jordan 36 Posted July 16, 2004 Why do you put that _ below the last 9? It signifies that whichever digit or digits are underlined repeat in that order forever. It can also be a line above the digit(s). 0 Share this post Link to post Share on other sites

JaKiri 13 Posted July 16, 2004 I suppose you could come up with ways to prove just about anything... If it's true, then you're likely (but not certain) to be able to prove it. If it's not true, then by the definition of true it will not have a valid proof. If you don't believe it, disprove it please. 0 Share this post Link to post Share on other sites

Gauss 10 Posted August 7, 2004 Just to spoil the fun... it wasn't 0.999' date=' aeroguy. It was 0.99[u']9[/u], or 0.999~. That means 0.9999999999999999... and so on forever. Theoretically, that does equal one. 1.\bar0 does not equal 0.\bar9 they are two distinct different numbers. 0 Share this post Link to post Share on other sites

Gauss 10 Posted August 7, 2004 1.\bar0 does not equal 0.\bar9 they are two distinct different numbers. [math]1.\bar 0[/math] does not equal [math]0.\bar 9[/math] they are two distinct different numbers. 0 Share this post Link to post Share on other sites

Gauss 10 Posted August 7, 2004 It's pretty obvious really. [math]0.\overline{0}1 = \lim_{n\to\infty} 10^{-n} \sum_{i=1}^{n} 0\cdot 10^{-i} = 0[/math] This equation is not correct. All you have done is constructed a statement and stated the LHS equals the RHS. 0 Share this post Link to post Share on other sites

Kedas 21 Posted August 7, 2004 Just because we can represent them/it different in mathematics doesn't mean they aren't equal for using it in the real world. But when 'designing' mathematics I would suggest not swap representations just to be on the safe side. I guess it comes down to believing how much you think nature is linked to mathematics or the other way around. mathematics did/can help find somethings we didn't know were there but that doesn't make it the bible. 0 Share this post Link to post Share on other sites

Dapthar 10 Posted August 7, 2004 It's pretty obvious really. [math]0.\overline{0}1 = \lim_{n\to\infty} 10^{-n} \sum_{i=1}^{n} 0\cdot 10^{-i} = 0[/math] Well' date=' you're supporting the correct result, but with incorrect arguments. To illustrate, let's calculate a couple of values of your series. At [math']n=2[/math] we get [math]10^{-2} \sum_{i=1}^{2} 0\cdot 10^{-i} = 10^{-2}(0 \cdot 10^{-1}+0 \cdot 10^{-2})=0[/math]. At [math]n=3[/math] we get [math]10^{-3} \sum_{i=1}^{3} 0\cdot 10^{-i} = 10^{-3}(0 \cdot 10^{-1}+0 \cdot 10^{-2}+0 \cdot 10^{-3})=0[/math]. Continuing on in this manner, (I omit the inductive argument, for the sake of brevity), one can see that your series always produces [math]0[/math], and not [math]0.\overline{0}1[/math]. I believe what you meant to write was [math]0.\overline{0}1=\lim_{n\to\infty}10^{-n}[/math]. For this expression, at [math]n=1[/math] we get [math]10^{-1}=.1[/math]. At [math]n=2[/math] we get [math]10^{-2}=.01[/math]. At [math]n=3[/math] we get [math]10^{-3}=.001[/math], and so on. Using Calculus, it can be shown that [math]0.\overline{0}1=\lim_{n\to\infty}10^{-n}=0[/math], as desired. As a final note, I would simply like to mention that [math]0.\overline{0}1[/math] is not correct mathematical notation, but the explanation of why would most likely spawn some side debate about [math]\infty[/math], so I'll leave that point unaddressed for now. 0 Share this post Link to post Share on other sites

jordan 36 Posted August 7, 2004 Please do adress why it's not correct notation. I can't guarentee I'll understand what you're saying, but now I'm interested. 0 Share this post Link to post Share on other sites

