matt grime Posted September 15, 2004 Share Posted September 15, 2004 Sorry, apologies for sounding insulting. I have no problem with you, though i have a problem with answers that don't use the properties of the objects in the question. Link to comment Share on other sites More sharing options...

matt grime Posted September 15, 2004 Share Posted September 15, 2004 An olive branch: yes, having to just deal in basic definitions is tedious and dull, no argument there, and often it is unnecessary to make explicit reference to them all. However *this* particular question exactly boils down to people not knowing what the definitions are (complete totally ordered field, or completion of the rationals in the euclidean norm, or the set of all dedekind cuts, whatever tickles your fancy). Link to comment Share on other sites More sharing options...

Dave Posted September 15, 2004 Share Posted September 15, 2004 Fair enough - I'll admit that I haven't really studied the matter of decimals in any significant form, and I don't understand them and their notation as much as I should. I also apologise for my outburst - I'm a little tired, been working for about 11 hours today and have driven ~200 miles so I'm a bit tetchy Link to comment Share on other sites More sharing options...

Gauss Posted September 17, 2004 Share Posted September 17, 2004 So what has an inductive statement about a finite decimal expansion . The statement is not about a "finite decimal expansion". What is it that you do not understand about "for all of n in [math]\mathbb{N}[/math]. The set [math]\mathbb {N}[/math] has never been finite and will never be finite. (hence only apllicable to the rationals) got to do with the real numbers in generality. Rationals are a subset of the reals. Therefore any number that is rational is also a real number. Oh, wait you're not about to conclude the "infinite" case follows from the finite cases inductively.... You may do that, but I will never deduce that. The proof states what it states. 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. I agree with you unreservedly on this. It is very bad language to be used by any mathematician. I would also state that 0.999... expressed as such is also very bad language (but thats my opinion). It is not rigorous enough, but everybody uses it that way, so by general agreement I also use it. Doing operations on this form leads to problems. there have been several worrying expressions of dislike of definitions (esp by a maths moderator). worrying because maths IS its definitions. Mathematics is only valid according to the definitions (and proofs) used and hence is only valid in that particular context in which it is used. Link to comment Share on other sites More sharing options...

BrainMan Posted September 17, 2004 Share Posted September 17, 2004 The problem is that if you think .999... is not the same as 1, then any proof using a limit is question begging. The very concept being questioned by those that think the numbers are not equal is that a number getting closer and closer to some other number, x = .9 x = .99 x = .999 ... never actually reaches that number, even after an infinite number of iterations. An infinite limit is presumed to ignore minute differences, so using it as a proof is not valid if there might be a miniscule difference between .999... and 1. In other words, the definition of a limit is really what is at question here (along with all other operations with infinite iterations to be completed). [Nonstandard analysis, in fact, defines limits as operations that ignore infintesimal differences by mapping them onto the closest real numbers, which conforms with the intuitions of those that think .999... is not equal to 1.] Link to comment Share on other sites More sharing options...

BrainMan Posted September 17, 2004 Share Posted September 17, 2004 there have been several worrying expressions of dislike of definitions (esp by a maths moderator). worrying because maths IS its definitions. The question is, COULD math have developed differently (remaining consistent) such that .999... does not equal 1, but is only very close to it? Link to comment Share on other sites More sharing options...

matt grime Posted September 21, 2004 Share Posted September 21, 2004 The statement is not about a "finite decimal expansion". What is it that you do not understand about "for all of n in [math]\mathbb{N}[/math]. The set [math]\mathbb {N}[/math] has never been finite and will never be finite. Rationals are a subset of the reals. Therefore any number that is rational is also a real number. You may do that' date=' but I will never deduce that. The proof states what it states. I agree with you unreservedly on this. It is very bad language to be used by any mathematician. I would also state that 0.999... expressed as such is also very bad language (but thats my opinion). It is not rigorous enough, but everybody uses it that way, so by general agreement I also use it. Doing operations on this form leads to problems. Mathematics is only valid according to the definitions (and proofs) used and hence is only valid in that particular context in which it is used.[/quote'] the point was, what the hell did your post have to do with anything? it dealt with an infinte set of finitely long decimal expansions. so? Link to comment Share on other sites More sharing options...

matt grime Posted September 21, 2004 Share Posted September 21, 2004 The question is, COULD math have developed differently (remaining consistent) such that .999... does not equal 1, but is only very close to it? of course it could, since these are just notations for objects with a formal set of properties. we could have given them any meaning we wanted. in fact in base 11, 0.999... doesn't equal 1. the thing about being very close to it indicates that you want a system where the archimidean axiom is not true, that isn't a very useful system really but you may define it, and use it. note that in that system you cannot prove almost any theorem that you take for granted in analysis (even limits are no longer unique). Link to comment Share on other sites More sharing options...

jcarlson Posted March 1, 2005 Share Posted March 1, 2005 Yes, but in base 11, assuming a is the 11th digit, .aaa... does equal 1. Link to comment Share on other sites More sharing options...

