Schmoe: It is not that I do not read what you post. I disagree with some of your statements and proofs. I wonder if you read and understand what I post. If you do not read this entire post, check out the last two paragraphs. You claim the following. Fact 1. Multiplying a real number by ten has the effect of shifting the digits to the left one, You even provide what appears to be a valid proof, although perhaps something has been omitted. Suppose x = 5.444 and you are required to compute 10x - .666? The first step seems to be: 10x = 54.44. moving the decimal point to the right as you have proven. Next line up 54.44 with .666, making the decimal points match. Gee whiz: I must treat 54.44 as 54.440 to do the subtraction. Your proof is valid, but what about stating that the addition of a zero at the end of a decimal fraction does not change the value? Do you want me to say 10 x = 54.449? How about amending your proven rule to be Move the decimal point right one position and append one or more zeros? Now I have difficulty inspecting the last few digits of an infinite string of digits. How can you know what is out there? If I multiply a decimal fraction by 10, I do not add a nine at the end, I add a zero to do alignment. If I multiply an infinite string of digits by ten, I see no reason to assume that a zero has not been appended for alignment prior to doing a subtraction. As noted earlier, the proof using strings of decimal digits seems to prove that recurrent nines is equal to recurrent nines. At least that is what it proves for all the strings of nines that I can inspect, and I see no reason to assume that the proof fails for an infinite number of nines. You have seen the proof severl times. I will not repeat it here. Now that I think about this issue, my memory is telling me about a requirement for proof of convergence prior to allowing various operations on infinite series. It has been a long time since I took a course relevant to this subject matter. One of your earlier proofs used an infinite series rather than decimal notation. What about the following? x = 1 + 10 + 100 + 1000 + 10000 . . . 10x = 10 + 1000 + 10000 . . . 10x - x = -1 9x = -1 ------>>> x = -1/9? Before you manipulation an infinite series, you must prove that it converges, else you get into trouble. As mentioned before, I do not dispute the conclusion. I only dispute the validity of the proof. Now it seems that you must prove that infinitely recurring nines converge prior to using the proof you are so fond of. Using an infinite string of nines or an infinite geometric series, you must first prove convergence. Once convergence to one has been proven, your favorite proof is just some extra work to validate the already proven fact. Of course, proof of convergence of a general geometric series validates the method posted here to show that recurrent .142857 is equal to 1/7 It seems that the method for evaluating recurrent decimal fractions is not a proof, only a method validated by a general proof of convergence of geometric series. Working with infinite series (or infinite sets) has some subtle pit falls. You should take a course in the relevant subject matter. You might learn some interesting concepts.
Don't be thick. Appending a zero is not valid for an infinite series. 9.999...0 is not a number. You don't even have to append a zero when multiplying 54.44 by ten, because 54.44 is more fully written as 54.44000... Shift the decimal over by one. End of story. Don't be thick. The product of two finite numbers is finite. The series representation of the product always converges. The proof involves a simple application of the binomial expansion. Do you need Weierstrass limit theory to calculate derivative of x<sup>3</sup> - 5x<sup>2</sup> - 10x + 4? Of course not. Doing so is far too tedious. Just because people have not supplied tedious expansions and justifications of each individual step does not mean the proof is invalid. It simply means the work lacks complete rigor. Argggh! For crying out loud! Have you not read any of the links Shmoe provided? You are dangerous. The absolute convergence of any series of the form sum a<sub>n</sub>b<sup>-n</sup> (n=1 to infinity) where 0<=a<sub>n</sub><b, b>1 is obvious. The proof: Case 1. a<sub>n</sub> > 0 for all n>N. In this case, the infinite series is identically equal to the finite series sum a<sub>n</sub>b<sup>-n</sup> (n=1 to N). The sum of a finite number of finite numbers is finite. Case 2. The series doesn't terminate. Denote the set {m<sub>k</sub>} as the indices of the non-zero elements of {a<sub>n</sub>}. Omitting the zero terms in the series, the series becomes sum a<sub>m<sub>k</sub></sub>b<sup>-m<sub>k</sup> (k=1 to infinity}. Examine the ratio of consecutive terms in this series: (a<sub>m<sub>k+1</sub></sub>b<sup>-m<sub>k+1</sup>)/(a<sub>m<sub>k</sub></sub>b<sup>-m<sub>k</sup>). Since a<sub>m<sub>k+1</sub></sub>/a<sub>m<sub>k</sub></sub> < b and b<sup>-m<sub>k+1</sub></sup>/b<sup>-m<sub>k</sub></sup> <= 1/b, the ratio is strictly less than one. The series converges. I think Shmoe teaches the course.