Dapthar 10 Posted August 7, 2004 Please do adress why it's not correct notation.Sure. The "bar notation", as I call it, indicates that the sequence under the bar repeats forever, like [math]1/3=.\overline{3}[/math] or [math]1/7=.\overline_{142857}[/math]. So, technically, writing [math]0.\overline{0}1[/math] is meaningless, since one cannot write something after the zeros. The whole point of writing [math].\overline{0}1[/math] is because some people, in an attempt to refute the fact that [math]1=.\overline{9}[/math], state that [math]1-.\overline{9}=.\overline{0}1[/math], and what dave was trying to show was that [math].\overline{0}1=0[/math], thus showing that [math]1=.\overline{9}[/math]. In fact, the main reason that some people have a problem grasping this equality is that the proof relies heavily upon limits, a tool regularly used in Calculus. In this case, one's intuition runs counter to reality, and thus, a problem arises when people attempt to "intuitively" understand it. 0 Share this post Link to post Share on other sites

jordan 36 Posted August 7, 2004 So nothing can ever come after the bar and still be considered correct? Saying something like "[math].\overline{9}8[/math] is the closest you can come to 1" is nonsense? 0 Share this post Link to post Share on other sites

AL 11 Posted August 8, 2004 So nothing can ever come after the bar and still be considered correct? Saying something like " is the closest you can come to 1" is nonsense? It is nonsense because there is no such thing as a "real number that is closest to 1" and not equal to it. It has to do with a property of the real numbers called Completeness. Suppose there is a number that is "closest" to 1 but not equal to it. We'll call it x. Since x is "closest" to 1, this implies that there is no such number y that is between x and 1 so that x < y < 1. Now set y = (x+1)/2. Then we get that x < (x + 1)/2 < 1, which implies there IS a y such that x < y < 1. Contradiction. Thus there is no such thing as a "closest" number in the set of reals. 0 Share this post Link to post Share on other sites

Gauss 10 Posted August 8, 2004 I believe what you meant to write was [math]0.\overline{0}1=\lim_{n\to\infty}10^{-n}[/math]. [math]0.\overline{0}1[/math] is another way of stating or trying to represent [math]\lim_{n\to\infty}10^{-n}[/math]. Which is just an identity equation or equivalence statement. Dave's statement' date=' stated [math']0.\overline{0}1[/math] = 0 0 Share this post Link to post Share on other sites

Dave 246 Posted August 8, 2004 I just used the definition of decimal notation. 0 Share this post Link to post Share on other sites

Gauss 10 Posted August 11, 2004 x = 0.999... Eq (1) 10 * x = 10 * 0.999... x+x+x+x+x+x+x+x+x+x = 0.999... + 0.999... + 0.999... + 0.999... + 0.999... + 0.999... + 0.999... + 0.999... + 0.999... + 0.999... Eq(2) Eq (2) - Eq (1) x+x+x+x+x+x+x+x+x = 0.999... + 0.999... + 0.999... + 0.999... + 0.999... + 0.999... + 0.999... + 0.999... + 0.999... Since addition or multiplication is normally carried out from Right to Left and since this is an area of contention of what occurs or happens after the repeating dots (ellipses) then addition or multiplication in this instance is carried out from Left to Right. 9*x = 8.999... [math] x = \frac {8.999...}{9} [/math] [math] x = 0.999... [/math] 0 Share this post Link to post Share on other sites

rbp6 10 Posted August 11, 2004 Gauss, your, proof fails because all you did was prove that .99999....=.9999999.... Unless I'm missing something. All you need to do is use infinite summation with partial fractions and you'll get the right answer. I don't even understand where the debate is, anyone who has taken calc 2 should know the answer instantaneously. Or anyone who understands limits for that matter. Its a fairly simply idea I don't really see why anyone would have trouble grasping it unless they have very limited mathematical and conceptual capabilities. 0 Share this post Link to post Share on other sites

