Haha, I just realized how much of an idiot >>119 really is.
>>It can be described as a limit LIM(n -> infinity)[.99~~]-> 1 where n is the decimal places
That's actually correct, but he thinks that's an argument against .999~ = 1.
Guess what? "LIM(n -> infinity)[.99~~]-> 1 where n is the decimal places" is exactly the same as .999~. This equals 1. Congratulations, now go back to high school math.
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Anonymous2006-07-04 5:09
1 is better than 0.999... because 0.999... has to converge to be 1, but 1 doesn't have to converge to be itself.
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Anonymous2006-07-04 5:49 (sage)
>>124
0.999... doesnt have to converge to be 1. It IS 1 already.
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Anonymous2006-07-04 6:21
>>125
Exactly. 0.999... IS 1, there's no fucking limits or convergence involved here. So stop saying that, you goddamn idiots. The convergence of a limit is a tool to prove that this decimal expansion is correct, but it doesn't actually have anything to do with the expansion.
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Anonymous2006-07-04 17:24
Well then what you are doing is starting with 1 and then partitioning it indefinitely, not starting with 0.9 then adding 0.09 then adding 0.009 and so on to magically create 1 without benefit of limit. You cheater.
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Anonymous2006-07-04 22:22
I can't believe this thread is still going on. 0.9999... = 1
The End
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Anonymous2006-07-07 12:53
now this is very simple:
0.999999999....=0.9999999999...
1=1
simple mathematics
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Anonymous2006-07-08 3:41
if 1 = .999...999
then .999...999 = .999...998
then .999..998 = .999..997
...
then .999...067 = .999...066
then 1 = .999...066
then 1 = .999...XXXXXXXXXXXXXXXXXX
etc.
so if 1/infinity doesn't matter at all, like the notation 1 = .999... indicates, then an infinite number of infinities wouldn't matter. Thus, if 1 = .999..., then 1 = .5, or anything else.
CORRECT ANSWER AHEAD:
1 does not equal .999...
But in all practical mathematics useage, they both work the same, and can be interchanged. But they are NOT the same.
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Anonymous2006-07-08 4:02
It's not cause it's getting rounded off to a higher number?
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Anonymous2006-07-08 17:14
It's not that complicated.
.9999~ = .9999~ while 1=1.
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Anonymous2006-07-08 18:05
Ok, let's get serious here. 0.999... = 1. An "easy" way to see it, that does not involve any of the proofs shown above, is this:
Theorem: 0.999... = 1
Proof: Let R be the set of real numbers. R is know to be an Dedekind-complete ordered field. Since R is a field, addition and multiplication are defined with their usual properties; since a field is a group under addition, it follows that R is an ordered group, and therefore it defines an uniform structure. An uniform structure is complete, and therefore for every two members a,b in R, there exist infinite members (denoted by s) s such that a < s < b. Since there is no real number s such that 0.999... < s < 1, it follows that 0.999... = 1. QED
The idea behind the math: if two numbers are different, one can always find a number in between them. You cannot find a number in between 0.999... and 1. Therefore, 0.999... = 1
PS: (1 + 0.999...)/2 does NOT does the trick; if you evaluate the division correctly, you find 0.999... once again.
Following that logic
1 = .999...
and
.999...9 = .999...8
therefore 1 = .999...8
But the number .999...9 is between 1 and .999...8
So 1 can't = .999...8
But if A = B and B = C, but A does not = C, then there has to be a problem with the logic system, no?
Well, the reason that there's nothing between 0.999... and 1 is because 0.999... gets infinitely close to 1, ie the limiting value is 1. It doesn't have to follow that 0.999... means the same thing as 1.
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Anonymous2006-07-09 4:18
>>135
Fuck you're stupid. Writing .999...8 makes no sense. You can't tack an 8 on the end, THERE IS NO END. STAB YOURSELF IN THE FUCKING FACE PLEASE KTHX, GOD YOU IDIOTS PISS ME OFF.
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Anonymous2006-07-09 15:43
First of all, just a quick note: the transitivity of equality (A = B and B = C -> A = C) DOES hold in the reals, but it does NOT in many other sets.
>>138
No, that's the WHOLE difference between the rationals and the reals. In the reals, "be infinitely close" is the same as "be the same". That follows from the completeness of the uniform structure.
>>135
Explain me exactly HOW do you intend to equate 0.999... to 0.999...8. in crude terms, you can't really put something "after an infinity of 9's". However, I'm willing to give you the chance: how do you prove that 0.999... = 0.999...8 ?
Really guys, I'm make it a little clearer now: in the reals, there is NO SUCH thing as the "last number before a number". Between any two different numbers, there is an infinity of other numbers. Think a little about the following related question:
Which is the smallest number greater than 0?
Remember that 0 = 0.000... Is any of the guys talking about "0.999...8" willing to say that the smallest number greater than 0.000... is 0.000...1?
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Anonymous2006-07-09 16:50
what the fuck are you talking about? equality is transitive by definition
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Anonymous2006-07-09 18:21
>>140
Remember that 0 = 0.000... Is any of the guys talking about "0.999...8" willing to say that the smallest number greater than 0.000... is 0.000...1?
