By using "a < c < b", you are talking about whether a set is "densely ordered". R is indeed densely ordered. Q is not "densely ordered" because the set S={x in Q such that x^2<=2} does not have a least upper bound: Q has "holes". "Dense" requires comparing the set in question to some superset that contains the set in question against some topological definition of neighborhood. Q is dense in R in the normal sense of distance because every neighborhood of a point in R contains a point in Q. For example, every neighborhood in R of sqrt(2) contains an infinite number of rationals. Z is nowhere dense in R.
DH, When non-locality and locality are logically defined (as can be seen in http://www.geocities.com/complementarytheory/TOUM.pdf ) then 0.999...[base 10] < 1
The definitions of locality and non-locality are fatally flawed. Your document defines local as "x in A xor x not in A". This is a tautology (the expression is always true). The definition of non-local is a contradiction (the expression is always false).
Schmoe: Thanks for giving me a citation that did not require my going to the library. As I expected, the citation does not use the proofs posted here. Erdos is a well known, reputable mathematician. His proof dealt with manipulations of infinite series. He did not use the proofs posted here based on arithmetic with infinite strings of decimal digits. I wonder why he did not use the simpler easier to follow proofs posted here. Perhaps he did not use the proofs posted here because he was too dumb to devise then. He seems smarter than that to me, so I do not vote for this option. Perhaps he did not use them because they did not seem as fancy and erudite as his more complex proofs. I would expect Erdos to favor a simple, easier to understand proof over a more complex one. Perhaps he did not use the proofs posted here because he knew they were invalid. My vote is for the last reason. As I have been posting, the conclusions reached at this thread are correct, but the proofs posted here are not valid. In particular, the following proof is not valid. x = .99999 . . . 10x = 9.99999. . . 10x - x = 9 9x = 9 x = 1 Conclusion correct, but proof invalid. As posted by me earlier, the proof fails for any finite number of recurring nines and does not magically become valid for an infinite string of nines. For x = .999: 10x - x = 8.991, leading to x = .999 For x = 99999: 10x - x = 8.99991, leading to x = .99999 For x = 99999999: 10x - x = 8.99999991, leading to x = .99999999 Et cetera, proving that any string of recurrent nines is equal to itself. The above proof does not magically work for an infinite number of recurrent nines when it fails for every finite number.
The infinite series are exactly what justify the manipulations of the digits. The decimal notation is just a shorthand for these series. The whole argument they are referring to "the decimal part" and the "digits" of the numbers while writing them in series form. What do you think a decimal is if not a series? Why do it this way instead of writing as decimals? the obvious motivation for me is so students can see how you go about doing algebra with decimals, i.e. convert them to the corresponding series. You said this earlier, I was trying to be polite and give you a chance to redeem yourself. You may be 'familiar' in the sense that you've seen lots of proofs, but you are not what I would consider 'competent'. Anyone who understands limits/series beyond at say the level required in Apostol or Spivak's Calc. books should cry if they they aren't able to prove these basic facts about decimals: 1) multiplying by 10 shifts the digits by 1. 2) subtracting decimals is equivalent to subtracting the corresponding digits, provided no "borrows" are required. It's especially sad if they can't follow this despite seeing my post #206 which shows explicitly the series involved since there's really nothing more to it (likewise with the reference I gave). anyone familiar at a 'intro real analysis' level, say baby Rudin, should be beaten if they can't do the above and should cry if they can't justify 2) when "borrows" are required ("borrows" will require rearranging an infinite series, this will be justified by absolute converge but not normally seen in a calc class. edit- being morning and more awake, coming out of a calc class you have what's needed to deal with "borrows" as well. It can, and probably should, be done by moving only finitely many terms at a time. Still, probably beyond what i'd expect from even an advanced calculus student). At one point you also wrote, ".999999. . . .*10 -.99999. . . = 8.9999. . . 1". Anyone familiar with even intro. calc. should know better, and their teacher should cry if they see a student write this. I've laid out the details, given you reference that uses the same arguments I did. There's not much more I can do for you. I'll leave you with this link to look at. Notice how they manipulate the digits of 1/7. I expect you will complain that they are starting with 1/7=0.142857... but just concentrate on the operations they do on the digits. Multiplying by 10^6 gets 142857.142857... which they then say is equal to 142857+1/7. Shocking? Again, essentially the same manipulations I did, and one step closer to the proof you are complaining about since for some reason you want to see these kinds of manipulations on decimals and not the series (despite them being the same damn thing). ps. Another link. Short version of the proof you are objecting to, but with 0.6363... I had naively believed something like this link and the link above from Stillwells book would be inadequate for the reasons that I usually raise when someone tries a proof like this- namely that it'ss valid provided you've already shown these manipulations are justified, which can be done via infinite series. Maybe you just wanted to see it printed in a textbook whether they take the time to justify it fully or not.
