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Mathematically.
The way *I* understand the problem is that 0.999.... approaches 1 but never reaches one. Ergo they aren't equal.
However I was told this was an incorrect way to view it, so I appeal to you guys bc yer smart and stuff :)
Oh and this isn't a debate of whether 0.999.... = 1 or not, because it isn't equal. :)
I remember this thread. xD
I've never involved myself with maths enough to give a damn. But I'd go with your interpretation of it PG.
this is just low,
even by my standards.
why my god....I remember you from WAY back!! Haven't seen you post in forever PG. Didn't you have like a pikachu sig or somethin??
anyways the decimals probably keep repeating neverending so your right it doesnt equal one. But maybe they round???
I was abused in chat for not taking part in this thing.
So yeah, its good.
:mad2:Quote:
Originally Posted by Samuraid
Yes, that website covers it better than I could have explained it to you.
Oh, now I remember why I dropped maths.
For something so concrete and provable, it sure is counter-intuitive sometimes.
Quote:
Originally Posted by Samuraid
It's valid, it works, and my head hurts now. Thanks Samuraid.
He might be a mathematical genius, however, the guy is also a total MORON because in trying to make math seem awesomesauce he states;
True, most likely. But Elvis, The Beatles, Johnny Cash, are still massive sellers 40, 50 years later.Quote:
In a year's time, every record currently in the UK Top 40 Singles Chart will have dropped out.
In five years' time, most of those artists will have also vanished without trace.
Yeah, which is why people still wear stuff from the 50's and 60's and 70's and 80's.Quote:
In fifteen years' time, everything you think you know about fashion will be laughably incorrect.
James Dean.Quote:
In fifty years' time, any movie star you can name today will be forgotten.
Probably true. And I'll let you know when that means Alexander Graham Bell never existed.Quote:
In a hundred years' time, every single piece of technology you own will be obsolete.
Yes, I had to look on Wikipedia for some leaders from 200+ years ago, so I suppose given that he's right and it's pointless to mention George Washington, King Henry VIII, Topiltzin Ce Acatl Quetzalcoatl, Augustus Caeser, Alexander The Great, Hammurabi, or Qin-Shi Huang.Quote:
In two hundred years' time, prominent world leaders like George W. Bush will be as unfamiliar to students of the day as President Zachary Taylor (1784-1850) is to us.
This has to be irony. Please tell me it's irony.Quote:
In a thousand years' time, Shakespeare will be just another piece of ancient history, as relevant as the Iliad.
Yes, because the Pacific Ocean has only been around for about four millennia. And the stars, we're not looking at things billions of years old there. Human civilization is at least a couple hundred thousand years old in a recognizable fashion, and ten thousand years plus of what we typically consider civilization has existed. In another two or three thousand years we'll reach a point where we can name individuals who lived ten thousand years ago.Quote:
In ten thousand years' time, everything you see will be gone, everything you know will have changed beyond recognition, and everyone you've ever heard of will have been forgotten for ever.
Anyway, to end my rant - maths is important, of course. It's a fundamental facet of the universe. But that's no call to make statements like these (He says "Fine, fine, I generalise, I make sweeping statements, I am melodramatic. But seriously. If you really, and I mean REALLY want to live forever, there is only one way to do it. A mathematical equation stands forever." apparently not realizing that there's a big difference between generalization/melodrama and outright lies. And that by his own logic Newton and Einstein will be forgotten, and only their equations will be remembered - which presumes our species will survive forever - which isn't immortality at all. Either that or he's thinking that mathematical equations do NOT in fact exist absent Humanity.) and he's so daft in these statements that I'm half-tempted to discount his mathematics.
God, PG, yes .999... = 1. >=o
For any distinct, real numbers A and B where A < B, there is a number x such that A < x < B. What number is between .999... and 1? Here's other ways of looking at it.
Don't be as dumb as Daniel now and argue about it for three pages.
Well not for three pages, but bleh.
I swore 0.999... < 1
:(
Well you're wrong, snookums.Quote:
Originally Posted by Pureghetto
.9999999.... = 9/9 = 1.
Reading through that last math thread made me want to kill someone. Daniel demanding Unne prove that 1/9 = .1111.... for example. That one actually made me chuckle, though.
You guys all need to go back to high school math. :mad:
Math does not exist. Its current set has too many issues and flaws. 1 is != .99999 simply becuse they look nothing alike. Make sence.
I am infallable.
Bipper
Learnt this in Maths C :D
let x = .9999
10x = 9.9999
9x = 9
x = 9/9 = 1
I didn't read through that website that Samuraid posted, if its in that, sorry.
EDIT: Damn, it did. Oh well! I am smart in my own mind!