Gauss 10 Posted August 11, 2004 Four points to consider First point: The proof simply states that x = 0.999... and the method used is smiliar to the one that is used to prove otherwise. The calculations are exact. Second point: Take another example: x = 3 Eq(1) 10*x = 10 * 3 Eq(2) Eq(2) - Eq(1) 9*x = 27 [math]x = \frac {27}{9}[/math] x = 3 Third Point: There is no reason whatsoever to use the formula for the sum of an infinite sequence in this case. I know the exact sum of: [math]S = \frac{9}{10} + \frac{9}{10^2} + \frac{9}{10^3} + ...[/math] Its [math]S = 0.\bar9[/math] However I would use the sum of an infinite sequence in the following case: [math]S_n = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... + \frac{1}{2^n} [/math] Obviously, as [math]n \rightarrow \infty \mbox{\, } S_n \rightarrow 1[/math] Fourth point: This is the case for 2/9 x = 0.222... Eq (1) 10 * x = 10 * 0.222... x+x+x+x+x+x+x+x+x+x = 0.222... + 0.222... + 0.222... + 0.222... + 0.222... + 0.222... + 0.222... + 0.222... + 0.222... + 0.222... Eq(2) Eq (2) - Eq (1) x+x+x+x+x+x+x+x+x = 0.222... + 0.222... + 0.222... + 0.222... + 0.222... + 0.222... + 0.222... + 0.222... + 0.222... Eq(2) Since addition or multiplication is normally carried out from Right to Left and since this is an area of contention of what occurs or happens after the repeating dots (ellipses) then addition or multiplication in this instance is carried out from Left to Right. 9*x = 1.999... [math]x = \frac {1.999...}{9}[/math] [math]x = 0.222...}[/math] 0 Share this post Link to post Share on other sites

rbp6 10 Posted August 11, 2004 I'm sorry, I don't know how to use that whole forum math thing so bear with me. .99999......=9/10+9/100+9/1000+9/10,000..... 9/10+9/100+9/1000+9/10,000.....=infinite summation(9/10^n)where n goes from 0-infinity infinite summation(9/10^n)=(9/10)/(1-1/10) (geometric series summation) (9/10)/(1-1/10)=1 Proven 0 Share this post Link to post Share on other sites

Dave 246 Posted August 11, 2004 I don't really want to get drawn into this debate because as far as I can see, it's just a matter of definition, and I really, really hate to argue over such minute and tedious things. Please don't let this stop you though 0 Share this post Link to post Share on other sites

matt grime 10 Posted August 18, 2004 Since addition or multiplication is normally carried out from Right to Left and since this is an area of contention of what occurs or happens after the repeating dots (ellipses) then addition or multiplication in this instance is carried out from Left to Right. There is no contention there at all. Arithmetic operations can and are easily defined on infinitely long decimal representations. 0 Share this post Link to post Share on other sites

paganinio 10 Posted August 31, 2004 has this been posted before? 0.999...=0.333...+0.666...=1/3+2/3=1 0 Share this post Link to post Share on other sites

Gauss 10 Posted August 31, 2004 Note the following pattern: [math] n = 0, \mbox { }then \mbox { } 1 - \frac {1} {10^0} = 0 [/math] [math] n = 1, \mbox { }then \mbox { } 1 - \frac {1} {10^1} = 0.9 [/math] [math] n = 2, \mbox { }then \mbox { } 1 - \frac {1} {10^2} = 0.99 [/math] [math] n = 3, \mbox { }then \mbox { } 1 - \frac {1} {10^3} = 0.999 [/math] [math] n = 4, \mbox { }then \mbox { } 1 - \frac {1} {10^4} = 0.9999 [/math] [math]\mbox{Then }1 - \frac {1} {10^n} = 0.9_19_29_39_4,...9_{n-1}9_n[/math] [math]Proof[/math] Let P(n) be the statement that [math] 1 - \frac {1} {10^n}[/math] has n 9's after the decimal place. [math]\forall_n \mbox { where } n \in \mathbb{N} [/math] [math] n = 0, P(0) = 1 - \frac {1} {10^0} = 0 \mbox {, thus P(0) is True.}[/math] [math] n = 1, P(1) = 1 - \frac {1} {10^1} = 0.9 \mbox {, thus P(1) is True.}[/math] [math] n = 2, P(2) = 1 - \frac {1} {10^2} = 0.99 \mbox {, thus P(2) is True.}[/math] [math] n = 3, P(3) = 1 - \frac {1} {10^3} = 0.999 \mbox {, thus P(3) is True.}[/math] [math] n = 4, P(4) = 1 - \frac {1} {10^4} = 0.9999 \mbox {, thus P(4) is True.}[/math] Assume that P(n) is true that [math] 1 - \frac {1} {10^n}[/math] (Basis of Induction) has n 9's after the decimal place. We shall show that P(n+1) is true. (Inductive Step) [math]P(n+1) = 1 - \frac {1} {10^{n+1}}[/math] [math]P(n+1) = P(n) + a_{n+1}[/math] [math]P(n+1) = 1 - \frac {1} {10^n} + \frac {10^1 - 1}{10^{n+1}}[/math] [math]P(n+1) = 1 - \frac {(10^n - 1)(10^{n+1})} {(10^n)(10^{n+1})} + \frac {(10^1 - 1)(10^n)}{(10^{n+1})(10^n)}[/math] [math]P(n+1) = 1 - \frac {(10^{2n+1} - 10^n)}{10^{2n+1}}[/math] [math]P(n+1) = 1 - \frac {1} {10^{n+1}}[/math] [math]\mbox{Therefore P(n+1) is True}[/math] [math]\mbox{Therefore P(n) is True}[/math] [math]\mbox{Therefore }1 - \frac {1} {10^n} = 0.9_19_29_39_4,...9_{n-1}9_n[/math] [math]\mbox{Therefore }1 = 0.9_19_29_39_4,...9_{n-1}9_n + \frac {1} {10^n}\mbox{ }\forall_n \mbox { where } n \in \mathbb{N} [/math] 0 Share this post Link to post Share on other sites