No, the definition of non-locality it true (see the truth tables) whether x is nothing (nor connective is used) or something (and connective is used). You limit yourself only to locality (xor connective) and by forcing this limitation you cannot conclude anything about the logical basis of non-locality. Locality and non-locality true values cannot be defined by each other or in other words, they are two different categories. The objects of my system are the result of the associations between these categories.
Shadmi, I am using your definition of non-locality, "x in A and not x in A". This is never true. Either x is in A, or it is not, and in both cases the boolean expression is false. You are very quickly falling into the realm of crackpot. If you want to invent a new number system, fine, have fun. But do not invoke the the reals in doing so. You are, after all, rejecting the reals. In particular, you cannot simply use the continuum as if it already exists. You need to develop a method to construct numbers consistent with the way you define numbers. Perhaps you will find something new. It had better be useful or it will wallow in the dust bin of other invented systems. The reals, including the counter-intuitive result that 0.999...=1, are very useful. P.S. I suggest you learn how to use something other than Word to write a math paper.
This is never true if you force a xor_only connective on the system. The rest of your reply is nothing but your inability to grasp the membership concept beyond the xor connective. One thing is for sure, you don't know what a true value is. Here is the relevant part of my work: The Membership concept needs logical foundations in order to be defined rigorously. Let in be "a member of ..." Let out be "not a member of ..." Definition 1: A system is any framework which at least enables to research the logical connectives between in , out. Let a thing be nothing or something. Let x be a placeholder of a thing. Definition 2: x is called local if for any system A, x is in A xor x is out A returns true. The true table of locality is: Code: [i][b]in[/b][/i] [i][b]out[/b][/i] 0 0 → F 0 1 → T ([i][b]in[/b][/i] , [i][b]out[/b][/i] are not the same) = { }_ 1 0 → T ([i][b]in[/b][/i] , [i][b]out[/b][/i] are not the same) = {_} 1 1 → F Let x be nothing Definition 3: x is called non-local if for any system A, x is in A nor x is out A returns true. The true table of non-locality when x is nothing is: Code: [i][b]in[/b][/i] [i][b]out[/b][/i] 0 0 → T ([i][b]in[/b][/i] , [i][b]out[/b][/i] are the same) = { } 0 1 → F 1 0 → F 1 1 → F Let x be something Definition 4: x is called non-local if for any system A, x is in A and x is out A returns true. The true table of non-locality when x is something is: Code: [i][b]in[/b][/i] [i][b]out[/b][/i] 0 0 → F 0 1 → F 1 0 → F 1 1 → T ([i][b]in[/b][/i] , [i][b]out[/b][/i] are the same) = {_[u]}[/u]_ Let system Z be the complementation between non-locality and locality. The true table of Z is: Code: [i][b]in[/b][/i] [i][b]out[/b][/i] 0 0 → T ([i][b]in[/b][/i] , [i][b]out[/b][/i] are the same) = { } 0 1 → T ([i][b]in[/b][/i] , [i][b]out[/b][/i] are not the same) = { }_ 1 0 → T ([i][b]in[/b][/i] , [i][b]out[/b][/i] are not the same) = {_} 1 1 → T ([i][b]in[/b][/i] , [i][b]out[/b][/i] are the same) = {_[u]}[/u]_ Not at all. The reals are the particular case of local-numbers of my system (each local-number has an exact location along the real-line), where any number of the form 0.xxx... is a non-local number (it does not have an exact location along the real-line). The real-line of my system is not a collection of infinitely many distinct and local objects, but it is a non-local ur-element (see http://en.wikipedia.org/wiki/Urelement). I agree with you that if some system is considered as non-useful, it will not be used. My system is useful because: 1. The standard system is a particular case of it (the case of locality). 2. By defining non-locality logically (as I did above) a new mathematical space, which is beyond a local-only system (the standard one), can be researched. 3. The new mathematical space is directly based on the Symmetry concept itself, and from a general point of view of the Symmetry concept, the standard number system is the particular case of a total broken symmetry (as it is found by Peano or ZF). 4. In my system, the number system is any object that exists between full symmetry and total broken symmetry. 5. By using symmetry, my system can define concepts like Redundancy and Uncertainty as its first-order properties (we have to understand that if our mathematical system is based on a totally broken-symmetry as its first-order property, it cannot be used as a proper tool to deal with non-broken states that have Redundancy and Uncertainty as their inherent properties). 6. A number system that is limited to a particular case of symmetry as its first-order property, cannot fully deal with the real complexity that can be found among living organisms. 7. I think that the suggested extended number system (which is based on the Symmetry concept itself as its first-order property) is the right way to develop a mathematical system that can deal much better with real complexity. If you understand the above, you are invited to look again (with an open mind) at http://www.geocities.com/complementarytheory/TOUM.pdf .
I've been meaning to come back and respond to this bit, but it would be a pain to go into the detail you require without latex, so here it is. I just discovered you can't embed more than 3 images at a time, so this is chopped into 3 posts, read them as if they were one. Let <img src="http://www.forkosh.com/mimetex.cgi?x_n=\sum_{i=1}^{n}9/10^i"> and call the limit of this x, so x=0.999... You are saying <img src="http://www.forkosh.com/mimetex.cgi?10x_n-x_n=8+\sum_{i=1}^{n-1}9/10^i+1/10^n"> is true for all n. I agree. You're then trying to say ( with your 'et cetera') that <img src="http://www.forkosh.com/mimetex.cgi?9x=10x-x=\lim_{n\rightarrow\infty}(10x_n-x_n)=\lim_{n\rightarrow\infty} \left(8+\sum_{i=1}^{n-1}9/10^i+1/10^n\right)=8+\sum_{i=1}^{\infty}9/10^i=8.99\ldots"> At least that's what I hope you are trying to say and you've abondened the "8.999...1" idea. this I agree with. So if you manipulate the sequences this way your end result is 9x=8.999..., and dividing by 9 gives only x=0.999... I have no problems with this. (I'll point out it's strange that you have no problems claiming you can divide 8.99... by 9 to get 0.99..., I'm presuming you're using the usual rules taught in high school? Did you worry about justifying that part?)