In that thread i took the position that 0.9999 = 1 too.
*sigh* do I get dumber as I age?
i always hated math
A couple of years ago I had to proof that 0,999999..... = 1 with "the real proof" It's not that difficult though if you look at it for an hour or so :-)Quote:
Originally Posted by Samuraid
I seriously suck at maths. Stuff that gets as complicated as this is way out of my understanding. I'm only doing Intermediate GCSE maths! I think by what I understand at the moment that some of these guys are madly and nerdly correct!
If 1 = 0.9999..., then why when I have an apple and take another apple I have 2 apples and not 1.9999... apples?
Thank you people. My brain just imploded. The one thing I know is 0.9 and 0.9999 have the same value, as the 9 is recurring, so I see where PG came from.
Quote:
Originally Posted by Maxx Power
you will have 1,99999999..... apples because
1 apple + 1 apple = 1 apple + 0,999999999 apples = 1,999999999.... apples
Ow yes I luv maths :-)
Oh my god. I hated that thread. I hate this thread too. 0,999whatever is not 1. :mad2:
Im going to go with common sence on this one and say that 0,9999...<1, proof or no proof. :p
If you mean 0.999 requring ( if thats how its spelt ) it will never reach 1. It will just keep going forever and ever and ever...
The guy who wrote this article is a total jerk, so it's a shame that he's also correct. .9 recurring is equal to 1, because there is no number in between the two. And I suck at maths.Quote:
Originally Posted by Samuraid
Just because there's no number in between doesn't make it equal to it. >:/ Then .9[insert 9s here]8 would be equal to .9[insert 9s here]9 who's apparantly equal to 1. So in the long run 1 would be equal to 9. And that just ain't right, girlfriend.
This is decimals, Levian. It's a totally different and screwed up world of numbers then the basic 1, 2, err, 3, 4, 5, 6, 7, 8 and 9 (yay I got it right!)
But the number 5 falls between 1 and 9, so the two can't possibly be equal. The idea of adding an 8 onto the end of .9 recurring doesn't really work, because there is no end to the sequence of nines for the 8 to be added to (since a recurring figure is repeated endlessly).Quote:
Originally Posted by Levian
The easiest way to think about it is to think of .999 recurring as 3 x .333 recurring. Since .333... is equal to one third, three times that amount is equal to 1. 3 x .333... = .999, so .999 must be equal to 1.
This is a sucky explanation, and the reasoning is probably flawed, but that's the way I've always seen it. *shrug*
Does math make you frustrated? It should.
Having thought about this with some intensity for the last few hours, it's making a lot more sense right now than it was last night.
It still doesn't make sense that because there's no number between the two, that they are not distinct numbers though. I understand the rest, but I really don't get that. :confused:
So is it safe to say that there is an issue with decimals as we know them? Should we revamp math!? Please, I want some new math, the old stuff is so out dated.
0.999999999 infinty doesnt = 1
but the little workaround so it can be expressed properly is 0.99' the little apostraphe is the (actually its a dot but hey this is a computer) expression for an infinate number
I think it would equal 1; if you have 0.9999 recurring to infinity, you can't very just add 0.00000000000000001 or 0.0000000000000000000000000000001 to it, because neither number will reach the end of infinity, if it makes sense. If you add 0.0!1 to it somehow, then it would be more than 1. 1 puts the stopper onto the 0.000, and hence it won't reach infinity.
I don't know if this is right or wrong or makes sense or doesn't. It sounded logical to me in my warped mind at time of writing so please don't rip. xD
Quote:
Originally Posted by Resha
i like this post, except that you'll never have 0.000!1
I miss math. I should start taking some classes again,
Exactly! :greenie: 0.0000!1 won't be ! anymore.
I always thought ! was for factorial not infinity but whatever.
Not entirely sure about .999~ = 1.
But if it did it would be because the repeating decimal at the end got hit by a one, then it would knock all the 9's to a one, crossing the decimal into the one's place. But since it would be a never ending decimal place, which would make it very hard to see what happens -after- al the 9's that you still see, It would almost be like trying to find -every single- decimal place to Pi, It's never ending.
So who knows, it might actually end up 1 at the end of the day. :D
But what if the .99999~ isn't infinitive, and we just think it is. =O
There's so many things wrong with that, I don't know where to begin. :pQuote:
Originally Posted by Levian
First off, not only does any two distinct real numbers have to have a number between them, any two distinct real numbers have to have an infinite amount of numbers between them. 1 and 2 have an infinite amount of numbers between them. There exists not even ONE real number between .99999... and 1.
