*Originally posted by ryans *

**Lethe settle the arguement.**

i would very much like to. i am greatly vexed that this issue is still around.

I have no doubt that 0.999. is an infinite regression of 9, but does that constitute it being rational, in light of the defintion of an irrational number I gave above.

this definition: "

**A number which cannot be expressed as a fraction for any integers p and q.**"? this implies that 0.9999.... is rational only if you believe that 0.9999.... =1. if you don t believe this fact, then i will have to convince you of that fact first. and frankly, i m not inclined to do so.

the other criterion for irrationality of a number, namely that its decimal expansion be nonrepeating, needs to be proved to be equivalent to your more fundamental definition. to do so requires that you believe in the correctness of the decimal expansion. basically it requires that you believe 0.9999.... =1.

Is this definition correct?

yes.

As for your arguement that 0.999...=1, would that not mean that every element in the continous number line would be equal to the next element,

next element? what does that mean? between any two real numbers, there are an infinite number of other real numbers. how do i define the "next" number after 0, for example? this notion only makes sense for well ordered sets, which the real numbers are

**not** (at least, under the standard definition of <. the axiom of choice claims that the reals are a well ordered set, but such an ordering cannot be constructed explicitly)

let me try to extract what you actually mean with this statement. every number of the form 0.abcdef9999999.... where a, b, c, d, e, and f are digits is equal to 0.abcde(f+1). is that what you mean by every number is equal to the "next" number? if so, then it is true. but rather uninteresting.

and as such, each element equal to every other element.

nonsense. every number is represented by an

**equivalence class** of decimal representations. two decimal representations are equivalent if they converge to the same thing. so 0.99999.... and 1 converge to the same number, but 0.5 and 1 do not. therefore they are different numbers.

I have constructed a primitive proofusing the sum

0.999....=(9/10)SUM(10^-n) where the sum is from n = 0 to infinity.

a proof of what? that every number is equal to every other number? that 0.9999.... is irrational? that 0.99999... != 1? let me see your proof. if you prove something that is false, then your proof is incorrect, i can tell you right now.

Also if 0.999....=1, then this raises the issue of series convergance. Take for example the sMaclaurin series of 1/(1-x) which only converges for values of x less than or greater than 1 or -1 respectively. For 0.999... this series converges, for one it does not.

uhh.... for 0.99999.... this series converges to 1. for 1 it also converges. observe: 1=1.000000000.... so the ratio of the series is 0. 1/(1-0)=1. this statement is also nonsense.

since, by definition, a decimal representation carrys a factor of 1/10 (which is less than 1), there are no issues of convergence, in any decimal representation.