What has been omitted then? Alright, I'll state it. If you have a terminating decimal, adding zeros after the last digit doesn't change the number. I also have difficulty inspecting the "last few digits of an infinite string of digits" as there are no "last few digits". Let me ask again how did you divide 8.99... by 9 to get .99... ? You believe this is where the proof leads, so justify this please. Yep. I stated convergence but didn't prove it. You didn't ask so I figured you were ok with it. More below. That's divergent, so no problems- I'm not saying anything about divergent series. Fine, agreed to this more than once already. Convergence can be done witout resorting to geometric series, for example for 0.99.. we need to consider the sequence 0.9, 0.99, ... which is 9/10, 99/100, ... correct? Each term is strictly less than 1, and it's an increasing sequence (you could prove this part with a geometric sum with a finite # of terms if you like, but induction would work just as well without ever mentioning the word "geometric"). Therefore it converges to the supremum of the set {0.9, 0.99, ...}, see any analysis text. likewise for any decimal number. The main point though, is this manipulation of the digits is perfectly valid. Done and done in this order: calc course based out of Edwards and Penny, Analysis course based out of Pffaffenberger (similar to Rudin), then out of Royden (measure theory etc.), and passed Analysis comprehensive exam and am currently doing a phd in Analytic number theory.
I had to cut short before I finished my post earlier and wanted to expand on a couple of things. That's fine, but you don't give indication on what in the proofs you thought was wrong, you'd just repeat your own versions, which I had already pointed out the problems of of. Tough to discuss anything when you won't address large and important chunks of my posts directly. Yes I have. If there's any point you've tried to make and don't feel I've responded to, let me know and I'll try to go into more detail. I want to stress again that in the case of a non terminating decimal, there is no place to append zeros. In either the terminating or non terminating case, you just shift to the right by one, if you want to add some zeros after in the terminating case, then go ahead though there's no harm in thinking of having infinitely many zeros fllowing in the terminating case to begin with (5.44400000000...). Note also my "Fact 2" is making no claims about subtracting decimals when you have to do any 'borrows', so doesn't say anything about your example (though I claim everything is still fine, it was just simpler to prove that limited case in this limited medium). It is absolutely critical that you realize in a decimal with infinitely many digits there is no "last digit(s)". This is just basic limit stuff, there's no last term in an infinite sequence. Equivalently, there is no largest natural number. This was pretty hasty and I figure a possible source of contention so I wanted to go into more depth. Consider the decimal 0.a<sub>1</sub>a<sub>2</sub>... I'll prove the sequence of partial sums of the corresponding series is bounded and increasing (restricing to a 0 to the left of the decimal is no problem, adding an integer to this won't affect the covnergence so this will handle the general case as well). Let S<sub>n</sub>=sum(a<sub>i</sub>/10<sup>i</sup>,i=1,n) be the nth partial sum. I claim S<sub>n</sub><=1-1/10<sup>n</sup> for all n. This will imply that the sequence is bounded by 1. Base case: S<sub>1</sub>=a<sub>1</sub>/10<1-1/10 as a<sub>1</sub><=9. Suppose for some n we have S<sub>n</sub><=1-1/10<sup>n</sup>. Then S<sub>n+1</sub>=S<sub>n</sub>+a<sub>n+1</sub>/10<sup>n+1</sup> <=1-1/10<sup>n</sup> + 9/10<sup>n+1</sup>=1-10/10<sup>n+1</sup> + 9/10<sup>n+1</sup>=1-1/10<sup>n+1</sup> where we've used the condition a<sub>n+1</sub><=9 and our induction hypothesis on the "<=" part. So our claim is proven and the sequence is bounded. Increasing is pretty much free from S<sub>n+1</sub>=S<sub>n</sub>+a<sub>n+1</sub>/10<sup>n+1</sup> since the a's are all >=0. An increasing and bounded sequence converges, so our sum is convergent. The reason to avoid the ratio test here is just to escape your objection about using geometric series results, the ratio test proof that comes to mind is the goemetric series + a comparison test. That's still fine though, it's not so much that we can prove 0.99..=1 this way that started all this, it's your claim that we can't manipulate the digits this way. Use a geometric series to justify convergence of a general decimal if you prefer on your way to the digit manipulation proofs.