Also, .99999...8 isn't a number. That's saying "an infinite number of nines that never ends, but when they do end, tack an 8 on the end." It doesn't work like that. 0.9999... means "a decimal followed by an infinite number of nines.
Hmm, I would probably take a bit harder way and solve this through series.
0.999... = 0.9 + 0.09 + 0.009 + ... = 9*10^(-1) + 9*10^(-2) + ... = 9 * [ 10^(-1) + 10^(-2) + ... ] = 9 * Σ (n=1...∞) 10^(-n) := 9*S
S is clearly a neverending geometric series, with its first term (n=1) being 0.1. q = 1/10 and |q| < 1, so the series converges. S = 0.1 / (1-q), so
0.999... = 9*S = 9*0.1 / (1-1/10) = 0.9 / 0.9 = 1.
Ok, MILFy, let me try to explain.Quote:
Originally Posted by I'm my own MILF
Take the numbers 1 and 2. How many numbers are between 1 and 2? The answer is infinite. You have 1.1, 1.01, 1.001, etc.
Now, take 1 and 1.00000001. How many numbers are between those two? The answer is again, infinite. 1.0000000000001, etc.
Between any two distinct, real numbers there exists an infinite number of other distinct, real numbers, no matter how close those two numbers are. It's called the Density Property, and is a fundamental concept of algebra.
There exists no number between 0.999... and 1, and both are real numbers, therefore, they are equal.
the real proof is kind of idiotic because it involves the limit argument which is, escencially, an aproximation. the algebraical proof (x = 0.99999....) is way more better if you ask me. the only problem that i find is how you pass from 0.9999... to 1.000...01. maybe my brain isn't abstract enough.
Limits aren't an approximation. It's only somewhat non-intuitive. The limit proof is simply carried out into infinity, which some people have trouble grasping.Quote:
Originally Posted by radyk05
EDIT: what I had below this was wrong. :(
You people need to go outside once in a while.
well, the way i was taught, the limit is when you come close enough to a number/point but never to the number/point (or something along those lines). then again, that was only a calculus class and not a real analysis class.
Yes it does!!!Quote:
Originally Posted by Old Manus
No no, limits are for as x approaches, but never reaches, a number. Limits have definitive answers.Quote:
Originally Posted by radyk05
For instance, the limit as x approaches infinity of (1/x) = 0. x gets closer and closer to infinity, but obviously never gets there. However, the limit has a definite answer.
Except 1 - (1/10) - (1/100) - ... isn't 0. :p It's 0.888... (= 8/9). Proof:Quote:
Originally Posted by Raistlin
1 - (1/10) - (1/100) - ... = 1 - [ 1/10 + 1/100 + ... ] = 1 - [ 10^(-1) + 10^(-2) + ... ] = 1 - ∑(n=1...∞) 10^(-n) := 1 - S.
Here, the first term is 0.1 and q = 0.1, so you'll get: 1 - S = 1 - [ 0.1 / (1-0.1) ] = 1 - [ 0.1 / 0.9 ] = 1 - [ 1/9 ] = 8/9.
Err... yeah. xD It's been a while.Quote:
Originally Posted by Sephiroth1999AD
I guess I was thinking about what I put in my last post: the limit (x -> infinity) (1/x) = 0.
:kaolaugh:
http://mathpwned.ytmnd.com/
There's an 8 at the end of the infinite nines. Now that wasn't so hard, now was it?Quote:
Originally Posted by Raistlin
Technically speaking, 0.9999 etc. does not equal 1. But if you round it off, it is 1 for all mathematical purposes.
There... an explantion so concise yet informative that even I am impressed. Even better... no long, drawn-out mathematical equations were used!
Then the series of nines isn't infinite. :p "Infinite" means, obviously, that there is no end. Therefore, there can't be any other numbers in the series of infinite nines, nor can there be something on the end, because an end doesn't exist.Quote:
Originally Posted by Levian
EDIT:
There is no rounding involved. 0.999... literally equals one.Quote:
Technically speaking, 0.9999 etc. does not equal 1. But if you round it off, it is 1 for all mathematical purposes.
Then think of this. There is a glass of water. 10/10ths of it would make it full. 0.99/10ths of it would make it almost completely full, but 0.01 of it would not be full.Quote:
Originally Posted by Raistlin
That 0.01 can apply to any number of 0.99s. By this I mean that 0.9999 has 0.0001 empty, etc., so on and so forth.
That changes when you make an infinite series of nines. There is no such thing as an infinite series of zeros with a 1 at the end.Quote:
Then think of this. There is a glass of water. 10/10ths of it would make it full. 0.99/10ths of it would make it almost completely full, but 0.01 of it would not be full.
