If you actually go to the wikipedia page, there's a whole section about why lots of maths students don't believe that .999... = 1. Doesn't change the fact that mathematically, they are the same number.

Haha, I know limits

There's lots of debate about infinite and the sorts, and this kind of falls under that category. Maybe our mathematics system proves that they are the same number. But theoretically, I'm on the side that says they're not the same number.

Wow. All of this just from my half-assed attempt at contradicting people. Do I get to be ranked with Kame now?

Kame makes fun topics and amuses us. You just waste time.

Well, or do useful stuff. You do useful stuff. But your trolls are just bad.

Quote from Centreri

But your trolls are just bad.

They work, though. Look at what a couple minutes and remembering something my 6th grade math teacher showed me did.

What about binary?

01001101011000010111010001101000001000000110100101110011001000000110000100100000011011000110000101101110011001110111010101100001011001110110010100101110

this actually means something.

It was an invalid proof as it made the assumption to prove it. Your teacher steered you wrong.

Which is equal to zero, thus making the two quantities equal.

I think you you're wrong here. This is what the first number of the range ]0...1] would be.

The culprit is the assumption that 0.999... would be off by 0.000[infinite zeros]0001 which is wrong because the 9s/0s go infinitely. There's no "space" so to speak for the 0001 to be placed at the end.

This is hard to understand for a human being as infinity is something we cannot imagine.

1 does equal .9...

What's your point?
No, the ellipsis means 8 repeats ad infinitum, which implies that it is exactly 8/9.
In real life applications, you don't write .999..., you write 1. Because they're the same number.
There's no debate.

To be distinct real numbers, there has to be an infinite amount of other real numbers between them. That's one of the definitions of a real number.

Name one, let alone prove an infinite amount, between .999... and 1.

Do you consider 1.0... and 1 the same number?

Post has been edited 1 time(s), last time on Mar 25 2010, 11:09 am by JaFF. Reason: Little by little, but you got there.

Yeah, you might be right... there's a reason I don't do math

Leeroy, I think the error you're making is, you're assuming that an infinitely long series of numbers cannot add up to a finite number. Infinite sequences can have finite values, however.

http://www.staredit.net/218644/
Maths > Physics, period.

Math > physics because if math = i, then there is a vertical asymptote at x = physics (without math, physics is nigh useless).
So sure, math > physics, but did you learn how to derive in one day?

In physics, I did. In math, however, I still haven't even gotten around to getting taught derivatives, and I got points taken off because I used one in a presentation I had to do in math. She pulled me aside after class and basically went on a rant that she's the teacher, and that when she wants people to use stuff they won't learn until next school year, she will let me know. After she cooled down, she mentioned that she took off points for writing lim a → 0 only once, and said that the explanation was alright.

No but really. It's an impossible number to have recorded anywhere. So what makes it real? It's a theoretical statement saying if this were to go on forever then it would equal 1. But forever doesn't exist...

SImply because we can't write out the number .999... in decimal notation does not mean we can not record it. We can simply say, let X be a representation for the real number in decimal notation which consists of infinitely repeating 9's. It's no different from saying "let X be equal to 3" or let "3" be the quantity which contains one plus one plus one objects. If you believe "3" is real, there's no reason not to believe "X" is real.

In math terms, it's easy to see that 1 - .999... = 0.

Here's how it works, we play the "epsilon" game:
Take any value for epsilon (I'll just call it e) greater than 0. e can be 1, 57, 100000000, .001, .23456789, .0000000000000001, .000000000000000000000001, ect.
Then we can define a sequence, such that a(n) = 9*(1/10^n). The nth term of a, is = .000(n-1 zeros)0000009 (nth digit is 9)
So, if we sum the first k terms of a(n), we have .9999(k 9's)9999
And obviously 1 minus .99999999(k times) = .000000001 (1 on the kth place).
So now we play the epsilion game. If you give me an epsilon, I'll give you the value of k such that 1 minus the sum from 1 to k of a(n) is smaller than your epsilon. And what you'll see, is that for any value of epsilon, no matter how small, I can give you a value for k. What this means, is that the difference between 1 and .999... is not a positive value. Distance isn't negative either, if it was negative we could just take the absolute value of it to make things easy, so the distance between .9999... and 1 must be zero.

Another easy way is to say 1/9 = .11111..., 9 *1/9 = 9*.111111... = .9999999... = 1