1.999 repeating = 2

Well, for this last Round I'll just respond to some of the comments that have been posted down. Lazarus Long indicated that two integers whose ratio give 1.9999... repeating are 4/2. The proof that he gave is the following. "Since 1.999... = 2, then 4/2 = 1.99999.... Simple, no?" The only simple thing I see in this proof is realizing how flawed it is. I'll just change one number so that people can realize how foolish this "proof" is: Since 500 = 2, then 4/2 = 500 Simple, no? Then he claims the following: "how else would you explain: 1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3 = 2 which is the same as 0.333... + 0.333... + 0.333... + 0.333... + 0.333... + 0.333... = 2, which, if we simply do the addition, IS 0.999... + 0.999... = 2 which, of course, requires that 0.999... = 1 (actually the same as the question at hand here), so Q.E.D. " Well, the way I already explained why the "proof" written above is wrong is because addition, substraction, multiplication, and division cannot be performed directly on repeating representations of numbers. I showed this on the Round 4 of the debate. Therefore, 0.333... + 0.333... + 0.333... DOES NOT EQUAL 0.999... Another comment written down was made by beem0r with respect to the points I made in Round 4 are the following. Beem0r wrote: "I think we should realize that it isn't the 8 that's repeating, it's an 18 that's repeating. The 1 in the tens place will always change the previous 8 to a 9. You can do addition and subtraction with repeating numbers, you just have to know what's repeating. .999... + .999... is 1.999..." The fact that he has to "realize" what is repeating proves my point that addition and all the other operations cannot be performed directly with repeating representations of numbers. I do understand what he is saying, so I'll rephrase it here. What he is saying is that the 8 at the end (end?) will constantly be changing to a 9 because of the 1 that is carried from the sum of the nines on the column that is directly to its right. So the number that we will have is 0.999...(infinite nines)...9998 with the eight constantly and forever changing to a 9. This is why addition and the other operations cannot be performed on infinite representations of numbers, because this operations are only DEFINED on finite representations of numbers. Therefore the only way to make algebraic operations on inifinite representations of numbers is to write them in a finite form. This could be done either by writing them as a fraction or representing them by a variable (or symbol, as is the case for Pi). Since 1.999 cannot be obtaind by dividing any two integers (which means it cannot be written as a fraction) then the only way to do algebra on it is by representing it with a variable. The proof that 1.999 repeating cannot be written as a ratio of any two integers is written in Round 1 and re-written in Round 3 of the CON side. Finally, I want to indicate that many people (including many mathematicians) don't really understand the concept of limits and infinity. For example, the limit of 1/x as x goes to inifinity equals zero. What this means is that as x goes to infinity the function will get closer and closer to zero, but it will NEVER equal zero. Many mathematicians make the wrong claim that, "at infinity the function will equal to zero". This statement is wrong because it assumes that there is a point where the function will actually equal zero. But then I ask, after it reaches zero then what? Does it remain being zero, does it become negative? The point that I am trying to make here is that infinity is not a point that can be reached, infinity just means that it will keep on going forever. Therefore 1/x will get closer and closer to zero forever, but it will never equal zero. The best we can do when dealing with infinity is to use limits to indicate what value some function will get closer to, but that is as good as it gets. This means that 1.999 repeating is by definition the number that is closest to 2 from the numbers that are smaller than 2.