That 0.01 can apply to any number of 0.99s. By this I mean that 0.9999 has 0.0001 empty, etc., so on and so forth.
No offense, but clearly you don't understand decimals then. It exists. 0.01, 0.001, etc. They are all real decimals. It represents a very small number.Quote:
Originally Posted by Raistlin
Anyone who knows math care to back me up on this one?
No, actually, Raistlin is right. If it's recurring, it is not 0.000001, or 0.000000000001, or anything of the like. It is infinite, and therefore there is nowhere to place the last 1. Infinite numbers are a bitch to grasp, but.Quote:
Originally Posted by Vincent, Thunder God
So. 0.999... must equal 1 for the following reason:
1 - 0.999... = 0.000... Now because there IS no end to 0.000 recurring, there can BE no 1 to make it up to a total of 1. And if 0.000... + 0.999... = 1, then 0.999... must equal 1. There's no other way for it. If there is a 1 in 0.000... it's not recurring, and therefore the matter is settled in another way (eq the numbers are finite.)
Yes, if it is infinite, you are definetly right. Thanks for pointing that out.Quote:
Originally Posted by I'm my own MILF
Quote:
Originally Posted by Vincent, Thunder God
*bows*Quote:
Originally Posted by me a couple of posts ago
Y'very welcome. Took me long enough to figure it out, might as well help others along :pQuote:
Originally Posted by Vincent, Thunder God
Have you people completely forgotten that .999... is an approximation?
Intergers, radicals, and fractions are exact values.
.999... vs. 1 is comparing an infinte concept (an approximation, at that) to an exact value. That is like asking someone to divide by 0.
Ok, so 1/3 == .333..., but .333... is still a approximation. You cannot have an exact value for 1/3 in decimal notation because it does not end. So stop saying that 1/9 * 9 == .999... == 1 or 1/3 * 3 == .999... == 1 is true. 1/7 * 7 is also an infinite set of numbers, yes, even though it does not equal .999..., but rather, a smaller number, it still would equal one.
Any number divided by itself is 1 (except 0/0). This the FIRST grand assumption that people miss/omit.
However, the "pick out a number between the two and you define them as different numbers", that one, I couldn't attack even if I were smart enough to explain it.
It's not an approximation at all. 1/9 (or 9/9, in the case of .999999...) is not an approximation. Any real number is not an approximation.
It's not an approximation. It's a method of describing a number as an infinite geometric series. So let's rewrite this without decimals.Quote:
Originally Posted by Tavrobel
Been a while since I've had calc, so forgive me if I mess this up. A decimal number
0.x1x2x3x4x5x6...xn
(those numbers are meant to be subscripts) is actually a way of writing
0 + x1/10 + x2/10^2 + x3/10^3 + x4/10^4 + x5/10^5 + ... + xn/(10^n)
It so happens in the case of .99999999 (however many 9's you'd like) that all of x1...xn all are 9.
9/10 + 9/100 + 9/1000 + ... + 9/10^n
Easier to write it like this:
SUM(k = 1 to n) of [9/(10^k)]
Or you could rewrite it equivalently as this:
SUM(k = 0 to n) of [(9/10) / (10^k)]
or
SUM(k = 0 to n) of [(9/10) * (1/10)^k]
This so happens to be a geometric series.
Now, imagine that we let n approach infinity. In the case of .999.... n does approach infinity. In this case we have an infinite geometric series.
Some infinite series add up to (converge to) a finite sum. It seems non-intuitive, but that's calculus for you. If it wasn't possible to add up an infinite number of numbers and (sometimes) get a finite result, we'd be screwed. See Zeno's paradoxes.
It so happens that this infinite geometric series is one of the types of infinite series that converges to a finite sum. But you have to use limits.
lim(n -> infinity) SUM(k = 0 to n) of [a * r^k] = a / (1 - r)
The proof of this is simple and can be seen here for example. This is true when |r| < 1, and clearly|1/10| < 1.
Substituting from above, a = 9/10 and r = 1/10, we have
.999...
= lim(n -> infinity) SUM(k = 0 to n) of [(9/10) * (1/10)^k]
= (9/10) / (1 - (1/10))
= (9/10) / (9/10)
= 1
So no approximations or decimals needed.
Unne: Yeah, that seems right, considering I did pretty much the same thing on the previous page, except I took 9 as a factor outside the sum and kept 10^(-n) inside the sum (where n goes from 1 to infinity).
http://pi.ytmnd.com/
:kaolove2: It's so...........hypnotic:kaodizzy:
Edit: http://whatismath.ytmnd.com/ :kaocheer:
mai gawd PG its cause the world is round! therefore we round! 9.5=10 .: .9999999...=1.
meh...