dougclayton Posted February 22, 2005 Report Share Posted February 22, 2005 Your mistake is the that you say: 3 * (1/3) = 0.99999999999999999 (repeating) 3 multiplied by 1/3 is 1. Not anything other than 1! That's the whole point. 1 IS 0.999~. This is not a contradiction--it is two different ways of expressing the same number (just like the earlier example of 1/2=2/4=0.5). Besides, if you disagree, what would you say 3 * 0.333~ is? I claim it is 0.999~--what do you think it is? 1 divided by 3 is not a number that can be accurately represented by a decimal. You can put as many 3s at the end of .333 until you drop dead, you still won't get exactly 1/3. By the same method you can put 9s at the end of .999 until you drop dead, you will never get exactly 1. You are not grasping the nature of infinity. 0.999~ is not "9s added until I drop dead"--it is "9s added without ever stopping". This number can never be written down complete with all the 9s, but it can be written precisely like this: 0.999~ (technically it is a bar over the last 9). Link to comment Share on other sites More sharing options...

Hal Posted February 22, 2005 Report Share Posted February 22, 2005 (edited) Your mistake is the that you say: 3 * (1/3) = 0.99999999999999999 (repeating) 3 multiplied by 1/3 is 1. Not anything other than 1! Yes, and 0.999~ is 1; they both refer to the same thing - we can use different labels to denote the same number. The infinite repeating thing here is just a quirk of the decimal system - as someone else said it's basically the same as saying that 1/2 = 2/4 (or that Samuel Clemens = Mark Twain). Edited February 22, 2005 by Hal Link to comment Share on other sites More sharing options...

Bryan Posted February 22, 2005 Report Share Posted February 22, 2005 There were a couple steps he glossed over: a + b = b substitute b = a b + b = b 2b = b divide by 'b' 2 = 1 Besides, you only start with 'a = b', how can you reach 'a = b = 0' from there? If a = b, the only way that a + b = b is if they are equal to 0. There is no other number that can be added to itself and get the same number. So at this point you know that b = 0 and only the final "magical" step that appears to make 2 = 1 is invalid. Those first four steps are valid, but this next one isn't: a+b = b This step is valid if a and b are 0. Link to comment Share on other sites More sharing options...

Bryan Posted February 22, 2005 Report Share Posted February 22, 2005 Another one is: x = x Since '^1/2' is another way of saying square rooted... x = (x^2)^1/2 Split up 'x^2' into '-x' times '-x'. x = ((-x)(-x))^1/2 Separate the two numbers under the radical. x = ((-x)^1/2)((-x)^1/2) You now have '(-x)^1/2' times itself, or '(-x)^1/2' squared. x = ((-x)^1/2)^2 If something is square-rooted, and you square it, they cancel out. x = -x Divide both sides by 'x' 1 = -1 Try to find the error in this one. ((-x)(-x))^1/2 is equal to positive or negative x (the absolute value of x). x = ((-x)^1/2)((-x)^1/2) is only equal to negative x, not positive x. While it is true that the absolute value of x does equal -x, this does not mean that x = -x. Link to comment Share on other sites More sharing options...

Hal Posted February 22, 2005 Report Share Posted February 22, 2005 If a = b, the only way that a + b = b is if they are equal to 0. There is no other number that can be added to itself and get the same number. But he didnt apply any conditions on what a, b were. He only managed to arrive at the a+b=b step because of the fallacy in the preceding line. Worked through with a=3: Let a = b a² = ab a² - b² = ab -b² (a-(a+=b(a- a+b=b 2b=b 2 = 1 Let 3 = 3 3^2 = 3*3 3^2 - 3^2 = 3*3 - 3^2 (3-3)(3+3) = 3(3-3) 3+3 = 3 the reason we have arrived at 3+3=3 is because of the (invalid) cancellation of (3-3) in the previous line. Link to comment Share on other sites More sharing options...