I know what xor does. I assumed not a member of ... is the logical complement if is a member of .... If that is what you intended, your silly truth tables have extraneous elements. Both a logical statement and its logical complement can not be simultaneously true or simultaneously false. Those lines in your truth table that have local(x,A) false and non-local(x,A) true are unreachable. You better explain exactly what you mean if not a member of ... is anything other than the logical complement if is a member of .... The reals are defined by the Cantor sequences. That 1=0.999... is tautologically true in the reals. 0.999... has a very specific location on the real-number line, You are rejecting the reals. So stop calling what you want to work with the reals. Learn how to communicate. Your words and your silly diagrams scream "crackpot" because of the way you write. Learn how to use LaTeX. Real mathematicians use it to the exclusion of all other documentation systems.
Ok. When I get the chance, I am going to post my book's take on 0.999... = 1. It uses definitions in number theory.
How about something like this... Given C is a real number and be a root to f(x) = x^n + a_(n-1)*x^(n-1) + a_(n-2)*x*(n-2) + ... a_1*x + a_0 where all a_0, a_1, ..., a_(n-2), a_(n-1) are integers and a_0 != 0. Then C is either an integer or an irrational number. So, set f(x) = x^2 - 0.999... x^2 = 0.999... = 9*10^-1 + 9*10^-2 + ... x^2 = 9*(10^-1 + 10^-2 + ...) = 9*(1/9) by the summation of a geometric series. x^2 = 9/9 = 1 x = +/- 1. Now, x^2 = x for x = 1. So, 1 - 0.999... = 0 and 1 = 0.999...
in is "a member of ..." out is "not a member of ..." I use the truth tables of the connectives themselves, as the logical basis of my system, and the truth table of the xor connective is: Code: 0 0 → F 0 1 → T 1 0 → T 1 1 → F In my system, the logical state of nothing is represented by 0 , and the logical state of something is represented by 1. So by a xor connective (where x is a placeholder of nothing or something) if in is 0, then out must be 1, and if in is 1 then out must be 0, and this is exaclty the turth table of the xor connective itself (which is the logical basis of Locality). In the case of the nor truth table (where x is nothing) both in,out memberships simultaneously must not exist (are both nothing) in order to return true. In the case of the and truth table (where x is something) both in,out memberships simultaneously must exist (are both something) in order to return true. nor or and are the logical basis of Non-locality. Again, in my system, Locality(based on xor connective) and Non-Locality(based on nor or and connectives) are based on the truth tables of the logical connectives themselves (as can be seen in http://en.wikipedia.org/wiki/Truth_table). The above can be reduced to these two membership's fundamental states: If in,out are the same, then membership is Non-local (logical and or nor truth tables are used). If in,out are not the same, then membership is Local (logical xor truth table is used). ZF is based only on a local membership (which is a particular case of my system). DH, if some one depends on a specific representation method in order to understand abstract notions, he cannot be considered as a real mathematician. Furthermore, my representation method can be understood only if you are opened for both serial (a step by step thinking) and parallel (not a step by step thinking) notions and notations. By your replies it is understood that you are limited only to the serial (a step by step thinking) notions and notations.