Newtonian Posted March 1, 2005 Share Posted March 1, 2005 Thats exactly what i was trying to convey earlier,but my math is crap! Link to comment Share on other sites More sharing options...

Johnny5 Posted March 2, 2005 Share Posted March 2, 2005 In case anyone here hasn't noticed' date=' there have been a lot of debates on this online between people on other forums, some with a good grasp of mathematics and some without. The debate is over whether 0.99[u']9[/u] does or does not equal 1. There are many different proofs to prove that it does, but because some people understand some proofs better than others, and some not at all the debates still go on. I've personally been involved in these sort of debates and I can think up at least five different proofs, all based on different sorts of logic, that show that they are equal. The more forms we can find, the better the chances are that someone will understand it. Well I would enjoy seeing just one of your five proofs, because they aren't equal. In fact, 1 is a number, and .99999... isn't a number, so your proof is guaranteed to contain an error, you will equate apples to oranges, which is similiar to adding apples and oranges, but not identical. I'm not playing semantical games either. In this particular problem, semantics happens to matter. Regards Link to comment Share on other sites More sharing options...

jcarlson Posted March 2, 2005 Share Posted March 2, 2005 [math].\overline{9}[/math] is most certainly a number. Not only that, its a real number, and also a rational number. It is merely a different representation of the same real number that 1 represents. Link to comment Share on other sites More sharing options...

Guest lamefif Posted March 8, 2005 Share Posted March 8, 2005 I think u guys r trying to make a hole in water. The real world is not precision perfect but fuzzy & mathenatics is one of the language the mortal humans r trying to make sense of it. But if u get to close to the edge u wont c it, is like going to the north pole and asking where exactly is it. Good luck Link to comment Share on other sites More sharing options...

matt grime Posted March 8, 2005 Share Posted March 8, 2005 Yeah, and some of us think that "u" isn't perfect for precision for "you".... Link to comment Share on other sites More sharing options...

mcoy Posted March 12, 2005 Share Posted March 12, 2005 ok...just consider the fact that 0.99999...99 cannot be expressed by a fraction, in which case i propose that 0.9999....999 is an irrational number....i might be wrong of course. and one more thing, where the heck will you use 0.999...999 for anyway? it has no use other than just being argued at..... Link to comment Share on other sites More sharing options...

J'Dona Posted March 12, 2005 Author Share Posted March 12, 2005 Well I would enjoy seeing just one of your five proofs' date=' because they aren't equal. In fact, 1 is a number, and .99999... isn't a number, so your proof is guaranteed to contain an error, you will equate apples to oranges, which is similiar to adding apples and oranges, but not identical. I'm not playing semantical games either. In this [i']particular[/i] problem, semantics happens to matter. Regards Okay, I'll post one of the proofs for you (although it's probably already been said two or three times in this thread in much better form than I could reproduce it). Let's say that 0.999... recurring is the sum of a geometric progression. This is pretty easy to demonstrate: if the first term, a, is 0.9 and the ratio, r, is 0.1, then the progression will be 0.9, 0.09, 0.009, and so on, and the sum will be 0.999 (which is just my way of writing 0.999... recurring). This has already been written mathematically in other posts by people like dave, but I can't figure out the LaTeX well enough to reproduce it without quoting and I don't fully understand the terminology anyway. Here's the formula for the sum of a geometric series: In the case of 0.999, n equals infinity, so the term r^{n} is actually 0.1^{infinity}, which is zero, and (1 - r^{n}) simplifies to 1. So the formula simplifies to this: Substituting the values of a and r in (0.9 and 0.1 respectively), we get this: and so if the math is all right it looks like the sum to infinity of the series, 0.999, is equal to 1. The only thing I can see wrong with this proof is the assumption that 0.1^{infinity} does equal zero and not just some ridiculously small number, though I'm not sure what else it might be, and falls into the "infinite zeroes means no endpoint" scenario again. I can't believe my first post after 2 months was in this stupid thread again... Link to comment Share on other sites More sharing options...

Dave Posted March 14, 2005 Share Posted March 14, 2005 Welcome back, J'Dona Yes, this thread has rather degraded; I don't know whether there's actually much discussion left in it anymore. That's why I stopped posting in it a while back. Link to comment Share on other sites More sharing options...

Daymare17 Posted October 3, 2005 Share Posted October 3, 2005 I think I might have found a way to debunk the simplest proof that 0.9 = 1. The proof goes like this: N = 0.9 10N = 9.9 10N - N = 9 N = 1 The debunking goes like this: Infinity is not a number but a concept. This is apparent from the fact that you cannot divide by infinity but only by successive approximations, i.e. continually increasing numbers towards a limit that equals infinity, yet never reaching infinity itself. 0.9 "in practice" has to be 0 followed by a set amount of decimal nines. It can be any amount but has to be a certain amount. It can be followed by a million nines, or a googolplex nines, or three nines, or five nines. Let's try five nines and see what we get. N = 0.99999 10N = 9.9999 - N = 8.99991 And then the whole proof falls to pieces. The same thing would happen no matter how many decimal nines we had. As we see, in order for the proof to be true we have to proceed from the supposition that N is two different numbers, namely 0.9 (followed by x decimal nines) and 0.9 (followed by x+1 decimal nines) When you proceed from contradictory premises it's no wonder the result is contradictory. I haven't found a way to debunk the proof that uses the infinite geometrical progression (let alone dave's first proof, which I don't understand at all ) but I imagine that it may have a similar flaw. Link to comment Share on other sites More sharing options...