Bryan Posted February 22, 2005 Report Share Posted February 22, 2005 But he didnt apply any conditions on what a, b were. He only managed to arrive at the a+b=b step because of the fallacy in the preceding line. Worked through with a=3: Let 3 = 3 3^2 = 3*3 3^2 - 3^2 = 3*3 - 3^2 (3-3)(3+3) = 3(3-3) 3+3 = 3 the reason we have arrived at 3+3=3 is because of the (invalid) cancellation of (3-3) in the previous line. (3-3)(3+3) does equal 3(3-3), they both equal zero. There is nothing invalid about that line. He did apply the initial condition that a = b. This being the case, the only way that a+b = b is if they are both 0. That is the line that is the "trick" in the "proof", not any other one. Link to comment Share on other sites More sharing options...

dondigitalia Posted February 22, 2005 Report Share Posted February 22, 2005 That proof is faulty. In order to reach "3+3=3," you need to divide both sides of the equation by "3-3." As Bryan pointed out 3-3=0. The result of dividing any real number by zero is not another real number; it is undefined. Edited to add: However, the equation itself is true (when, as Bryan pointed out, [a=b=0]). (3-3)(3+3) = 3(3-3) (0)(3+3) = 3(0) 0 = 0 Edit2: I just noticed that Hal already pointed out that the cancellation of (3-3) was invalid. You can take my post as an explanation of why. Link to comment Share on other sites More sharing options...

Nate T. Posted February 22, 2005 Report Share Posted February 22, 2005 The reason you need to define .999... as the limit of the sequence (.9, .99, .999, ...) is because there is no meaningful way to subtract 9.999... from 0.999... without this definition, and the proof given originally in this thread uses this as a step. Normally, if you want to subtract two decimals which terminate, you use an algorithm that everyone learned in grade school, involving subtracting along columns, with borrowing from the next column sometimes being necessary. The important thing is that this procedure is a right-to-left procedure, i.e., it begins at the rightmost decimal place and works its way backwards. However, you cannot do that with 9.999... and 0.999..., since there is no rightmost column. You have to define 9.999... - 0.999... to be that number to which the sequence 9 - 0 9.9 - 0.9 9.99 - 0.99 9.999 - 0.999 ... gets infinitely close to. Since each of the numbers above is just 9, it's clear that this sequence apporaches the number 9, and that's why 9.999... - 0.999... = 9. Link to comment Share on other sites More sharing options...

cdexter_89 Posted May 10, 2005 Report Share Posted May 10, 2005 (edited) See if you can figure out this one: Let a = b a² = ab a² - b² = ab -b² (a- b )(a+ b )=b(a- b ) a+b=b 2b=b 2 = 1 Lifted from absolutereason.com at the point a+b=b please follow then, a= b - b ... look carefully that is, a = 0, and b = 0 so 0 = 0 Edited May 10, 2005 by cdexter_89 Link to comment Share on other sites More sharing options...

LauricAcid Posted May 10, 2005 Report Share Posted May 10, 2005 (edited) In the abstract sense, .999~ is the largest number that is less than 1. What is the abstract sense? In the mathematical sense, there is no greatest real number less than 1. Whatever real numbers are - whether unspecified members of a set satisfying the axioms for the reals, Dedekind cuts, or equivalence classes of Cauchy sequences - for any real number x, there is no real number that is greater than every other real number less than x. Edited May 10, 2005 by LauricAcid Link to comment Share on other sites More sharing options...

jrs Posted May 11, 2005 Report Share Posted May 11, 2005 (edited) (a- b )(a+ b )=b(a- b ) a+b=b The error here is that you cannot divide by (a- because it is zero in this example. ... there is no real number that is greater than every other real number less than x. Correct. In practice, decimal fractions are approximations. When you measure a rod with a ruler, you might find that the length, L, of the rod lies in some range, say: 99/100 < L < 101/100 With a better ruler and a microscope, you might find that: 99,999/100,000 < L < 100,001/100,000 So you cannot directly show that two reals are equal. The most you can say is that they are not known to be distinct, i.e. we neither know that X < Y nor that Y < X. Edited May 11, 2005 by jrs Link to comment Share on other sites More sharing options...

Alfred Centauri Posted May 28, 2005 Report Share Posted May 28, 2005 My brother posed me this proof that says that .9999999 repeating = 1. Doesn't this violate the law of identity? <snip> If a = b, then a - b = 0 = 0.000... Let a = 1 = 1.000... Let b = 0.999... a - b = 0.000... ==> 1.000... = 0.999... = 1 ;>) Link to comment Share on other sites More sharing options...

source Posted June 13, 2005 Report Share Posted June 13, 2005 (edited) Let a = b a² = ab a² - b² = ab -b² (a-(a+=b(a- a+b=b 2b=b 2 = 1 After the line with emphasis, there should be 0 = 0. You can't cancel out (a - because a - b = 0. Zeros don't cancel themselves out, because 0 / 0 = nonsense. THIS is the error, not a + b = b. That's when the error has already been made. Edit: Who invented this smiley? He must not have been a mathematician. Edited June 13, 2005 by source Link to comment Share on other sites More sharing options...

