[split] 0.999... equals 1

[Violet]

Edit: I should note before i begin, that there are some people who do not think you can add or subtract an infinite (see

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As in real analysis, in complex analysis the symbol , called "infinity", denotes an unsigned infinite limit. means that the magnitude | x | of x grows beyond any assigned value. A point labeled can be added to the complex plane as a topological space giving the one-point compactification of the complex plane. When this is done, the resulting space is a one-dimensional complex manifold, or Riemann surface, called the extended complex plane or the Riemann sphere. Arithmetic operations similar to those given below for the extended real numbers can also be defined, though there is no distinction in the signs (therefore one exception is that infinity cannot be added to itself). On the other hand, this kind of infinity enables division by zero, namely for any complex number z. In this context is often useful to consider meromorphic functions as maps into the Riemann sphere taking the value of at the poles. The domain of a complex-valued function may be extended to include the point at infinity as well. One important example of such functions is the group of Möbius transformations.

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a few math teachers I know personally [and discussed the concept of infinity with] among them. Even the concept of infinity is not accepted by every mathematician as having the same definition or even in being possible... so how could all mathematicians be on board with the idea of .9^=1? Be aware that not everyone even agrees on infinity's concept in math... let alone on a specific concept. For the full context of the wikiquote: http://en.wikipedia.org/wiki/Infinity

1.79, 1.799, 1.7999,... We could observe that the numbers are "approaching" 1.8, the limit of the sequence.

I did read it... and I do not see how it is equal to 1.8. The point of the article is simply this: .9^ approaches 1... but never reaches it. It is http://en.wikipedia.org/wiki/Infintesimal far away from 1. Proof that infinitesimals exist (according to the wiki):

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Proving or disproving the existence of infinitesimals of the kind used in nonstandard analysis depends on the model and which collection of axioms are used. We consider here systems where infinitesimals can be shown to exist.
In 1936 Maltsev proved the compactness theorem. This theorem is fundamental for the existence of infinitesimals as it proves that it is possible to formalise them. A consequence of this theorem is that if there is a number system in which it is true that for any positive integer n there is a positive number x such that 0 < x < 1/n, then there exists an extension of that number system in which it is true that there exists a positive number x such that for any positive integer n we have 0 < x < 1/n. The possibility to switch "for any" and "there exists" is crucial. The first statement is true in the real numbers as given in ZFC set theory : for any positive integer n it is possible to find a real number between 1/n and zero, but this real number will depend on n. Here, one chooses n first, then one finds the corresponding x. In the second expression, the statement says that there is an x (at least one), chosen first, which is between 0 and 1/n for any n. In this case x is infinitesimal. This is not true in the real numbers ® given by ZFC. Nonetheless, the theorem proves that there is a model (a number system) in which this will be true. The question is: what is this model? What are its properties? Is there only one such model?
There are in fact many ways to construct such a one-dimensional linearly ordered set of numbers, but fundamentally, there are two different approaches:
1) Extend the number system so that it contains more numbers than the real numbers.
2) Extend the axioms (or extend the language) so that the distinction between the infinitesimals and non-infinitesimals can be made in the real numbers.
In 1960, Abraham Robinson provided an answer following the first approach. The extended set is called the hyperreals and contains numbers less in absolute value than any positive real number. The method may be considered relatively complex but it does prove that infinitesimals exist in the universe of ZFC set theory. The real numbers are called standard numbers and the new non-real hyperreals are called nonstandard.
In 1977 Edward Nelson provided an answer following the second approach. The extended axioms are IST, which stands either for Internal Set Theory or for the initials of the three extra axioms: Idealization, Standardization, Transfer. In this system we consider that the language is extended in such a way that we can express facts about infinitesimals. The real numbers are either standard or nonstandard. An infinitesimal is a nonstandard real number which is less, in absolute value, than any positive standard real number.
In 2006 Karel Hrbacek developed an extension of Nelson's approach in which the real numbers are stratified in (infinitely) many levels i.e, in the coarsest level there are no infinitesimals nor unlimited numbers. Infinitesimals are in a finer level and there are also infinitesimals with respect to this new level and so on.
All of these approaches are mathematically rigorous.

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As you can see, infinitesimals exist as much as infinity... therefore my use of 0.0^1 (An infinitesimal number)... which you stated does not exist: essentially exists as much as .9^, which is an infinitesimal distance from one. Please do not insult me by suggesting I did not read the articles I used as evidence... esp. since I found them to be rather fascinating (therefore insulting what I find fascinating).

Also, I do not think wikipedia is an entirely trustable source... http://en.wikipedia.org/wiki/Wikipedia#Reliability

Do you not think that if this were some kind of disproof, the 0.999 page at Wikipedia would mention it? No. Instead it shows 2 elegant proofs (which I repeated here with the addition of another) of how 0.9... = 1.

So yes, I think it is possible that a disproof might not be mentioned. I was not convinced by these pages of proofs, and so I will actively hunt down each one and disprove it. In previous posts, i just suggested that the notion of this inequality being equal was laughable... now I am giving evidence to back up my previous laughter.

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I'll start with the first one.

No it doesn't. 1 is provably 0.999... (reoccuring)

Here is my favourite proof:

x = 0.999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9
x = 1

So x = 0.999... = 1

Fine the error in my math if you want, but you won't since this thing has been known by mathematicians for years

You removed the infinite from the equation when you subtracted it from itself. Essentially, you have cancelled out the effect of infinity. Your problem becomes this:
x=.9^
10x=9.9^
9.9^-.9^=9 (You will notice that you have cancelled out infinity in this step... therefore infinity is no longer a part of the equation.)
9(new x)=9
New x=1

Essentially... this 'proof' is invalid, because you have changed the value of X by canceling out the effect of infinity upon the number. As in, you are no longer proving that the first value of X (.9^)is equal to one... but that the new value (1) is. You changed the value of X when you canceled out part of the equation. Simply, 1-1=0... just as infinity - infinity=0. You negated its effect in your calculation, so that you could solve it without infinity's effect.

You accept that 2*10=20, 20-2=18, no? Then how come when we take 2*9, it is ≠ 18? Essentially, that is what is being done in this 'proof'. Registering different answers for 20-2 and 2*9 is a fallacy under finite circumstance... and is only (and according to some people, not even then) possible when eliminating an infinite or infinitesimal or other undefined number. For example, 10x2.2^-2.2^ would not be equal to 9x2.2^.

x=.9^
9x=8.9^1
8.9^1/9= .9^

Once you nullify the infinite part of x, then you are no longer proving .9^. You are proving a finite by canceling out the infinite. By doing so, you get a different answer for 10x-x, and for simply 9x. If this were a finite number to begin with: you would not get different results from those two normally equivalent expressions. I do not accept this 'proof' of an infinite equalling a finite... for it is only by the canceling of the infinite concept that the inequality can 'solved'.