Yes, 1 = 1.000...
In the context of real numbers, this is more accurate and meaningful, because the characteristics of real numbers are completeness and infinite divisibility.
edited 9th Sep '11 8:32:11 PM by abstractematics
Now using Trivialis handle.Rounding has nothing to do with this. Limits are about exact values when dealing with infinite operations.
edited 9th Sep '11 8:36:24 PM by abstractematics
Now using Trivialis handle.and 1=0.999_ isn't.
Very big Daydream Believer. "That's not knowledge, that's a crapshoot!" -Al Murray "Welcome to QI" -Stephen Fry
x3 1 = 0.9 is rounding. Not the case here.
edited 9th Sep '11 8:39:23 PM by abstractematics
Now using Trivialis handle.That's what I meant. You ninja'd me. :P
I did the integrating of 1/x thing back in calc. I know how .999_ is 1.
EDIT: Ninjas! Ninjas everywhere!
edited 9th Sep '11 8:40:26 PM by Enkufka
Very big Daydream Believer. "That's not knowledge, that's a crapshoot!" -Al Murray "Welcome to QI" -Stephen FryNo, it's a limit. You see, .99999... actually means the limit* of the infinite sequence 0.9, 0.99, 0.999, 0.999, and so on. Because you can get as close to 1 as you want, the limit is 1.
- It's actually a little more complicated than that but you don't want me to drone on about Cauchy sequences.
The problem is that most of the times the arguments don't get to leave the basic presumptions. If a mathematical argument is given for one side, there should be explicit reasons to reject the argument.
Now using Trivialis handle.I learned something new today!
This is amazing. Math is amazing.
Also, Erock, you can't have math without logic. If it's a fight between mathematical logic and your logic...math wins.
"You fail to grasp the basic principles of mad science. Common sense would be cheating." - NarbonicA decimal number — any decimal number, terminating, repeating, or altogether irrational — is equivalent to an infinite geometric sum of its digits divided by successive powers of 10.
It just so happens that when all of those digits are 9, that sum as you approach infinite terms is very easy to calculate. And it's 1. Exactly.
edited 9th Sep '11 9:32:40 PM by Pykrete
It's objectively true that 0.999... is exactly equal to 1. (There's no room for argument about that; mathematical facts are absolute like that.* ) It's simply a different representation of the same number. It might seem a little counterintuitive, but it makes perfect sense if you understand how the real numbers are constructed.
Here's one fairly simple way of proving it: Given any two (different) real numbers, one can find a third number in between them. But there's no number between 0.999... and 1, so they must be equal.
The more formal argument, as some people have mentioned above, involves Cauchy sequences and is rather technical, but the essence of it is this: the decimal representation of a real number is shorthand for the limit of an infinite sequence. The decimals 0.999... and 1.000... represent the sequences (0, 0.9, 0.99, 0.999, ...) and (1, 1.0, 1.00, 1.000, ...), respectively.
You can subtract one sequence from the other to obtain a new sequence: (1, 0.1, 0.01, 0.001, ...). As you can see, elements of this sequence get closer and closer to 0; in fact, they get arbitrarily close to 0, which means that the limit of the sequence is 0.
Because of how the real numbers are constructed (which is, as I said, a little complicated), this implies that (0, 0.9, 0.99, 0.999, ...) and (1, 1.0, 1.00, 1.000, ...) are two sequences representing the exact same number. So 0.999... and 1.000... are just two ways of writing the same number. It's really no different than how "one", "1", and "1.0" all mean the same thing.
Does that make sense? If not, consult one of the more intuitive proofs. I'm just trying to convey the formal reasoning behind it.
edited 9th Sep '11 10:02:43 PM by Enthryn
today is cirno day? what?
Very big Daydream Believer. "That's not knowledge, that's a crapshoot!" -Al Murray "Welcome to QI" -Stephen FryIt's basically "9's unite, we are one!"
The standard formal proof is 0.999... = lim n->inf sum [9/(10^n)] (from 1 to n) , where 9/10^n = 1 - 1/10^n, and n approaching infinity results in the limit of 1 - 0 = 1.
Now using Trivialis handle.Hey guys-
You're all forgetting something crucial to this argument, which you would learn or should have learned in calculus.
There is no such number as .999 repeated infinitely many times. There is no actual number with an infinite number of nines after it, because infinity itself isn't a number.
Rather, as the number of nines approaches infinity, the value of .9999 etc. approaches one.
The only way to even achieve something like .9999 infinitely repeated is to use limits, and yes, in limits, you would say it is approaching 1.
So it's not .999=1. It's .999 infinitely repeated approaches 1, which is a totally different mathematical concept.
edited 9th Sep '11 10:23:55 PM by deathjavu
Look, you can't make me speak in a logical, coherent, intelligent bananna.

The same is actually true for rationals like 0.333... because you can't write infinite 3's. However, because it has a consistent repeating pattern, we can use a bar to indicate it in one fell swoop.
In fact, every rational number (quotient of integers) is a repeating decimal.
Now using Trivialis handle.