matt grime 10 Posted September 15, 2004 So what has an inductive statement about a finite decimal expansion (hence only apllicable to the rationals) got to do with the real numbers in generality. Oh, wait you're not about to conclude the "infinite" case follows from the finite cases inductively.... and 0.00...01, an infinite number of 0s then a 1 makes no sense as a decimal expansion for anyone still under the impression it was meaningful. there have been several worrying expressions of dislike of definitions (esp by a maths moderator). worrying because maths IS its definitions. 0 Share this post Link to post Share on other sites

Dave 246 Posted September 15, 2004 I'm sorry for not living up to your exacting standards. I have not yet completed my degree - indeed, I am only just about to start my second year. Therefore I'm beginning to understand what a role definitions have to play in the world of mathematics, but as hard as I try, I find them tedious and boring - although necessary, I do not wish to spend my free time talking about them on an internet chat forum. I also know that I'll never become anything of any great significance to the mathematical community, basically because I'm not intelligent enough. I do (moderate) this because I enjoy doing mathematics in general and discussing it with other people. Unlike you, I am not a research mathematician and do not have the experience of such. I guess what I'm trying to say is: we're not all research mathematicians, so cut me a little slack, will you? 0 Share this post Link to post Share on other sites

matt grime 10 Posted September 15, 2004 sorry, but being a research mathematician has nothing to do with it, being a mathematician does (i don't use real numbers in my research, none of my colleagues uses them in research either). if you don't actually pay attention to the definition of any of the terms in such a question as "why is 0.999.. equal to 1" then you've not answered the question since the truth of the statement *follows* from the definitions and the principal reason people do not understand this particular question is because they do not understand the definition of the terms. Those two objects are certainly different *representations* of the same real number but they are definitely the same *real* number. The reals are a basic object that we all use, though often without knowing what's going on. asking this question is a good sign of curiosity, understanding the (correct) answer a good sign of aptitude. http://www.dpmms.cam.ac.uk/~wtg10/decimals.html you do not need to discuss them, they are not up for discussion, they simply *are* if you will. 0 Share this post Link to post Share on other sites

Dave 246 Posted September 15, 2004 sorry, but being a research mathematician has nothing to do with it, being a mathematician does. I'm not going to go into it here because I don't want to get off-topic, but I don't believe this statement to be correct. The reals are a basic object that we all use, though often without knowing what's going on. asking this question is a good sign of curiosity, understanding the (correct) answer a good sign of aptitude. If you have a problem with me (which, considering the general language you used in your post), then take it up with me in PM instead of trying to insult me. 0 Share this post Link to post Share on other sites