[split] 0.999... equals 1

[Tiberius]

So i take it, if .3^*3=.9^... and 3*3=9... then 3*3= ten? ^_^ lol ^_^

No. How on earth does that follow?

If you can't get any closer without overshooting... it means that you cannot evenly distribute the 1 into that many pieces. 3's and 4's, 6's, 7's 8's, and 9's do not perfectly go into 10... and I do not see why they should perfectly go into one either. What you do not understand is that this 'proven' math is complete nonsense.

No it doesn't. It means your value is equal to the target. 3,4,6,7,8,9 don't go perfectly into 10, but they do go into 10. Perfection isn't what we are arguing about here. I never claimed that 3 went into one perfectly (firstly because it's impossible since 3 is larger than 1) but it does go into 1...0.3r times.

You tied a concept into a number, I tied a concept into a number. Your .9^ is no more valid than .0^1. This number exists only because .9^ is not one until you add a number with the = # of 0's as you have nines... and then put a one following them. If you have limitless nines... you have limitless zeros. But no matter how many nines you get, you will always need that minute .0^1 to reach one. By the definition of infinite nines: your nines will never reach one.

Your concept is invalid. You cannot have a 1 on the end of an infinite amount of 0s. An infinite amount of 0s has no end for the 1 to go on.

By the definition of infinite nines, there can be no possible gap between 0.9r and 1, and numbers that have no gap between them are equal.

There is no contradiction. There is a limitless amount of zeros, one zero less than the amount of nines, every digit of the way. And wherever your ever-moving last 9 in .9^ is: there is a 1 occupying the equivalent place in the 0^1. You are always adding your digits on at the end of your equation... but I am inserting my infinite zeros from the beginning.

"one zero less" is a limit, thus nulling your argument that this is a limitless amount of zeros. I'm not always adding digits. You can't add digits to an infinite number...it's already infinite.

You mock my understanding of infinity... when you keep rounding infinite nines to one ^_^ Infinite nines will infinitely never reach 1, just as infinite .3^ will infinitely never reach .34. (The point, is that the gap is NOT finite... the gap is ENDLESS! )

I'm not rounding at all. This has nothing to do with rounding. I'm mocking your understanding of infinity because you keep making amateurish mistakes when dealing with the concept, including saying your example is limitless and then proceeded to put a limit on it.

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So does it converge?

Yes, in the same way 2 converges with 2 by its very definition. 0.9 approaches one, as you add more 9s, the number gets closer to one. 0.9r is 1, since there is no gap between them. As I've said before, numbers that have no gap are the same, because there is nothing you can add to them to make them closer. 0.9r + 0 = 1 in other words, which simplifies to 0.9r = 1.

I agree (how can't I?) I can see nothing would go on the end. Only 9's are possible. But this is surely illustrating the weakness of the digital representation of a whole. 1 is simple. 1/3 is simple. 0.999... & 0.333... are inaccurate representations of those.

I never said it wasn't simple, but it's accurate. I fail to see any reason why a complex number is an inaccurate one. Pi is a complex number, but is perfectly accurate, more so than any fraction. Fractions try to represent numbers by comparison to other numbers (i.e. 1/3 is a third of 1, comparing 3 to 1). Decimals state their actual values as a representation of digits.

I get that it's a valid number and works for maths. I think it's the convergence thing I have a problem with logically. If 0.999... & 1 never converge how are they the same?

Like I said, they converge because they are the same thing.