ed84c Posted October 3, 2005 Share Posted October 3, 2005 Hmm, this means 0.99999999999... is infinitley close to 10. And we know 1/infinity= 0 and surely hence this means 9.9....=10?? Link to comment Share on other sites More sharing options...

ed84c Posted October 3, 2005 Share Posted October 3, 2005 After making a cup of tea i have concluded on the following: If a mathematical proof requires real numbers then 10 != 9.999........ However If a mathematical proof can be conceptual, using infinity (not defining it, just using it) Then 10=9.9999...... Link to comment Share on other sites More sharing options...

AndrewMech Posted October 3, 2005 Share Posted October 3, 2005 Try this, A frog has 2 meters to travel, On its first leap it can jump 1 meter, Every leap after its first it can only leap half the distance of the previous leap. How many leaps before it travels 2 meters? Link to comment Share on other sites More sharing options...

timo Posted October 3, 2005 Share Posted October 3, 2005 http://redwing.hutman.net/~mreed/warriorshtm/necromancer.htm Link to comment Share on other sites More sharing options...

Dave Posted October 4, 2005 Share Posted October 4, 2005 Infinity is not a number but a concept. We're not using it as a number anywhere in the thread; certainly not in my posts This is apparent from the fact that you cannot divide by infinity but only by successive approximations, i.e. continually increasing numbers towards a limit that equals infinity, yet never reaching infinity itself. 0.9 "in practice" has to be 0 followed by a set amount of decimal nines. It can be any amount but has to be a certain amount. It can be followed by a million nines, or a googolplex nines, or three nines, or five nines. Let's try five nines and see what we get. But we're not concerned with this. "0.999..." doesn't mean "have a finite number of 9's in the expansion". We're only concerned with the limit, so none of this counts. I haven't found a way to debunk the proof that uses the infinite geometrical progression (let alone dave's first proof, which I don't understand at all ) but I imagine that it may have a similar flaw. I'm betting you won't find one either. After considering this thread for a bit, I thought of something a little more interesting. If you have a problem accepting 0.999... = 1, then it follows that you must have a problem with things like exponentials, and more specifically, exp(). After all, we define e using an infinite sum (or equivalent limit): [math]e = \sum_{n=0}^{\infty} \frac{1}{n!}[/math] Really guys, what's so difficult to accept about this? Whilst I accept that it may not be completely intuitive, that's pretty much true of a lot of mathematics out there. Link to comment Share on other sites More sharing options...

myrkle Posted October 11, 2005 Share Posted October 11, 2005 Well I'm gonna try to do this like my calculus teacher told me: 1/9 = 0.111111 2/9 = 0.222222 3/9 = 0.333333 etc... so 9/9 = 0.999999 therefore 0.99999 is equal to 1 Link to comment Share on other sites More sharing options...

Zanthra Posted October 11, 2005 Share Posted October 11, 2005 I think of it by the fact that any number with a repeating decimal is a rational number. In the case of .333 it is equal to 1/3. In the case of .131313 it is equal to 13/99. For such numbers there is an easy algebraic way to turn it into a fraction. Take X to be the repeating decimal: x = 3.723969696 Break X into two parts the repeating portion and the nonrepeating portion: x = y + z y = 3.723 z = .000969696 First find the fraction of z: z = .000969696 : Multiply both sides untill you have one complete repetion to the left of the previous repetition, equal to 10 times the number of digits in the repetition, in this case 100. 100z = .0969696 : Subtract one of the variable from both sides. On the left use the representative variable, and on the right use the repetitive value. 99z = .096 : Now that you have gotten rid of the repetition, you can find a fraction by simplyfying. z = .096 / 99 z = 96 / 99000 Now go back and simplify the same fraction for the origonal nonrepetitive portion: y = 3.723 y = 3.723 / 1 y = 3723 / 1000 Multiply them to have the same denominator: z = 96/99000 y = 368577/99000 x = y + z : Put them back together and simplify. x = 368673/99000 x = 122891/33000 122891/33000 = 3.723969696 This works for any number with a repeating decimal, including .999: x = .999 x = y + z y = 0 z = .999 10z = 9.999 9z = 9 z = 1 x = y + z x = 0 + 1 x = 1 .999 = 1 There is no number with a repetative decimal that does not work in this method. There are also no numbers with repetative decimals that do not exqual an integer ratio. Meaning that .999, 1.999 etc. all are integer ratios 1/1, 2/1 etc... Link to comment Share on other sites More sharing options...

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