I don’t know if this will convince anyone who isn’t already convinced at this point, but consider 7/7. That equals “1”, right? But 7/7 can be broken down into: 7/7 = 2/7 +5/7 2/7 = 0.285714285714285714285714285714 repeating 5/7 = 0.714285714285714285714285714285 repeating Which (surprise!) add together to give 0.999… It works with any fraction that has a repeating decimal representation. 2/55 + 53/55 = 55/55 = 1 2/55 = 0.03636363636363636363636363636363 53/55 = 0.96363636363636363636363636363636 or 2800/10543 + 7743/10543 = 10543/10543 = 1 2800/10543 = 0.2655790571943469600682917575642 7743/10543 = 0.7344209428056530399317082424357 Does the fact that this can be done with any fraction that has a repeating decimal representation perhaps at least suggest to anyone that 0.999… = 1?
Schmoe: This thread started with the following proof. x = 0.999...10x = 9.999... 10x - x = 9.999... - 0.999... 9x = 9 x = 1 0.999 . . . = 1 That looks like ordinary arithmetic taught in grades 1 to 5 and some elementary algebra taught in grades 8 or 9. Those topics do not include the axioms, definitions and other requirements of analysis, calculus, Cantor Transfinite number concepts, et cetera. Without using those bigger guns, the proof does not seem valid although the conclusion can readily be proven using slightly more sophisticated methods. You ask for objections to the above proof. First, consider a very similar proof. x = .99999 . . . 4x = ? Perhaps 4x = 3.999999 . . . now 4x - x = 3 —->>> 3x = 3 & x = 1 The above is essentially the same proof as the original. Can 4x = 3.99999 . . . . be justified using the notions of ordinary arithmetic, without using analysis, limits, calculus, Cantor transfinite number concepts? If we multiply .99999 by 4 we get 3.99996 Similarly, if we multiply .9999 9999 9999 by 4 we get 3.9999 9999 9996 I do not know of a defined algorithm for multiply infinitely recurrent nines by 4 or any other number. Ordinary arithmetic defines multiplication with a single digit multiplier as working from the right to the left of the multi-digit multiplicand. This requires starting with the last digit of an infinite decimal fraction, but you folks claim that there is no last digit. That is a reasonable claim if we consider concepts beyond those used in ordinary arithmetic. It is obvious that for say 10<sup>500</sup> nines, multiplication by 4 results in a final digit of 6. Is it unreasonable to suppose, that the same is true for an infinite number of nines? If not, how can it be shown that the multiplication results in an infinite string of nines? The original proof multiplied by ten, which seems to me to hide the problem of defining multiplication involving an infinite decimal fraction. For a multiplier of ten, we bypass the arithmetic definition and use the shortcut of shifting the decimal point, allowing us to not analyze the final digit, if any. If the proof using 4 as the multiplier invalid, while the proof using ten is okay? It is not difficult to prove that recurrent nines is equal to one, using metods only slightly more sophisicted than arithmetic and elementary algebra. Once this is done, the above proofs (uesing ten or 4 as the multipier) are justified by higher level methods. I still remember encountering the notion that the set of all even integers has the same cardinality (I think this is number of members) as the set of all integers. At age 18 or 19, my intuition insisted that there were twice as many integers as even integers. Since then, I do not trust my intuition relating transfinite sets and arithmetic on infinite strings of digits in the absence of some Cantor-like pairing process. BTW: I once thought I had a proof that the set of all real numbers was Aleph<sub>1</sub>. The set of all subsets of Aleph<sub>0</sub> is Aleph<sub>1</sub>. Each member of the set of all subsets of Aleph<sub>0</sub> can be represented by an infinite string of binary digits, where a one bit indicates the member is included in the subset and a zero bit indicates that it is not included. Hence each member of Aleph<sub>1</sub> corresponds to some infinite string of binary digits. Each member of the set of all real numbers also corresponds to an infinite string of binary digits. Hence, Aleph<sub>1</sub> is the number of members of the set of all real numbers. I forget why the above proof is invalid. I am not sure that I understood the refutation provided by my math professor at the time, but I trusted him.