source Posted June 13, 2005 Report Share Posted June 13, 2005 10X - X = 9X 9X = 9 This proof has a flaw. And it's here. From 10X - X = 9X does not follow 9X = 9. Nor does that follow from any of the above forumulae. Besides, you are obviously forgetting that you said that X = 0.999999999~. In that case, 9X = 8.99999999~, not 9. In other words, there is no violation of the law of identity. Link to comment Share on other sites More sharing options...

rei ichi Posted June 23, 2005 Report Share Posted June 23, 2005 In order for .999... to equal anything other than one, there would have to be a number - even an infinitely small number - between .999... and one. And yet, were we to start at the number one and move backwards towards .999..., we would immediately find there to be no number between the two - which means that they are the same number. And besides that: IF 1/9 = .111... 8/9 = .888... AND IF .111... + .888... = .999... AND ALSO 1/9 + 8/9 = 9/9 = 1. THEN .999... = 1. This series of repeating decimals is just a semantic limitation of decimal notation, is it not? Link to comment Share on other sites More sharing options...

source Posted August 24, 2005 Report Share Posted August 24, 2005 This series of repeating decimals is just a semantic limitation of decimal notation, is it not? Right. Using other notations (binary, octal, hexadecimal) shows this, but they have different limitations. Link to comment Share on other sites More sharing options...

shakthig Posted August 25, 2005 Report Share Posted August 25, 2005 (edited) X=.99999999999 repeating 10X = 9.9999999999 repeating 10X - X = 9X 9X = 9 X = 1 The basic error here is that the concept "repeating" (In the way you have used it) cannot mean the same thing for both X and 10X. That error led to the wrong result 9X = 9. The correct value for 9X can be expressed as: 9X = 8.9999....repeating....9991 Consider X = 0.9999.......repeating. What the words "repeating" means in this case is that 9 is repeated continuously without end. Now can the word "repeating" mean the same thing in 10X = 9.9999...repeating? The number of times the digit 9 is used has to be the same in the numeric value of X and in the numeric value of 10X. In the case of X all the digits of 9 are to the right of the decimal point. In the case of 10X one of the digits of 9 are to the left of the decimal point and the rest of the digits of 9 are to the right of the decimal point. Hence, the numeric representation of 10X will have one digit of 9 less to the right of the decimal point than the numeric representation of X. So if X = 0.9999....repeating Then 10X = 9.9999....repeating(once less than the number of repetitions above for X) However many digits of 9 there are to the right of the decimal point in X, 10X will have one digit of 9 less than that to the right of its decimal point. This issue can be illustrated effectively by "harmonizing" the way the word "repeating" is used in the case of X and 10X. This can be done by shifting the word "repeating" from the end of the numbers to the middle in such a way that the left most digits are also emphasized. Also the following examples should be taken into consideration when deciding how the representation of 10X is to be determined. If Y = 0.99 then 10Y = 9.90 (10Y is not equal to 9.99) Also, if Z = 0.9999 then 10Z = 9.9990 (10Z is not equal to 9.9999) Therefore if X = 0.99999....repeating....9999 ___then, 10X = 9.99999....repeating....9990 As in all cases where a number is multiplied by 10, all digits are shifted to the right by one position and the right most non zero digit is replaced by a 0. Now, the value of 9X can be calculated as given below - _9.9999....repeating....9990 - 0.9999....repeating....9999 ---------------------------------------- _8.9999....repeating....9991 Therefore 9X = 8.9999....repeating....9991 When you divide 8.9999....repeating....9991 by 9 you get exactly 0.9999....repeating....9999. Hence X = 0.9999....repeating....9999. (Please note that the underscore character was used 3 times to achieve the desired alignment.) Edited August 25, 2005 by shakthig Link to comment Share on other sites More sharing options...

shakthig Posted August 25, 2005 Report Share Posted August 25, 2005 Therefore if X = 0.99999....repeating....9999 ___then, 10X = 9.99999....repeating....9990 As in all cases where a number is multiplied by 10, all digits are shifted to the right by one position and the right most non zero digit is replaced by a 0. This is a correction of an error in my previous post. The quotation above contains that error. The first occurence of the word "right" should be replaced by the word "left". Then the corrected form will be given below - Therefore if X = 0.99999....repeating....9999 ___then, 10X = 9.99999....repeating....9990 As in all cases where a number is multiplied by 10, all digits are shifted to the left by one position and the right most non zero digit is replaced by a 0. Link to comment Share on other sites More sharing options...

