I didn't even respond to you. But since you told me to shut up, I guess I will now. If you're bad at understanding infinity, then don't bother trying to argue about it.
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I believe he was talking to One Fish Down -.-
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oh lol, he responded right after I posted and there hadn't been posts before me for a while, so I figured he meant me.
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That will teach you to make assumptions without facts.
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I usually don't, but that seemed too weird that he made a response right after I posted that could have been directed at me for all I knew.
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"i don't think you know what the laws really are, so you wouldn't know if he was using them incorrectly. an accusation like that isn't very convincing when you don't explain his mistake or correct him."
I did explain his mistake. Sorry if it's too complicated for you. I'll sumarize it right here. IF IT'S A CONSTANT THEN 9.9R/10 = .9R BUT THAT WOULD HAVE ONE LESS 9. IF IT ISN'T A CONSTANT THEN IT GOES TO INFINITY AND YOU WOULDN'T BE ABLE TO SUBTRACT IT FROM ITSELF AND GET A REAL NUMBER. If you don't get that point by now then don't bother responding. I knew round off error for some people was bad, but this whole thing is silly. | |
Kator, infinity does exist.
Take the set of all integers. It has infinite size. An actual infinity, not a potential infinity. It's infinity the number - Better to call it aleph-1. IF IT'S A CONSTANT THEN 9.9R/10 = .9R BUT THAT WOULD HAVE ONE LESS 9. Not quite. let x = 9+0.9+0.09+0.009... x = a constant. This is provable. For an infinite geometric series, the sum to infinity = u1 / (1-r), where |r|<1. r is the common multiplier of the series, u1 is the initial term. the series 9+0.9+0.09... has u1=9, r = 0.1. Sum to infinity = 9/0.9=10. If u1 was 0.9 instead, as for 0.9R, then it becomes 0.9/0.1=1. Lummox brought this up earlier - It's an alternate proof. Attack that, if you can. | |
There isn't much to attack. You're just presenting a formula that can allow you to plug in sets and get rounded off answers.
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They aren't 'rounded off'. They're exact.
The derivation of the sum to infinity formula is fairly simple. The sum of the first n members of a geometric series = u1(1-r^n )/(1-r). Note that if n=0, the result is zero - as expected. If r=1, the formula fails, but that's just simple multiplication. Let |r| be <1. Let n go to infinity. As n goes to infinity, r^n goes to zero (As |r|<0). Therefore, when n=infinity, r^n=0, and s(infinity)=u1/(1-r). It's really quite simple. It works. You don't get 'rounded off answers' - They are the answers when you let the number of figures be infinite. Exactly. Correctly. Just because you aren't comfortable with the concept of infinity doesn't mean it doesn't work - Any mathematician would agree that 0.9R=1. | |
"Just because you aren't comfortable with the concept of infinity doesn't mean it doesn't work - Any mathematician would agree that 0.9R=1."
Hehe, not even i'm that ignorant to make a statement like that. Many mathmeticians don't agree. I'm not comfortable with infinity being a constant, because infinity isn't a number, it's the idea of something that never ends. Your little formula is flawed because it's assuming that r < 1, and r^inf = 0. Again, this isn't exactly 0, it would be the concept like .0R1. That number doesn't exist, but it's simply the concept of what you need to add to .9R to make 1. It would come out to 0.9R / 0.1 = 9.9R | |
r < 1 Actually, it assumes |r|<1. That is fine - Otherwise, there is no sum to infinity. In this case, r is less then one. It is r/10. r^inf = 0 It is zero. Not just infintesmal - negligible, but zero. I'm not comfortable with infinity being a constant, because infinity isn't a number, it's the idea of something that never ends. How many integers are there? Infinity. The size of the set of integers in infinity. It is a constant. In this case, aleph-1. And, funnily enough, the proof relies on it. EDIT: Actually, this is incorrect. Sorry. The proof doesn't rely on it. And the size of the set of integers is aleph-0. Hehe, not even i'm that ignorant to make a statement like that. Many mathmeticians don't agree. I don't see them anywhere. Any mathematician that disagreed with Lummox's proof (Which I prefer to the other one) is just not being rational. It would come out to 0.9R / 0.1 = 9.9R Look at the damned formula. It is proven. It works. It gives the limit of the series as it sums to infinity. The limit of 0.9R as gets infinite digits is 1. That means, with infinite digits, it hits one. That's how limits work. | |
"It is zero. Not just infintesmal - negligible, but zero."