Get a life, Dinosaur. Did you take any course in math beyond calculus? For example, real analysis, and algebra (NOT baby algebra; I mean the algebra where you learn what numbers truly are). These are typically upper-level undergraduate / lower level graduate college math courses. If you had done so, you would know Why those tricks you learned way back when are indeed valid. Why there is no six on the end of 4 * 0.999... Why this works so nicely with 10 (hint: 0.999... is in base 10). That multiplication distributes over the sum of a countable number of terms.
No kidding stuff beyond grade 9 is needed to justify the steps. I did say that the steps in the above proof are not justified even for a high school student. I've been saying they can be justified using calculus once you know the definitions behind the decimals (which they won't deal with in high school) and the theorems contained in even a first year calc class. Of course not. You can't even define what 0.99... means with the arithmetic you see pre-calculus, so you have no chance at proving whther any manipulations with it ar valid or not. Err, where are these relevant here? You don't need to know anything at all about transfinite numbers for this, at least nothing beyond a basic knowledge about the natural numbers having no upper bound. Okie. I do. It's not terribly hard. Just use the series definition of 0.99...: 4*0.999...=4*sum(9/10<sup>i</sup>,i=1,infinity)=sum(36/10<sup>i</sup>,i=1,infinity) =sum(30/10<sup>i</sup>+6/10<sup>i</sup>,i=1,infinity)=sum(30/10<sup>i</sup>,i=1,infinity)+sum(6/10<sup>i</sup>,i=1,infinity) =sum(3/10<sup>i</sup>,i=0,infinity)+sum(6/10<sup>i</sup>,i=1,infinity)=3+sum(3/10<sup>i</sup>,i=1,infinity)+sum(6/10<sup>i</sup>,i=1,infinity) =3+sum(3/10<sup>i</sup>+6/10<sup>i</sup>,i=1,infinity)=3+sum(9/10<sup>i</sup>,i=1,infinity) =3.999... Where I've used the usual theorems for convergent series all over the place. I consider them "usual" at least, and I think I've explicitly stated the theorems used in the manipulations like the above that You can look back at if you need. Crickey, if there's a last digit, what's the largest natural number? The digits to the right of a decimal are a sequence of integers from 0 to 9, a last digit would be the last in this sequence which would be the largest natural number. So what is it? Yes, totally unreasonable. You are completely missing the point of limits, sequences, etc. There's no last digit, no last element of a sequence, etc. Shown above how getting an infinite string of 9's can be justified. Multiplying by 10 just makes it a little simpler. You can define multiplication on two non-terminating decimals. Heck, you can take the approach of defining the reals as being the decimals, then defining the operations on them and showing they satisfy the axioms we like instead of the typical dedkind cuts, or cauchy sequences, or whatever else you like. I'm going to say it again for the 3rd or 4th time this post, and the nth time this thread. I'm even going to include a swear word for emphasis because this is so darn important. There is no fucking final digit in a non-terminating decimal. (for a terminating decimal, "last/final digit" is sometimes used to mean what is more properly called the "last/final non-zero digit") This sentence doesn't make sense, but in any case the proof with the 4 is ok as well. You don't have to show 0.99..=1 first to justify the above. You don't have to prove anything about geometric series either. Your mistake was believing you had any intuition at all. How could you? How many infinite sets had you worked with on a daily basis? They aren't something you meet in the playground (compare with everything you do pre-calculus though, arguably you have some intuition for that stuff). You certainly shouldn't trust your intuition. Where's the fault in the proofs I've provided though? Did you have something else beyond wanting to add zeros on the "end"? Is this all you are stuck on? there being no "end"? Why are you assuming that the power set of the naturals is Aleph<sub>1</sub>? The reals do have the same cardinality as the power set of the naturals.