TomL Posted August 25, 2005 Report Share Posted August 25, 2005 I just had an encounter in my IRC channel that pertains to this, so I had the idea of posting a log of the discussion. Enjoy. [12:46] <woolcut> I'm a PhD candidate in mathematics [12:46] <@TomL> I see [12:46] <woolcut> I saw the post about .99999... = 1 and I was incited to respond [12:46] <@TomL> I haven't read that one myself [12:46] <@TomL> it seems rather obvious to me [12:46] <woolcut> They say it contradicts the law of identity [12:46] <@TomL> .99999... does NOT equal 1 [12:47] <woolcut> But it does equal one [12:47] <@TomL> it approaches 1 [12:47] <@TomL> not the same thing [12:47] <woolcut> Hrmm... [12:48] <@TomL> without grasping that, you can't understand differential calculus [12:48] <woolcut> No, you see a real number is often defined to be an infinite decimal expansion [12:48] <@TomL> same thing: dividing by 2 over and over again approaches 0, but even done infinitely you cannot ever EQUAL zero [12:48] <@TomL> that would be peachy if it were true [12:49] <woolcut> But thats not what an infinite decimal is [12:49] <@TomL> in reality, an integer does not represent the concept of an infinitely repeating expansion [12:50] <woolcut> why not 3 = 3.000.... [12:50] <woolcut> ? [12:50] <woolcut> Why is that not an infinite decimal expansion? [12:51] <@TomL> 3.0000... carries a different meaning than just 3 [12:51] <woolcut> The problem here is a conflict of terms [12:51] <@TomL> the problem is equivocation [12:51] <woolcut> You don't believe that "infinite decimals" represent anythiing but a process [12:51] <@TomL> in a strictly arithmetic context, 3 = 3.00000.... [12:51] <woolcut> But they can be used fruitfully under the appropriate equivalence relation [12:52] <@TomL> but outside of arithmetic, 3 does not equal 3.000... [12:52] <@TomL> they ahve different meanings [12:53] <woolcut> I understand, but they can be equated by an injective homomorphism that preserves order [12:53] <@TomL> I have no idea what you just said [12:53] <@TomL> blah blah blah [12:53] <@TomL> there is a philosophic basis upon which the idea of "3" exists [12:53] <woolcut> i.e. there is not more structure to 3.000... than there is to 3 [12:53] <woolcut> or vice versi [12:54] <@TomL> yes there is [12:54] <woolcut> Prove me wrong [12:54] <@TomL> 3.000... tells you more than 3 [12:54] <woolcut> How so [12:54] <@TomL> integers represent the law of identity: this thing as existing apart from that thing [12:54] <@TomL> so I might say I have 3 apples [12:55] <@TomL> as opposed to your 3 oranges [12:55] <@TomL> it would make no sense whatsoever to say I have 3.00000 apples [12:55] <@TomL> 1 apple is not necessarily the same as another [12:55] <@TomL> so 1 does not equal 1 if we're talking about apples [12:55] <woolcut> Okay [12:56] <woolcut> If you wish to make a philisophic difference between the two [12:56] <woolcut> But as I pointed out before, there really is no difference [12:56] <woolcut> (structurally) [12:56] <@TomL> there really is [12:56] <woolcut> However, the point of the argument then is that [12:57] <@TomL> you cannot use philosophy to form a concept and then throw the basis for the concepts existence out the window so you can focus on one aspect of its existence [12:57] <woolcut> .99999.... = 1.00000... [12:57] <@TomL> in this case, the arithmetic aspect [12:57] <@TomL> there is a reason and purpose for numbers [12:57] <@TomL> which exists apart from the numbers themselves [12:57] <@TomL> numbers do not exist apart from philosophic premises [12:58] <@TomL> just as 3.0000.... apples makes no sense, so does an accleration due to gravity of 3 for a given object make no sense [12:59] <@TomL> acceleration due to gravity is not a collection of like things [12:59] <woolcut> Okay, you're making a distinction between 3 and 3.0.. [12:59] <@TomL> of course [12:59] <woolcut> i.e. the distinction between an natural number and a real number [12:59] <@TomL> there's a reason there's a difference [12:59] <woolcut> And these are different entities of a sort [13:00] <woolcut> No one disagrees [13:00] <woolcut> The point then is that people disagree that: .999... = 1.000... [13:00] <woolcut> As _real_ numbers [13:00] <@TomL> if .9999... is arithemtically equivalent to 1.000.... fine.. I'm no math expert. But 1.000... is not the same concept as "1", which is what represents the law of identity [13:01] <woolcut> fine [13:01] <@TomL> if that is the claim, then they are equivocating arithmetic with philosophy [13:01] <woolcut> But that clearly is not the point of dispute [13:01] <@TomL> I don't know what the point of dispute is, I haven't read the thread [13:01] <woolcut> Well then you made a false assumption [13:02] <woolcut> Bad mojo for an objectivist [13:02] <@TomL> huh? [13:02] <@TomL> I have assumed nothing [13:02] <@TomL> I clearly prefaced my statement with "if" [13:02] <woolcut> [13:02] <woolcut> fair enough [13:03] <woolcut> but anything following an "if" is an assumption [13:03] <woolcut> So the starting point wasn't the one that I intended [13:04] <woolcut> And I lack the ability to scroll back up in this conversation because its in this weird java window [13:04] <woolcut> o_O [13:04] <@TomL> bummer [13:05] <woolcut> I'm actually very happy people make the distinction between real and natural numbers [13:05] <@TomL> I do, as I've demonstrated [13:05] <woolcut> But it really burns me when people disagree that .999... = 1.000.. [13:06] <woolcut> But you see, no one writes 1.000... [13:06] <woolcut> they just write 1 [13:06] <@TomL> which is wrong [13:06] <woolcut> Well, its a matter of context [13:07] <@TomL> .999... = 1.0000... (maybe, I don't know), but .999... != 1 [13:07] <@TomL> heck, even 1.000... != 1 [13:07] <woolcut> If you really wanted to acurately represent a real number you would need 4 lines on a page to do it [13:07] <@TomL> depends on how accurately, I suppose [13:07] <woolcut> Because 1.0000 and 3.1415926535... have no meanings besides a sequence of integers [13:08] <@TomL> well, they do [13:08] <woolcut> No, you see people have developed the real numbers from the natural numbers [13:08] <woolcut> And proven that starting from that base, we can successfully model all of the relevant properties they have [13:08] <@TomL> I certainly know what you mean by 3.14159265358... when you write it [13:09] <@TomL> even if all the digits aren't there (and can't possibly be) [13:09] <woolcut> But its a matter of technical representation [13:09] <woolcut> Technicaly a real number is just a function from the set of natural numbrs to the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} [13:09] <@TomL> uh [13:09] <woolcut> Or if you like binary the set {0, 1} [13:10] <@TomL> yes, i suppose so [13:10] <woolcut> Its really a matter of representation [13:10] <woolcut> And then we _equate_ particular representations [13:10] <woolcut> (Which is done in some artifical, but not arbitrary manner) [13:11] <@TomL> which is only valid in a particular context [13:11] <@TomL> jsut like concept formation [13:11] <woolcut> I suppose [13:11] <@TomL> like I was saying, 1.000... might equal 1 arithmetically, but not in any other context [13:12] <@TomL> it is an error to equate them in one context validly, and then switch contexts while maintaining the equation [13:12] <@TomL> when it was the context that made the equation valid in the first place [13:12] <woolcut> Sure, but for the purpose of analysis the only significant structure IS its arithmetic properties [13:12] <@TomL> no [13:13] <@TomL> for the purpose of arithmetic analysis [13:13] <@TomL> for the purpose of philosophical analysis, one considers more than arithmetic [13:13] <woolcut> granted [13:13] <@TomL> [13:14] <@TomL> thus, it is erronous to state an arithmetic truism and then conclude that some philosophic premise is in conflict [13:14] <woolcut> Well, thats the point! [13:14] <woolcut> [13:14] <@TomL> good, as long as we're straight on that [13:15] <woolcut> The topic started (stuff we've been discussing) then ("doesn't this break the law of identity?") [13:15] <@TomL> it doesn't even bring the law of identity into questeion [13:15] <@TomL> ridiculous [13:15] <@TomL> here's another ironic part [13:16] <@TomL> if real numbers are a function of natural numbers {0, 1} , you could also say: [13:16] <@TomL> that real numbers are a function of the law of identity [13:16] <woolcut> And I was totally floored [13:16] <woolcut> So thats why I'm so desperate to clear things up [13:16] <woolcut> However, I admit that I've come to objectivism by way of mathematics [13:16] <woolcut> (and philosophy in general) [13:16] <woolcut> So when I say "analysis" I really mean "real analysis" [13:16] <woolcut> Just because thats the lingo of my trade [13:16] <@TomL> ugh, lag [13:16] <woolcut> s'all right [13:16] <woolcut> Thats interesting [13:16] <@TomL> would you agree that real numbers are a function of the law of identity? [13:17] <@TomL> "1" is a mental concept representing it almost directly [13:17] <woolcut> If you are interested, I could send you a reference to the way we build the real numbers out of the natural numbers [13:17] <@TomL> and thus in turn, all math is a function of the law of identity [13:17] <woolcut> But I have to go to class in a couple minutes [13:17] <@TomL> so then to say that math somehow shows the law of identity to be false means that all math is false [13:18] <@TomL> and thus the math that proves the law of identity is also false [13:18] <@TomL> rather hilarious [13:18] <woolcut> [13:18] <woolcut> agreed [13:18] <woolcut> Would you like a reference to the construction? [13:18] <@TomL> not particularly, but thanks for the offer [13:18] <woolcut> Fair enough [13:19] <woolcut> It is interesting though, that we can construct the real number line from only a few definitions and 5 axioms [13:19] <woolcut> (Mathematical axioms) [13:20] <woolcut> But maybe only interesting to me, the mathematician [13:20] <woolcut> So, off I go to my (real) analysis class [13:20] <woolcut> best regards Link to comment Share on other sites More sharing options...