This is a similar thing. It gets infinitely smaller, not just canceling out or whatever the hell you do to it. "How many integers are there? Infinity. The size of the set of integers in infinity. It is a constant. In this case, aleph-1. And, funnily enough, the proof relies on it." Oh, i'm sorry. I guess that means that infinity can just be made x or something now. So what, multiply infinity by 2 and get 2inf. divide by 2 and get 1/2inf. Is that how it works for you? "I don't see them anywhere. Any mathematician that disagreed with Lummox's proof (Which I prefer to the other one) is just not being rational." I suppose if your version of rational translates to not using your head and blathering out the same thing over and over and over without thinking, then you .9R = 1 people are the most rational people ever. "ZOMG DA FORMULA WORKS K BIATCH STFU I SAID SO LOLZ" Sorry, I already proved that it didn't. "LIMITS R COOL, METHINKS THEY LET ME DO W/E I WANT EM 2 DO AND SAY THAT SOMETHING THAT GETS INFINITELY HIGHER IS AN ACTUAL UNCHANGING NUMBER. DUDE HOW SHWEET IS DAT" Sorry, limits represent what something approaches, it never says it actaully equals it. | |
This is a similar thing. It gets infinitely smaller, not just canceling out or whatever the hell you do to it. No, it is zero. Exactly zero. That's maths - Live with it. Oh, i'm sorry. I guess that means that infinity can just be made x or something now. So what, multiply infinity by 2 and get 2inf. divide by 2 and get 1/2inf. Is that how it works for you? Aleph number is actually a little different. http://en.wikipedia.org/wiki/Aleph_number I suppose if your version of rational translates to not using your head and blathering out the same thing over and over and over without thinking, then you.9R = 1 people are the most rational people ever. Look. At. The. Damned. Formula. Sorry, I already proved that it didn't. Funny, then. What's the sum of the infinite series 1 + 1/2 + 1/4 + 1/8 + 1/16...? It is two. Exactly two. Try it out - Add as many numbers as you like. It approaches two - Therefore, with an infinite number of digits, it IS two. That checks out. That fits the formula. Look at the DERIVATION of the damned formula. IT CHECKS OUT. IT IS USED. And writing AOLspeak in for what I actually said is quite interesting. I've never used AOLspeak - I am in no way connected to it. And your representation of my argument - As well as Lummox's and Fishy's - is really, really bad. Are you literate at all? I'm afraid even Wikipedia disagrees with you - And I haven't edited the page, before you accuse me of fraud. http://en.wikipedia.org/wiki/ Recurring_decimal#The_case_of_0.99999... Infinity really does work that way. | |
IF IT'S A CONSTANT THEN 9.9R/10 = .9R BUT THAT WOULD HAVE ONE LESS 9. we've explained (and i think even you have mentioned it) why this is wrong. there are an infinite number of nines after the decimal point in 0.9R. 0.9R * 10 would give 9.9R, which would have infinity minus one nines after the decimal point. infinity minus a constant is still infinity. how many integers are there? infinity. excluding the number seven, how many integers are there? still infinity. you can exclude any finite amount of integers from the set, but there is still an infinite amount of integers left. Sorry, limits represent what something approaches, it never says it actaully equals it. 0.9R is a constant. it has no limit, and its value does not approach anything; it has a constant and definite value. its value can be represented as an infinite series. an infinite series has infinite terms, but its sum can be a definite, constant value. the series that people have used to represent 0.9R does approach a value as you add more terms to it. the value of 0.9R is the sum of the series with infinite terms. the series has infinite terms, but that does not mean that its value is infinity. when that series has infinite terms, it will equal 0.9R (which also equals 1). but 0.9R does not have terms, and its value isn't changing. it already has infinite decimal places, so its value isn't approaching something, its value *is* something. if any of this sounds repetitive its because i've probably said it several times before, but you still don't seem to get it. solbadguy, there are two reasons why we stick by what we say instead of going by what you say: the ideas we use are based on real mathematical concepts, your's aren't. you can question how a series can have infinite terms. you can question how a series with infinite terms can have a finite sum. you can question how you can calculate the value of a series with infinte terms. you can question everything right on down to the very base of mathematics, but it all works. we didn't invent math, and there are plenty of resources out there if you want to learn about it, so we're not going to explain everything to you. solbadguy, there are a lot of ideas here that you're coming up with, but you hardly explain them. when i ask you a question, you either ignore it or complain about the questions. mathematics works, and you can read all about how it works in text books and on the internet. the ideas you've come up with don't work, and you offer no explanation of how they work. how can two numbers be so close that there are no numbers in between them, but they aren't equal? how can a number like 0.0R1 exist? how is 0.9R not a constant? instead of explaining those things you just act stupider and stupider, which makes me think that these ideas are getting pulled from deeper and deeper parts of your ass. | |
Oh great, there's two of you, well at least you both have similar arguments, so it'll make it easier.