Well, I guess I would. Although I might argue that 0.000...1 isn't even an actual number, and is more of a concept, if even that.
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Anonymous2006-07-09 19:17
>>142
Why, you could call that concept the 'infinitesimal'. Ingenious!
Perhaps, if you added these to the real line, you could formulate a sort of 'non-standard' analysis...
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Anonymous2006-07-09 20:51
0.999~<1
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Anonymous2006-07-09 21:21
0.999~>1
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Anonymous2006-07-10 3:20
>>143
Just because you can name a concept doesn't mean it can actually be concieved
Hay guys
This guestion has no answer. It is both = 1 and < 1, depending on what you are trying to do with it
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Styrofoam2006-07-11 2:43
God damn it, you stupid fucks are all god damn retarded. You know what? Shut the god damn fuck up about infinitesimals, the number line, and analysis, even though those are all valid ways of figuring out that 0.999~ = 1. Shut the fuck up, just for a second, and read this. Read it slowly.
Let:
x = 0.999~
Multiply both sides by 10.
10x = 9.999~
Subtract 0.999~ (which is equal to x) from both sides.
9x = 9
Divide both sides by 9.
x = 1
We began by stating that x = 0.999~, therefore
0.999~ = 1
There. Motherfucking algebra. No analysis, no calculus. Just motherfucking algebra. Go home and cut yourself.
.9999999 does not equal 1. It equals .9999999.
0.9(Dash above the 9) on the other hand, DOES equal 1. 1/3=0.3(Dash above the 3) times that by 3 and you get 0.9(Dash above the 9) And 1/3x3 equals 1. Think of the same rules as infinity, except with decimal places. But, .9999999999999(Put as many 9s as you want) does not equal 1, because .999+0.001 equals 1. It needs the dash ontop of the 9 to indicate that it has infinite decimals.
Hey, re-re. Algebra laws fall apart when you are dealing with infinities. Pretty much all laws fall apart when you are dealing with infinities. Don't they teach this shit in middle school anymore? What the hell. This is why the problem has no real answer, because you are trying to compare a simple number to an abstract concept. Just give it up already.
But it is debated. Why? Because there are really no definate laws when dealing with infinities. 1 * 2 = 2 is definate. It is not debated. I propose that the reason .999... = 1 is accepted is because there are no math problems where a 1/infinity amount is going to make a difference.
But conceptually, .999... and 1 are very different things. And in at least that respect, they are not the same number.
If you read the fucking proof, it doesn't matter that it's an infinitely long decimal. The first operation performed is multiplying 0.999~ by 10. This clearly produces 9.999~ Then, you subtract off the quantity 0.999~. That's it.
It's debated? That's fucking news to me. Oh wait, you mean it's debated by little 14 year old shits on the internet who know nothing about mathematics. Oh, and there are plenty of definite(gb2/spelling/) laws about infinities. Learn Calculus.
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Anonymous2006-07-12 2:20
>>156
Again, you are trying to solve a problem with infinite places with algebra. I'll say this again, you simply can not do that. Algrebra breaks down when dealing with infinity.
Let me put it this way. We are told in our first math class that infinity is not a number. If you don't accept this, you are a fool. Need more? A 1/0 slope is undefined. Why? Because you can't define an infinite slope. Because normal rules don't fucking work with infinity.
Since the .999 is followed by an undefinable number of 9s, there is no way you can apply math operations to it. This includes .999... * 10, or anything else you can think of.
Thinking about calculus? Well, you have a point. Or would, I suppose, if it weren't for the fact that calculus is, by it's very nature, an estimation (who the hell doesn't know this?). Why do you think they word it like "the LIMIT of f(x) as x APPROACHES c is L"? Because you can't prove that f(x) = L for any x. Why? Because they never actually meet.
Thanks for proving you know nothing about mathematics. If you actually managed to take a Calculus course like you imply, I suggest you take another. Either you are a fool, or your teacher was, and either way you have lost all credibility in this conversation. Leave now
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Anonymous2006-07-12 2:41
>>157
0.99999… is finite not infinite so whatever the fuck you are talking about fails.
Calculus an estimate? Go back to learning more basic maths dumbass.
ONOZ IT HAS INFINITE DECIMAL PLACES LOOKS LIKE WE CAN'T DO SHIT ABOUT IT, SORRY
Except as a repeating decimal, it is a rational number. Get that, shithead? 0.999~ IS A RATIONAL NUMBER. CHRIST, YOU ARE AN IDIOT. OOPS, CAN'T DO OPERATIONS ON RATIONAL NUMBERS! Can't fucking multiply a rational number by 10, NO FUCKING SIR!
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Styrofoam2006-07-12 3:51
You know, if this idiot won't shut up about not doing operations on an infinitely repeating decimal, I whipped up a proof using an infinite series. Maybe that will shut him up.
0.999~ = SUM(n=1,inf,9/(10^n)) [The series is 0.9 + 0.09 + 0.009 + ...]