LauricAcid Posted August 26, 2005 Report Share Posted August 26, 2005 (edited) If Z set theory is consistent, then mathematics, such as real analysis, does not contradict that, for all x, x = x. There does not need to be confusion about the subject of this thread. A few points: 1. Most importantly, '...' is an informal notation. One cannot make calculations with the notation without taking into account that the notation only informally points to what is made rigorous in a formal language. In a formal language, the supposed problems commented upon in this thread (and at least one incorrect solution that recourses to nonsense notation) vanish, since the source of these misunderstandings is in taking informal notation as if it were literal. 2. It is true that the natural number 1 is not the real number 1.000... In general, in the most rigorous sense (which should be consulted in conversations in which confusions are arising), natural numbers are not integers, not rationals, and not reals. 3. However, as one of the interlocutors in the dialogue quoted in the post above tried to get across, the natural numbers with the operations of addition and multiplication are isomorphically embedded in the reals. For that matter, the natural numbers and the operations are isomorphically embedded in the integers and in the rationals. In this sense, as long as the context is understood, it's fine to informally refer to the real number 1. The real number 1 is what is mapped to by the natural number 1 by a certain isomorphism. 4. The real number informally named by '1.000...' is the same real number that is informally named by '.999...', which is the real number informally referred to as 1. It is not correct that .999... only gets closer and closer to 1.000. The sum of the terms in the sequence 9/10, 99/100, 999/1000... IS the real number 1.000..., which is informally referred to as the real number 1. 5. The set of real numbers is usually defined as the set of Dedekind cuts of rational numbers or the set of real numbers is defined as the set of equivalence classes of Cauchy sequences of rational numbers. In either case, the real numbers (with the operations and the ordering) is a complete ordered field, which is the motivation for whatever definition is given. Edited August 26, 2005 by LauricAcid Link to comment Share on other sites More sharing options...