What you are missing is that we can evaluate the limits in another way that corresponds to the proof you are objecting too. I'll just take a second to mention (dunno why one of the limits isn't rendering correctly, should be clear though), <img src="http://www.forkosh.com/mimetex.cgi?\lim_{n\rightarrow\infty}(10x_{n}-x_{n})=\lim_{n\rightarrow\infty}10x_{n}-\lim_{n\rightarrow\infty}x_{n}=10\lim_{n\rightarrow\infty}x_n-\lim_{n\rightarrow\infty}x_n=10x-x"> the first and second equalities are justified as all limits involved are convergent, the last from the definition of x. I thought it's worth pointing it out as you used it implicitly (at least I used it explicitly in my interpretation of the bit I quoted in my last post). For a reference, see "Principles of Mathematical Analysis", W. Rudin, third edition, page 49, theorem 3.3 property a) and b). If you don't have Rudin handy, let me know what you have easy access to and I'll try to accomodate. Rudin was within arms reach and is pretty common so i went with that. It should be bleedingly obvious that removing any finite number of terms from the begining of a sequence has no effect on the limit, but see Rudin page 51, just before theorem 3.6 he leaves it as an exercise to prove the limit of a subsequence is equal to the limit of the sequence (he asks something more, but it applies here since our sequence is convergent). Note x_2, x_3, x_4, ... is a subsequence of x_1,x_2,x_3,... so we have: <img src="http://www.forkosh.com/mimetex.cgi?\lim_{n\rightarrow\infty}10x_n=\lim_{n\rightarrow\infty}10x_{n+1}"> Applying this to the limits involved with 9.99..-0.99.., we have: <img src="http://www.forkosh.com/mimetex.cgi?9.99\ldots-0.99\ldots=10x-x=\lim_{n\rightarrow\infty}10x_n-\lim_{n\rightarrow\infty}x_n=\lim_{n\rightarrow\infty}10x_{n+1}-\lim_{n\rightarrow\infty}x_n"> I have no idea why that is rendering as though I typed "\in fty", it should be clear though. I can't recall now if you agreed 10x=9.99... as a decimal, this is easy enough to justify but I don't want to make it 4 posts and I've run out of images I can put in. Just ask and ye shall receive.
Apply theorem 3.3 from Rudin again to the sum of the limits on the right to get (no idea why the \infty isn't rendering properly): <img src="http://www.forkosh.com/mimetex.cgi?\lim_{n\rightarrow\infty}10x_{n+1}-\lim_{n\rightarrow\infty}x_{n}=\lim_{n\rightarrow\infty}(10x_{n+1}-x_n)"> I guess this could be considered the tricky part, that you can cut a term of the begining of one sequence and still line up the terms of the two sequences, this is like the Hilbert hotel 'problem', but it shouldn't be a problem if you take the time to look at the relevant sequences and have even a minor understanding of limits. We can evaluate the terms in the limit easily enough, <img src="http://www.forkosh.com/mimetex.cgi?10x_{n+1}-x_n=10\sum_{i=1}^{n+1}9/10^i-\sum_{i=1}^{n}9/10^i=\sum_{i=1}^{n+1}9/10^{i-1}-\sum_{i=1}^{n}9/10^i=9+\sum_{i=2}^{n+1}9/10^{i-1}-\sum_{i=1}^{n}9/10^i=9+\sum_{i=1}^{n}9/10^{i}-\sum_{i=1}^{n}9/10^i=9"> This is just shifting the index of a finite sum, and other operations of finite sums. No problems here? Hence, <img src="http://www.forkosh.com/mimetex.cgi?\lim_{n\rightarrow\infty}(10x_{n+1}-x_n)=\lim_{n\rightarrow\infty}9=9"> and we have justified the 9.99...-0.99...=9 step. Since 10x-x=9.99..-0.99.. this gives 9x=9, divide by 9 to get x=1. This is my last stab at it. You should be able to find the errors above if you think it's an invalid argument (likewise with my post #206 you ignored which does the above but in the language of series instead of limits, no real difference of course) In the end it seems you are dead set on taking the difference to of a limit as the difference of the limit term by term as they came at you. You can shift either limit by any finite amount you want and get the same result. It may "look" different, as with 8.99... and 9.000..., but shifting by any finite amount or chopping of terms at the begining of a sequence are all perfectly valid.
