Lucida Wrote:Let me rewrite it.
x = 1 + 2 + 4 + 8 + 16 + 32 + ...
2x = 2 + 4 + 8 + 16 + 32 + 64 + ...
x = 1 + (2 + 4 + 8 + 16 + 32 + 64 + ...) = 1 + 2x
x - 1 = 2x To get to the next line you'd have to subtract x on both sides. So you'd get 2x -x which you reduced to x. This does not work when x is something infinitely large. Infinity minus Infinity is undefined, thus the contradiction later.
-1 = x
1 + 4 + 8 + 16 + 32 + ... = -1?
But yes, you're right; the proof is fallacious because divergent series can't be algebraically manipulated like that.
8/2 is an integer because it hasn't been completely simplified to 4.
0.9999... is not an integer because you can see with your eyes that it is not. But, it is, for all intensive purposes, 3/3. Basically, the 3/3 thing is like putting a negative into a radical.
I feel everybody has overcomplicated things.
In terms of ARE THEY DIFFERENT NUMBERS, AS ON THE NUMBER LINE!!!, as in, the 1 2 3. YESSS GOD. 0.999... is NOT 1 BECAUSE IT DOES NOT SAY 1 IT IS 0.99999... IF IT WERE 1 0.999... should not even exist, but it DOES and it CAN.
Is 99.999...%, 100%? Noooooo, in terms of real life,. Even though technically, there is a difference of 0.000...! IT IS NOT 100% because ITS NOT. ITS 99.999...% BECAUSE IT FEELS LIKE IT.
I know what you're thinking, but.. 0.000... might as well make it 100%, BUT, IT IS 99.999...%, because it is not 100%! ARGHH, I CANT PROVE THIS POINT!!!
YOU KNOW WHAT I MEAN. FORGET ANYTHING RELATED TO MATHEMATICS!!!! DO YOUR EYES SEE 1!!!!!?????!!?!? NOOOOOOOOO THEY SEE 0.999....
WHY DOES 0.999... EXIST AT ALL, HMMMMMM?
In terms of MATHHH AND EVERYTHING RELATED, 0.999..=1 because there is no significant difference between the numbers! Are you really going to care? It goes on infinity and obviously you aren't going to go through the pain of adding or multiplying 0.999... when you could just use 1, which is in no real way different from 0.999... .
YOU GUYS ARE DOING THE EQUIVALENT OF TRYING TO FIND THE END TO INFINITY. STOP. YOUR HEAD WILL EXPLODE.
0.999...=1
In reality, 0.999... is about equal(estimated to)/not 1, BUT NO ONE GIVE A S*** SO WE JUST SAY 1.
Milelke Wrote:8/2 is an integer because it hasn't been completely simplified to 4. 0.9999... is not an integer because you can see with your eyes that it is not.
No, I can't see that. What's the difference between 1.000 and 1? 1 and 0.999...?
Wiki says Integers can be written without fractions or decimals. Using that, 8/2 is not an integer. It is a Real Improper fraction.
8/2 is an integer because it hasn't been completely simplified to 4.
0.9999... is not an integer because you can see with your eyes that it is not. But, it is, for all intensive purposes, 3/3. Basically, the 3/3 thing is like putting a negative into a radical.
I feel everybody has overcomplicated things.
In terms of ARE THEY DIFFERENT NUMBERS, AS ON THE NUMBER LINE!!!, as in, the 1 2 3. YESSS GOD. 0.999... is NOT 1 BECAUSE IT DOES NOT SAY 1 IT IS 0.99999... IF IT WERE 1 0.999... should not even exist, but it DOES and it CAN.
Is 99.999...%, 100%? Noooooo, in terms of real life,. Even though technically, there is a difference of 0.000...! IT IS NOT 100% because ITS NOT. ITS 99.999...% BECAUSE IT FEELS LIKE IT.
I know what you're thinking, but.. 0.000... might as well make it 100%, BUT, IT IS 99.999...%, because it is not 100%! ARGHH, I CANT PROVE THIS POINT!!!
YOU KNOW WHAT I MEAN. FORGET ANYTHING RELATED TO MATHEMATICS!!!! DO YOUR EYES SEE 1!!!!!?????!!?!? NOOOOOOOOO THEY SEE 0.999....
WHY DOES 0.999... EXIST AT ALL, HMMMMMM?
In terms of MATHHH AND EVERYTHING RELATED, 0.999..=1 because there is no significant difference between the numbers! Are you really going to care? It goes on infinity and obviously you aren't going to go through the pain of adding or multiplying 0.999... when you could just use 1, which is in no real way different from 0.999... .
YOU GUYS ARE DOING THE EQUIVALENT OF TRYING TO FIND THE END TO INFINITY. STOP. YOUR HEAD WILL EXPLODE.
0.999...=1
In reality, 0.999... is about equal(estimated to)/not 1, BUT NO ONE GIVE A S*** SO WE JUST SAY 1.