Hal Posted August 26, 2005 Report Share Posted August 26, 2005 (edited) As long as you accept that 3 = 3.000... and 0.9999... = 1 "in a strictly arithmetic context", thats all that really matters. Your distinction between 1.000 and 1 strikes me as pointless if you mean anything more than that one of the symbols is conventionally used to denote a real number while the other denotes an integer, but it lies outside the domain of mathematics so I dont think its relevant here. Edited August 26, 2005 by Hal Link to comment Share on other sites More sharing options...

Hal Posted August 26, 2005 Report Share Posted August 26, 2005 (edited) 4. The real number informally named by '1.000...' is the same real number that is informally named by '.999...', which is the real number informally referred to as 1. It is not correct that .999... only gets closer and closer to 1.000. The sum of the terms in the sequence 9/10, 99/100, 999/1000... IS the real number 1.000..., which is informally referred to as the real number 1. Debateable. We can make the decision within a formal language to use the sign '=' in the context of limiting processes, and to say that a series = its limiting sum. But problems arise when you 'translate' out of that formal language, and into English. If someone wants to say that the series "0.9 + 0.09 + 0.009 + ..." doesnt ever equal 1 although it gets arbitrarilly close, would there be any objection to this? He is still doing the same mathematics as us, and his real analysis will be identical to ours - he is just choosing different English words to describe what he is doing. And at the end of the day, it doesnt matter. As long as we all agree on the mathematics itself, who cares? A similar issue arises with people translating statements about the non-existence of bijections between certain infinite sets into English sentences like "there are different sizes of infinity". Again, there are no objections to the maths itself - the only problem is whether that English statement is a reasonable description of what is going on. Edited August 26, 2005 by Hal Link to comment Share on other sites More sharing options...