Schmoe: You are incredibly stubborn! First: I never claimed that recurrent nines could not be proven to be equivalent to one. I never disagreed with that conclusion. I have only argued that the proofs posted here using arithmetic on infinite strings of decimal digits are invalid. I have provided some very reasonable arguments to support my view. After some arguments back & forth you provided a citation which you claimed shows just such a proof presented by a reputable mathematician (Erdos) in a serious publication. Analysis of your citation makes it obvious that the Erdos proof uses manipulations of an infinite geometric series. It does not use arithmetic manipulation of infinite strings of decimal digits, even though the latter proof is much simpler and is easier to understand.. Your expert does not back you up. It is reasonable to believe that he does not back you up because the very simple proof is invalid. Id est: Your cited expert does not provide any support for the use of ordinary arithmetic on infinite strings of decimal digits. With your expert failing to back you up, you next make a series of somewhat complex posts using manipulations of infinite geometric series to prove that recurrent nines are equivalent to one. I am well aware of such proofs and consider them valid. However, they do not validate the flawed proofs using arithmetic on infinite strings of decimal digits. An invalid proof can lead to a correct conclusion, but that does not make it a valid proof. I clearly showed that for any value of n, the alleged proof shows only that n recurrent nines are equal to n recurrent nines. When a proof can be shown to lead to the same conclusion for all n as n grows without bound, it cannot be considered to prove some other conclusion when n reaches some magic value called infinity.
An issue related to this thread: Some apparently reasonable operations are not valid for divergent infinite sums, and can lead to erroneous conclusions. Consider the following series, which grows without bound as n grows without bound. Sum = 1 + ½ + 1/3 + 1/4 + 1/5 . . . + 1/n Sum = (1 + 1/3 + 1/5 . . .) + (½ + 1/4 + 1/6 . . .). This operation is invalid for a divergent series. Sum + (½ + 1/4 + 1/6. . .) = (1 + 1/3 + 1/5 . . .) + 2*(½ + 1/4 + 1/6 . . .) Sum + (½ + 1/4 + 1/6. . .) = (1 + 1/3 + 1/5 . . .) + (1 + ½ + 1/3 + 1/4 + 1/5 . . . ) Sum + (½ + 1/4 + 1/6. . .) = (1 + 1/3 + 1/5 . . .) + Sum ½ + 1/4 + 1/6. . . = 1 + 1/3 + 1/5 . . . But every term of the series on the right is greater than the corresponding term of the series on the left of the equals sign. The result is absurd and is due to the invalid operation early in the proof. The above proof seems valid except for its conclusion and is an example of subtle errors that can occur when using intuitive rather than formal methods. BTW: Using some sloppy semantics: Several of the above equalities are of the form Infinity + infinity = infinity + infinity
I never said you believed otherwise. I've explained what was wrong with your "8.99...1" attempt and the others. I went through in more detail what you were doing to get 9x=8.999... above. I agree if you manipulate the limits the way you did, this is the result, but this isn't representative of the proof you had trouble with. Once again, if you don't think a decimal is just shorthand for a series, what do you think it is? The Erdos book is doing the operations on the infinite series because this is exactly how you would justify it in terms of familiar limit theorems. Same ideas, different notations. See also the other links I gave a few posts back, they do EXACTLY the manipulations with the decimals that you want. The second one, "Elementary Number Theory in Nine Chapters" by James Joseph Tattersall writes: "For example, if n=0.6363..., then 100n=63.6363... . Thus 99n=100n-n=63. Therefore, n=63/99=7/11." Maybe you should write him for printing such invalid proofs? Likewise with Stillwell for using the invalid digit manipulations. Complex? That's a joke for someone who claims to be very familiar with these kidns of proofs. Nothing I did above invokes geometric series, it's all manipulating the terms of the sequence of partial sums, but you know that since you read it right? this is equivalent to manipulating digits of a decimal. Find the errors. I have to ask again since you keep avoiding the question (you are incapable of responding to direct questions it seems but I'll give it another go): What do you think a decimal is if not shorthand for a series? Why do you think converting to a series and using series results is not valid? Why do you think this doesn't justify the digit manipulations you were taught in high school? Furthermore, why doesn't dividing 8.99... by 9 to yield 0.99... bother you when other, simpler manipulations do? no, you didn't. What you did is not representative of the proof you have problems with. The proof you have problems with involves shifting the sequence by one before subtracting. I made this very clear above, you can't seem to find fault though? If you hope for a property to transfer from a sequence to a limit, you *must* prove this is the case. not everything does. Simple example, 1/n>0 for all n. The limit of this sequence is 0, yet 0>0 isn't true. Looks like the conclusion that's true as n grows without bound isn't necessarily true in the limit. Your crap about "8.99...1" falls in this category. You better believe I'm stubborn. I think you are dangerous to students who look in the wrong place on the internet. I cry whenever I see crappy maths on the internet, and worry that i may have to deal with the results of this kind of misinformation in the class room.
