of the sequence may not be used interchangably with 1/x This is a neat problem, and I wrote an actual proof that shows .9999... is equal to 1. Sorry, direct link above is wrong, below is correct: There is a flaw in your proof at step (3).

It should be:
It converges since |r| = 1/10 and -1 < |r| < 1 and |r| != 0, which is the requirement for exponential decay. The problem is an error with the decimal system. 1/3 =.3333... but you cant write an infinative number of 3's. Any one would agree that 3*(1/3) = 1 because it is written correctly but when you see 3 * 0.3333... the number 0.9999... comes to mind becuase of the way we write and think about numbers. Fine, but you haven't shown that the sum is a/(1-r), neither have you defined the term "sum of an infinite series". (Hint: is there really such a thing?) Q:What is 1-epsilon?
A: 0.999999999999999....

Since epsilon !=0, 0.9999999.... ~= 1.000000000000 I'm sorry but I have no proff of the my comment above and it is simpley what think. Please do not cite it but if any one has reason to beleave that it is righte or wrong please tell me. Pedantic:
If |r| < 1 as in my proof, then -1 < r < 1, maybe I'm misunderstanding?

Knut Arne Vedaa:
My intent was not to derive the formula for the sum of a geometric series, it was to show that the series converges and has a closed-form solution. What you are asking is outside the scope of the proof. In addition, it would make it more than one page long :) Your proof is a nice formalization of the basic idea behind the assumed equality of 0.999... and 1, however that idea is sort of trivial. It is easy to see that the limit of that series is 1. However, should we regard the limit of a series to be of the same _cathegory_ as a number, even though the value of the limit is the same as the value of the number? How much free time do you have? to be posting about a stupid rounding error you keep having? 1/3 and 2/3 are just estimates. They are not exact numbers. So the 1/3+2/3 =.999 is just a mathematical estimate. These estimates should not be taken as a proof. This whole site is flawed. I like how Polymath runs to Google groups and copy pastes his response here. :-) This discussion is worth some 2500 odd pages dating back to 10+ years.

And Where I might question this proof is in the exponential qualities.

2 x 1 = 2
2 x 0.999... = 1.9999999...8
4 x 0.000... = 3.9999999...6

Regardless of your theoretical limit on the repeating decimal, the end result will continue to get further and further from a whole number as long as you are not using one. 0.9999999999999999... = 1. there is no question.

Too many people are getting the representation of numbers and the numbers themselves mixed up. Just because a number's representation has a zero and a decimal point does not mean that it cannot be an integer. Integers are whole numbers, and so is 0.99999999999999999... Think beyond the representation.

For what its worth, this tiny debate has rekindled my interest in mathematics. I had a teacher in high school that understood algebra well, but higher algebra and calculus were lost on him, and they were therefore lost on me. Perhaps I should try these subjects again and find a good teacher... Zero, it doesn't work that way.

How can you have an INFINITE number of 9s followed by an 8? You can't.

2 x 0.99999999... = 2. Dear Polymath,

It is obvious to me that you do not understand the concept of infinity. Please brush up on it before you continue to teach math beyond an elementary school level. The problem with your logic is that .9 repeating is not an integer, it is an estimation of a number. While .9 repeating and 1 behave identical in any and all algebraic situations, the two numbers differ fundamentally by an infinitely small amount. Therefore, to say that .9 repeating and 1 are the same is not correct. As you continue .9999999... out to infinity, the number becomes infinitely close to 1, however it absolutely never becomes one, so your statement .999 repeating =1 is not correct. My first reaction is that I am very worried if you are a maths teacher.

None of your theories make sence and you obviously don't understand decimals and fractions and how to convert one to another.

Next you'll be telling us that 99.9999999999999999(infinity)% = 100%

I am very worried for any of your students! I'm writing in defense of .999... = 1.

I think there's some essential problems with the previous proofs. First, Polymath does not claim that .999... the real = 1 the integer. There is a real that is equal to 1, is there not?

Second, infinity is something that is hard to wrap your head around. It behaves in ways that I can only understand some of the time. For example, it blows my mind that the set of even integers is the same size as the set of all integers - but it's true due to infinity. That .333... = 1/3 is one of those cases. Same with the proof that involves 10x = 9.999... In both cases, you don't end up with inequality, or trailing 0s, because these numbers have an infinite number of trailing 3s or 9s. This argument bears a striking resemblance to Pascal's work on conic sections. In that essay (drafted in his late teenage years, discovered in a desk drawer posthumously), he asserts that when the observer's eye is placed at the apex of a cone, all sections created in that cone (circle, ellipse, parabola, hyperbola, antabola, and a single point) will, when the cone is extended infinitely, assume the appearance of a circle. It's a striking observation, but important to recognize that it's a conflation of certain sensible limitations with the more uninhibited exactitude of pure mathematics (eg, Descartes would probably dismiss Pascal's observations). 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.

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