LauricAcid Posted August 26, 2005 Report Share Posted August 26, 2005 (edited) But I don't think that denying that the sum equals 1 is the same mathematics as real analysis, since in real analysis it is a theorem that the sum does equal 1. It seems to me that to think that 'the sum does not equal 1' and 'the sum equals 1' can be reconciled is indeed to mock the law of non-contradiction. To excuse 'doesn't equal 1' is to be much too much forgiving of vague, undefined pseudo-mathematics best junkfiled as ignorant and confused cant. Don't get me wrong: If someone has mathematical axioms and a theory in which 'doesn't equal 1' obtains, then that's fine. But just to flail in the dark (as I read in different places, not your posts, of course) at mathematics is foolishness. And there is no intellectual advantage whatsoever in flinching from just saying that the formulations that provide mathematics include many that, at least at an informal conversational level, are most easily conveyed (though not necessarily asserted to have non-formal existence) as about infinite sets, infinite sequences and infinite summations (as these are rigorously defined in impeccably rigorous systems, as impeccable as, arguably more than, computer languages themselves, and indeed the theoretic basis and paradigm of computer programming) but to instead resort to hand-waving claptrap such as "potentially generated". We disagreed about formalization in another discussion. I think that this present context, as I mentioned in my previous post, is yet another example that points to the advantage (or, I and many others view, ultimately, the requirement) of formalization. But your point about translating into English is an interesting one, as is the question of translating from English. It is crucial to keep in mind that the semantics of first order languages are given by structures, which offer semantical rigor to go with the syntactical rigor of the theory. More fundamentally, roughly speaking, the mathematical meaning of formal systems comes from the relations of the terms, not their denotations in any particular structure. It's not as if, as too many people presuppose, mathematics names objects in the world and then talks about what those objects do among themselves. Rather, mathematics sets up systems in which one only has to observe the relations among the formulations to understand their analogy with the abstract relations that are the subject of mathematical thinking. And you've raised an excellent point about cardinality. Too many people who attack set theory, while knowing virtually nothing about it, don't understand that one may apply set theory without a commitment to platonist infinities, let alone a commitment to a physical infinity. Also, perhaps, and ironically, to argue that clashes among natural language mathematical formulations can be resolved by taking these formulations as being different ways of saying the same thing is to argue for a certain kind of realism. I don't deny that there can be a lucid theory of mathematical empiricism, but from naive proponents, at least in posting forums I've read, I've yet to see even an attempt to meet routine challenges such as: What is the empirical status of, say, the square root of -1, or what is the empirical status of irrational numbers, or transcendental numbers, etc.? So unless one can answer such questions to vindicate the demand that mathematics be empirical through and through, one can hardly say what it means for natural language formulations to have a common "meaning" about such mathematical objects even as simple as the square root of -1. Meanwhile, Objectivism claims not to be caught in the wedge between realism and formalism (loosely speaking, nominalism), since, for Objectivism, these are but two sides of a thousands year old bad penny. And that's easy for Objectivism to say, since Objectivism doesn't offer a mathematics, not even a developed philosophy of mathematics. Hey, anyone can scoff at a philosophical minefield...as long as one is content not to venture for the goods on the other side. Edited August 26, 2005 by LauricAcid Link to comment Share on other sites More sharing options...