base 10 has problems The problem is the base 10 number system. in a base twelve system 1/4 or .3 ="s .250 in base ten. 1/3 or.4 in base 12 ="s .33333 in base 10, .3333 does not adiquilly describe 1/3. .99 would be bettewr described in base 100. We would do better understanding math problems by using a base # system that has a common denomunator with the components wew are using.Please Register or Log in to view the hidden image!
Schmoe:I feel the same about you I cry even more if you are actually a teacher. Crappy math on the internet is not as dangerous as crappy math in the classroom.
Maybe a stand alone proof of some digit manipulations is in order? I hope against all hope that you willthrow off your tradition of ignoring what I write and actually read this post and respond to it and tell me exactly where you believe there are errors. This isn't difficult stuff. If you can find errors besides possible typos, I will eat my degrees. The latex workaround irritates me, so I'll use the following notations: a<sub>-k</sub>a<sub>-k+1</sub>...a<sub>-1</sub>a<sub>0</sub>.a<sub>1</sub>a<sub>2</sub>... is a decimal number where 0<=a_i<=9 for all i and k is some non-negative integer. This notation is shorthand for the series sum(a<sub>i</sub>/10<sup>i</sup>, i=-k..inf)=a<sub>-k</sub>10<sup>k</sup>+...+a<sub>-1</sub>10<sup></sup>+a<sub>0</sub>+a<sub>1</sub>10<sup>-1</sup>+a<sub>2</sub>10<sup>-2</sup>+... We're used to the lead digit being non zero, except possibly when k=0, so I won't go against that. Fact 1. Multiplying a real number by ten has the effect of shifting the digits to the left one, Proof. if x=a<sub>-k</sub>a<sub>-k+1</sub>...a<sub>-1</sub>a<sub>0</sub>.a<sub>1</sub>a<sub>2</sub>... then 10*x=10*sum(a<sub>i</sub>/10<sup>i</sup>, i=-k..inf). Multiplication distributes over a convergent series, so: 10*x=sum(10*a<sub>i</sub>/10<sup>i</sup>, i=-k..inf)=sum(a<sub>i</sub>/10<sup>i-1</sup>, i=-k..inf)=sum(a<sub>i</sub>/10<sup>i</sup>, i=-k-1..inf)=a<sub>-k</sub>10<sup>k+1</sup>+...+a<sub>-1</sub>10<sup>2</sup>+a<sub>0</sub>10<sup>1</sup>+a<sub>1</sub>+a<sub>2</sub>10<sup>-1</sup>+a<sub>3</sub>10<sup>-2</sup>... this series is the decimal a<sub>-k</sub>a<sub>-k+1</sub>...a<sub>-1</sub>a<sub>0</sub>a<sub>1</sub>.a<sub>2</sub>a<sub>3</sub>..., so when multiplying a real number by 10 we can just shift the decimal to the right by 1. Fact 2. Subtracting decimal numbers (special case where no "borrows" are needed and y has only zeros to the left of the decimal point). If x=a<sub>-k</sub>a<sub>-k+1</sub>...a<sub>-1</sub>.a<sub>0</sub>a<sub>1</sub>a<sub>2</sub>... and y=0.</sub>b<sub>0</sub>b<sub>1</sub>b<sub>2</sub>... and a<sub>i</sub>>=b<sub>i</sub> for all i from 1 to infinity, then x-y is given by the decimal a<sub>-k</sub>a<sub>-k+1</sub>...a<sub>-1</sub>.