To DH Dear DH, Please reply to http://www.sciforums.com/showpost.php?p=1206111&postcount=309 Thank you.
Schmoe: A little bit or research resulted in the answer to your following question. The above belief is due to erroneous memories of the Continuum Hypothesis and related topics from a course taken before you were born. Have we been really wasting time because neither of us noticed simple statements in each other’s posts. For example, you last post included the following (referring to the proof in the first post to this thread). I bolded a sentence in the above. I thought this what I have been claiming from my first post on. On its face, the proof is not valid without invoking some slightly more sophisticated methods. BTW: The following from your recent post seems more complicated than necessary. 4*0.999...=4*sum(9/10i,i=1,infinity)=sum(36/10i,i=1,infinity) =sum(30/10i+6/10i,i=1,infinity)=sum(30/10i,i=1,infinity)+sum(6/10i,i=1,infinity) =sum(3/10i,i=0,infinity)+sum(6/10i,i=1,infinity)=3+sum(3/10i,i=1,infinity)+sum(6/10i,i=1,infinity) =3+sum(3/10i+6/10i,i=1,infinity)=3+sum(9/10i,i=1,infinity) =3.999... Once you transform recurrent nines into a geometric series, the original proof in this thread becomes superfluous. The limit of the series is one, using the simple formula for the limit of a convergent geometric series. Simlarly once you multiply the convergent sereis by 4, you can skip to showing that 4 is the limit of the resulting series.
I've been reading what you've been writing. The part you bolded is not what you've been claiming, at least it's not what you've been writing. You've began with no mention of it being beyond high school, and when faced with my claims the steps could be justified with calculus (and proofs later) you've continued to argue they weren't justified. You went well beyond just saying it isn't justified in high school, which I agree with (exception of some advanced high schoolers), and claimed that it can't be justified at all. You seem to later accept that the steps could be justified after you've got the geometric series proof, claiming you needed this to justify convergnce of your series, this I proved to be an uneccessary requirement a few posts ago and you didn't comment on that. You made some terrible leaps from sequences to their limits (i.e. having 'last decimal places'), I don't see how I could have misinterpreted these when you repeated them over and over. You've made broad claims that these decimal manipulations aren't valid, and used this 'last decimal place' a critique of the proofs I supplied for these manipulations. You've made some odd choices of things you accept, like multiplying by 10 and shifting the decimal was a nono, but dividing 8.99... by 9 to get 0.99... was somehow ok (and you failed to respond when I pointed this oddity out, more than once). What have I misunderstood of your posts? From what I've seen you really don't understand limits and such. You probably took a first year university calculus class long ago, maybe even a multivariable one later on. You probably did well in it, and figured you were da bomb and understood everything. Nothing you did later on questioned your understanding beyond this level and you've gone many years believing you knew what's what. Shocking now that you are confronted with someone telling you otherwise. I've given you my background, after my first year of calculus I would have claimed I knew what was what. after taking "real" math classes, analysis and such, I would have told you I was just plain ignorant of my own ignorance after I finished the 'calculus' stream. A first year calculus student in a 'typical' calculus class understands next to nothing about limits. You just don't get it. I was proving that you can justify 4*0.99...=3.999... (without already having proven 0.99..=1) something you seemed to claim would invalidate the earlier proof. This just requried basic limiet/series theorems, you don't need geometric series at all here. The decimal manipulations that you have claimed time and time again to be unjustified appeals to intuition can be rigorously proven to be true and they work on *all* decimals, not just ones that are geometric series.
You're making this too complicated. Here's the best proof I've seen: 1/3=.33333333333333333333333333333333333 3 x .3333333333333=.999999999999999999999 but 3 x 1/3=3/3=1 so .999999999999999999999=3/3=1
well 2 =1 ... GG! let a = b Multiply both sides by a a² = ab Add (a² - 2ab) to both sides a² + a² - 2ab = ab + a² - 2ab Factor the left, and collect like terms on the right 2(a² - ab) = a² - ab Divide both sides by (a² - ab) 2 = 1