Hal Posted August 26, 2005 Report Share Posted August 26, 2005 (edited) But I don't think that denying that the sum equals 1 is the same mathematics as real analysis, since in real analysis it is a theorem that the sum does equal 1. It seems to me that to think that 'the sum does not equal 1' and 'the sum equals 1' can be reconciled is indeed to mock the law of non-contradiction. No, I'm talking about the difference between ordinary language terms, and technical terms. The symbol '=' in mathematics is not identical in meaning with the word 'equals' in English. Its similar to how the logical AND is not exactly identical to "and" in English, and '=>' differs from 'implies'. Its not that "the sum both equals and does not equals 1" - its that the sum equals 1 if you are talking about "equals/=" in the mathematical sense, but this is not necessarily so if you are using the word 'equals' in its everyday English sense. In mathematics, 2 things are defined to be equal if they can be made arbitrarilly close. This is not necessarily so when talking about equality in English. To excuse 'doesn't equal 1' is to be much too much forgiving of vague, undefined pseudo-mathematics best junkfiled as ignorant and confused cant.It isnt pseudo-mathematics because its not mathematics - its a description of mathematics in English. Here is a piece of mathematics For all eps > 0, there exists an integer N > 0 such that for all n > N, the nth partial sum of the series (0.9 + 0.09 + 0.009 + ...) is less than eps. Now, how do we translate from this formal language into English? We can say either that the series equals 1, or that it gets arbitrarily close to 1. Both statements agree with the maths, they just question the English translation. We disagreed about formalization in another discussion. I think that this present context, as I mentioned in my previous post, is yet another example that points to the advantage (or, I and many others view, ultimately, the requirement) of formalization.No matter what your formalization is like, youre still going to have to use a natural language to explain it to others. And then the potential for disagreements arises. Again, consider how strongly people can object to something as simple as translating => into 'implies'. More fundamentally, roughly speaking, the mathematical meaning of formal systems comes from the relations of the terms, not their denotations in any particular structure. It's not as if, as too many people presuppose, mathematics names objects in the world and then talks about what those objects do among themselves. Rather, mathematics sets up systems in which one only has to observe the relations among the formulations to understand their analogy with the abstract relations that are the subject of mathematical thinking.I disagree with this. I dont see any real benefits to this emphasis on formalization, and I would hold that the 'meaning' of mathematical terms is contained just as much in the images and ideas people have in their heads as their formal definition. Mathematicians often reason visually/geometrically - they dont manipulate strings of symbols. The purpose of formalization is to give a rigorous definition of ideas and concepts that already exist anyway. Mathematicians had no problems working with real numbers for centuries before Dedekind gave us a formal definition. And mathematicians today do not think of real numbers as being 'cuts' or 'cauchy sequences' while they manipulate them, unless perhaps they are teaching a class on foundations of mathematics. So unless one can answer such questions to vindicate the demand that mathematics be empirical through and through, one can hardly say what it means for natural language formulations to have a common "meaning" about such mathematical objects even as simple as the square root of -1.I wouldnt say mathematics is empirical, but its formalization is guided by empirical concerns. Lets consider the dirac delta function. Once this had proved useful in physics, it was always going to be accepted, no matter the price - there is no question of a useful mathematical tool being rejected because it violates the traditional foundations. Luckily, someone managed to sneak it in by treating it as a limit of a sequence of functions. But if noone had managed to do this, we would just have revised the definition of 'function' in order to include the delta function (ie we would have changed the formalization because it was useful to do so). Similarly, if the axiom of choice were necessary for physics, we would have no problems about including it in our formalization. And if science needed the continum hypothesis to be resolved one way or the other, we would do it. If Dedekind hadnt managed to produce his definition of real numbers, we would just include 'real numbers' as being an irreducible undefinable term within our formalization. Complex numbers became accepted because it was shown that you could do interesting things with them - by the time Hamilton (?) gave a formal definition in terms of ordered pairs, they had already become a standard part of the mathematicians toolkit. The motivations for our adopting a particular set of definitions and axioms are quite often empirical. Edited August 26, 2005 by Hal Link to comment Share on other sites More sharing options...

source Posted August 26, 2005 Report Share Posted August 26, 2005 (edited) Here's a mathematical proof that 0.999~ = 1. Let f(x) = Sum[n=0 to infinity](x / 10^n). When x = 9/10 => f(x) = 0.9 + 0.09 + 0.009 + ... = 0.999~. Also, f(9/10) = 9/10 * Sum[n=0 to infinity](1 / 10^n) Since 1/10 < 1, I can use Maclaurin formula to calculate the sum of the above series. Maclaurin formula states that: Sum[n=0 to infinity](y^n) = 1 / (1 - y) for each y < 1. In the above series, y = 1/10. So, f(9/10) = 9/10 * 1 / (1 - 1/10) = 9/10 * (1 / (9/10)) = 9/10 * 10/9 = 1 Now we have: f(9/10) = 0.999~ and f(9/10) = 1 therefore, 0.999~ = 1 Edit: This post earned me a promotion to "Advanced Member." Edited August 26, 2005 by source Link to comment Share on other sites More sharing options...

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