(a<sub>0</sub>-b<sub>0</sub>)(a<sub>1</sub>-b<sub>1</sub>)(a<sub>2</sub>-b<sub>2</sub>)..., i.e we can just subtract the digits of y from the digits of x (note that our conditions on the a's and b's will imply 0<=a<sub>i</sub>-b<sub>i</sub><=9) Proof. We write x-y in terms of the series corresponding to the decimals: x-y = sum(a<sub>i</sub>/10<sup>i</sup>, i=-k..inf)+sum(b<sub>i</sub>/10<sup>i</sup>, i=1..inf)= sum(a<sub>i</sub>/10<sup>i</sup>, i=-k..0)+sum(a<sub>i</sub>/10<sup>i</sup>, i=1..inf)+sum(b<sub>i</sub>/10<sup>i</sup>, i=1..inf) The second inequality comes from breaking off the first k+1 terms of the first series, justified by the usual limit theorems (i.e. the limit of (y+x_n) as n->infinity is equal to y + limit of x_n as n->infinity. Apply this k+1 times to the sequence of partial sums). The infinite series are convergent, so we may subtract them term by term to get: sum(a<sub>i</sub>/10<sup>i</sup>, i=-k..0)+sum(a<sub>i</sub>/10<sup>i</sup>-b<sub>i</sub>/10<sup>i</sup>, i=1..inf) = sum(a<sub>i</sub>/10<sup>i</sup>, i=-k..0)+sum((a<sub>i</sub>-b<sub>i</sub>)/10<sup>i</sup>, i=1..inf) this is the series form of the decimal expansion claimed, and we are done. You'll notice Fact 1 justifies saying 10*(0.99...)=9.99... and fact 2 justifies saying 9.99...-0.99...=9.00...=9 but try not to hold that against them and give them an honest evauluation for correctness.
I'm devastated. Multiple times in this thread I've told you how to justify the digit manipulations. Multiple times you've totally ignored this and been unable to offer any criticism whatsoever. I've tried to encourage you to work them out on your own, they aren't difficult. I've offered up references that show these manipulations in the same way I did. I've offered up references that use these manipulations without justifying them. My last post I have proven the two relevant ones we need for the proof you dislike. I think I've responded to pretty much every point you've made, I'm asking you to read my last post and find the errors. I will eat my degrees if you find any that aren't typos.
This thread reminds me of an interesting notational system: Balanced trinary. Trinary number systems were outlawed many years ago by prudes. They were afraid that the digits would be called tits in keeping with binary digits being called bits. Due to the prudes, you find very little published about trinary number systems. When the radix is an odd number, the ue of digits with a negative value is preferred. The following is considered superior to using the digit values: 0, 1, 2 Balanced trinary uses three digits: +, 0, -and is a radix three system. You count from zero to 13 as follows 000 00+ 0+- 0+0 0++ +-- +-0 +-+ +0- +00 +0+ ++- ++0 +++ The digit values are plus one, zero & minus one. The position weights are 1, 3, 9, 27 . . . as you move left of the radix point. They are 1/3, 1/9, 1/27 . . . as you move right of the radix point. The notation has many interesting properties, one of which is the following. +.- - - - - - . . . represents 1/2 in decimal notation. 0.++++++ . . . . also represents decimal 1/2 The above is analogous to 1.00000 . . . being equivalent to .99999 . . . when using decimal notation. Another interesting property is that there is no special symbol representing the sign of a number. Negative numbers have - as the most significant digit. Positive numbers have + as the most significant digit.
