kazzin

0.999 x 5 = 4.995 0.999... x 5 = 4.999...

Why use all these algebraic proofs when basic analysis of the reals tells us that it must be so?

No, because the majorities ruling does not matter to me; I meant detest, as in, I I disagree immensely with your idea that logic cannot be overturned. I agree, that may not be done here but that doesn't change my stance. I see and understand why you believe that .999... is equal to one and the theorems regarding rational and irrational numbers. I do not agree however, that .333 or .999 extending in any direction is equivalent to any fraction; the fraction being the more accurate of the two. This explain all of these proofs. The number extends to infinity with 0 in front of it. It eventually will reach a value of one in any given decimal place should this infinity cease. Which, although it doesn't, would be greater than zero. The number you are trying to visualize is, in fact, 0. The number at the end is greater than a value of zero. What end? There is no end to an infinite string of 0's.

The number extends to infinity with 0 in front of it. It eventually will reach a value of one in any given decimal place should this infinity cease. Which, although it doesn't, would be greater than zero. The number you are trying to visualize is, in fact, 0.

If 1/3 isn't equal to 0.333..., then clearly the argument is that 1/3 > 0.333... If that were true, then 1/3 - 0.333... > 0, that is that the difference between them is some positive real number. Define this number. You can't--it's zero.

How about instead of joining just to be an annoyance and a jerk you could actually pitch in the conversation? Oh, I'm so very sorry sir. I guess you missed all my other posts in this thread.

Consider the open interval (0,1) in R. 0.999... is obviously not in this set (0.999... is not an interior point), and thus 0.9... is greater than every number less than 1, and smaller than every number larger than 1. Ergo, 0.9.. = 1

If 3+3+3= 9, how is .333...+.333...+.333...= any different?Because .333... is 1/3, and 1/3 + 1/3 + 1/3 = 1No, it isn't. Any number of threes added vertically above each other will result in any number of nines. One third can only be represented as a fraction.Every real has an exact decimal representation. People just get confused by the fact that not every real has a unique decimal representation.Note the word "representation". 0.999... represents 1, it does not equal 1. That's just silly.

If 3+3+3= 9, how is .333...+.333...+.333...= any different? Because .333... is 1/3, and 1/3 + 1/3 + 1/3 = 1 No, it isn't. Any number of threes added vertically above each other will result in any number of nines. One third can only be represented as a fraction. Every real has an exact decimal representation. People just get confused by the fact that not every real has a unique decimal representation.

Seriously. If you don't think 1/3 = 0.3... then you probably should've paid more attention back in 4th grade when you were learning long division.

But it eventually doesn't stop at one. It rests when the number beings subtracted stops, which it doesn't. So then I don't see how that equals one, should both stop at any given point, its value will equal ...1. No idea what you're talking about. You may find it useful to attempt to define 1 - 0.9..., however. That is, define the difference between 0.9... and 1.

Look, the Reals form an ordered field, which most importantly has the property that given x and y from R, x is either less than y, equal to y, or greater than y, where x < y is taken to mean that y - x > 0. Obviously no one will argue that 0.999... > 1, so let's assume 0.999... < 1. Then 1 - 0.999... > 0 But what is 1 - 0.999...? Well, the greatest lower bound of the set {0.1, 0.01, 0.001,...} of course, which is clearly 0. That is, 1 - 0.999.. is SMALLER than any positive real number, and thus 1 - 0.999... = 0. Thus we conclude that 0.999... = 1, and we are done.

It can't? 1/3 is a third of 1. I assume you're suggesting that 1/3 != 0.333..., in which case I have to ask what exactly you think the decimal expansion of 1/3 is? And yes, every real has a decimal representation.

No. Your math teacher doesn't understand math very well, and that's why he's teaching high school instead of getting his PhD.