"there are an infinite number of nines after the decimal point in 0.9R. 0.9R * 10" then it isn't a constant, it's a limit as it approaches 1. "instead of explaining those things you just act stupider and stupider, which makes me think that these ideas are getting pulled from deeper and deeper parts of your ass." "Are you literate at all?" Oh no, the intelligence attack. Excuse me for not outlining every little detail for you ( OFD ) and for summarizing your argument out of frustration ( Jp ) Here, i'll read you the definition of a limit out of my own caluculus textbook. Let f be a function defined on both sides of a, except possibly at a itself. Then lim x->0 f(x) = inf means that the values of f(x) can be made arbtrarily large ( as large as we please ) by taking x sufficiently close to a, BUT NOT EQUAL TO a. It becomes larger and larger ( or increase without bound ) And as an added bonus for the other argument here it also says lim x->0 (1/x) does not exist. We use the notation lim x->0 1/x = inf. This does not mean that we are regarding inf as a number. Nor does it means that the limit exists. It simply expresses the particular way in which the limit does not exist Gee, what a coincidence. My book agrees with me on both of these things. I guess I can snip your little arguments about all the books and internet disagreeing with me. I think i'll trust an official calculus book over what some guy blurbs out on wikipedia. And your little formula u1(1-r^n )/(1-r) doesn't make it 1 either. something less than 1 taken to a power of infinity is not 0 as you assume it is in the equation. It approaches 0 getting infintely smaller. Just incase you don't believe that was from my book, the book is called Calculus James stewart 5e Early Transcendentals Single Variable. Websites are http://www.brookscole.com and http://www.thomsonlearning.com | |
<QUOTE>
Gee, what a coincidence. My book agrees with me on both of these things. I guess I can snip your little arguments about all the books and internet disagreeing with me. I think i'll trust an official calculus book over what some guy blurbs out on wikipedia. And your little formula u1(1-r^n )/(1-r) doesn't make it 1 either. something less than 1 taken to a power of infinity is not 0 as you assume it is in the equation. It approaches 0 getting infintely smaller. </QUOTE> That is the formula that is used when calculating the sums of infinite series, in all mathematics. Look up 'Sequences and Series' in your text book. Geometric series to be precise. That is how the formula is derived - The value does hit zero. You weren't summarizing my argument, you were just being a dickhead. It wasn't even a remotely accurate 'summary'. | |
All right, i'll just type this as I look through my book. Wow, Geometric Series is Chapter 11, no wonder I haven't seen it yet. I'll right, i'll quote it "In fact, by adding sufficidently many terms of the serires we can make the partial sums as close as we like to 1. So it seems reasonable to say that the sum of the infinte series is 1 and right
( sigma ) inf on top, n =1 on bottom 1/ ( 2 to the power of n ) = 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/( 2 exponent n ) + ... = 1 All right, you win, I hadn't gotten this far and didn't know you could do this, but fine, you're right and i'm wrong. *sigh* it hurts my pride to admit this but there's no other way around it. Eh, I was just going by what I had learned so far. Just let me admit defeat in quiet. I hope no one rubs it my face. | |
And as for the 5/0 thing, 5/0 is undefined but the lim x->0 5/x is inf.
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Good to see you can change your mind, too. I know I have, twice, over the contents of this thread.
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So stop trying to prove your Math Godlyness over me and Stfu.