Bryan,
You haven’t even told me what 0.999~ is yet, let alone that it’s a number not equal to one. In fact, if you assume that 0.999~ < 1, you can get a pretty big contradiction.
Suppose that 0.999~ < 1. Then e = 1 - 0.999~ is a positive number. Hence, there exists an integer N such that 1/10^N < e. Now it should be clear that 0.99…9 (N + 1 9’s) < 0.999~. But this implies that e = 1 - 0.999~ < 1 - 0.99…9 (N + 1 9’s) < 1/10^N <= e. Once again, we get e is not itself, which is a contradiction.
Not if N is infinity.
If you insist…
The way the real numbers are constructed from the rational are as equivalence classes of Cauchy sequences of rational numbers, equipped with termwise addition and multiplication. It’s a result in Analysis to show that this structure is an algebraic field and has the familiar order properties of the real line, as well as being complete.
…
This the same definition I gave using more complex terminology.
You’re right, to this extent: if you accept axiomatically the existence of an infinitely large number, you can build a system in which infinitesimal numbers exist, and in such a system the series 0.9 +0.09 + … will converge to a number less than one by an infintiesimal. You can even say which infintiesimal.
This is why .999~ < 1. If you put a cap on the largest number, whatever the sum of .9+.09.+.009+.0009… iterated the “largest number” of times, the final result will be less than 1.
But what is the ‘smallness’ exactly? Presumably it’s some number-- you appear to be postulating this ‘something’ as existence without identity here. If a number is larger than zero it has to be larger than zero by some amount. What amount is it?
The “something” is whatever units your are working with. The actual unit you use is arbritrary.
Okay, well let’s work with this. Let’s define f(x) = (1 + x)/2, as you’ve done above, and consider the sequence of numbers
f(x)
f(f(x))
f(f(f(x)))
…
You claim that the “final answer” a = f(f(f(…))) is equal to 0.999~, right? Okay, well clearly by the definition of a, we have f(a) = a. Therefore, (1 + a)/2 = a, whence 1 + a = 2a, so a = 1. QED.
No!!! The final answer is dependant the cap you put on the number of iterations. No matter what the cap it is, the final answer is less than 1!.
Such a number doesn’t exist. I gave you the proof of that already. If you can find an incorrect step in that proof, please show me.
The incorrect step that there is not an infinite number of steps. I understand that it can be demonstrated that there always a number that between two numbers, in fact there are an infinite number of numbers between any two numbers. This is beginning to fall into the dead horse beating catagory.
Nothing you have demonstrated shows that .999~ is equal to 1. Everything I have demonstrated shows that it is less than 1.