In end, 0.999.. is 1 and it is not 1.
Seriously? Essentially you are saying it is not one because you believe it isn't. .999... is only writing one in a different form. Is 4/4 1? Yes. Is 5/5 still 1? Yes. You are just writing the number 1 in a different form. Even if they look different to your eye doesn't mean they are actually different. Read the thread/proofs more till you understand.
I know nothing of math, and dislike it. Mind my ramblings.
3/3=1
0.999...=3/3
Sigh, if a=b and b=c, then a=c, therefore 0.999...=1
Anyway, I never said (in the last post, not including previous) it wasn''t.
I just said that we really shouldn't care.
... Wait, Cancambo, 0.999 is not written in fractions, you used fractions as an example, however, 0.999... is a decimal. But in fraction from its 3/3... but...
Argh, I don't know. I'll keep my opinions to myself.
It's correct, they're one and the same, it's just the representation of real numbers that people have trouble conveying. Instead of using integer as the base, it could easily be represented by 0 and 0.999... as the endpoints, then 1 itself wouldn't have existed, or like, counterargument for this again. It's like cutting a pie into pieces, when you sum the pieces with the lines, you get 0.999..., without the cuts, you get 1. They're one and the same. 1/9 = 0.111..., 2/9 = 0.222..., ... , 9/9 = 0.999... and 1.
Kabanaw Wrote:No matter how many figures you go to in any real life application, it will always round to 1. So for any practical purposes, .99999....=1.
It's not just for practical purposes. it IS equal.
Since the number line is ordered numbers can be compared. If a and b are two different real numbers, then there exists at least one real number between them (actually infinitely many of them). (a+b)/2 is the easy way to find such a number. Otherwise if the average of a and b is either a or b then a = b. Which is what occurs in the case: 1.99999.... divided by 2 gives you .999999999..... and it will never stop. Thus .999999....=1.
Or how about this, since there has to be another number in between two real numbers: Attempt to construct a real number bigger than .999..... but smaller than 1. It's impossible. Thus .99999......=1.
Lucida Wrote:0.999... is an alternate representation of 1, just as 8/2 is an alternate representation of 4. Visually different, but the same number.
One can also "simplify" 0.999... (which is really an infinite series, 9*10^-n from 1 to infinity) to 1.
It's tempting to say that 0.999... is an infinite geometric converging series, but it's actually not. It it, however, best explained as the result of the infinite geometric series ,
which is 1 (or 0.999...).
If you're stuck on the fact that 0.999... can't be equal to 1 because it forever gets closer and closer and eventually has to stop sometime, think of it this way.
You have a number which is infinite. It's a number that is forever increasing and never, ever stops increasing.
In a sense, what you're doing is stopping the number at 1,394,194,103 and saying, "See? It stops there. So it can't possibly get any closer to infinity." But it can get closer. By stopping an infinite number at a finite number, you are contradicting what infinity really is. The whole point of infinity is that it never stops increasing.
Horusmaster Wrote:Just realized in sets, it's different:
[1,0] is a closed set. In this set all real numbers between 1 and 0 are included.
(1,0] is an open set. In this set 1 is not included, but .999.... is.
find it through google.
Sorry for reviving this thread, but I'm baffled by this. My Calculus teacher, a Ph.D. in Mathematics, told me today that the closed set [0,1) has no upper bound. Specifically, I asked him the question, "what is the highest possible number that can fit in that set"? He replied,"there is no highest number". Then I proposed that the highest number might be 0.999...9! and he replied that it isn't, because it equals 1, and that trumps the definition of the closed set.
2147483647 Wrote:Sorry for reviving this thread, but I'm baffled by this. My Calculus teacher, a Ph.D. in Mathematics, told me today that the closed set [0,1) has no upper bound. Specifically, I asked him the question, "what is the highest possible number that can fit in that set"? He replied,"there is no highest number". Then I proposed that the highest number might be 0.999...9! and he replied that it isn't, because it equals 1, and that trumps the definition of the closed set.
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That's because he wasn't correct when he stated that .999999..... is in the open set [0,1).
.99999..... is 1. It's not in [0,1)
Also, what definition of upper bound do you guys have 0.o
In real analysis, an upper bound is any number bigger than or equal to everything in the set. So any number bigger than 1 is an upper bound for both [0,1] and [0,1).
The least upper bound would be the minimum of all upper bounds. Which in both cases would be 1.
[0,1
shouri Wrote:That's because he wasn't correct when he stated that .999999..... is in the open set [0,1).
.99999..... is 1. It's not in [0,1)
Also, what definition of upper bound do you guys have 0.o
In real analysis, an upper bound is any number bigger than or equal to everything in the set. So any number bigger than 1 is an upper bound for both [0,1] and [0,1).
The least upper bound would be the minimum of all upper bounds. Which in both cases would be 1.
[0,1
I was always taught that 0 <= x < 1 is equivalent to [0,1). So yeah, there's no largest number in that set.
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