mm-1027 === Subject: Ann: Ch 4.0 released Ch 4.0 is released. Ch is a C/C++ interpreter for cross-platform scripting, 2D/3D plotting, numerical computing and embedded scripting. * C/C++ interpreter Ch supports 1999 ISO C Standard (C99), C++ classes. Ch also supports many industry standards with over 8,000 functions including POSIX, socket/Winsock, X11/Motif, OpenGL, ODBC, C LAPACK, GTK+, Win32, and CGI. Functions in static or dynamical binary C/C++ libraries can be executed interpretively in scripting without re-compilation using Ch SDK. * 2D/3D plotting and numerical computing Ch supports 2D/3D graphical plotting, C LAPACK, high level numeric functions. Ch has built-in 2D/3D plotting support, generic mathematical functions and computational arrays for linear algebra and matrix computations, and advanced high-level numerical functions for linear systems, differential equation solving, integration, non-linear equations, Fourier analysis, curve fitting, etc. For example, linear system equation b = A*x can be written verbatim in Ch. Its numerical extensions to C makes Ch the best choice for numerical computing in C/C++ domain. With SoftIntegration graphical library (SIGL), the same program using 2D/3D plotting features can be executed in Ch or compiled using C++ compilers without any modification. Ch is ideal for rapid application development and deployment. For example, a Control System Toolkit for design and analysis of linear time-invariant control systems developed in Ch is available. * Embeddable scripting Ch 4.0 is optimized for embedded scripting. Embedded Ch allows users to embed Ch into other C/C++ application programs and hardware. It is ideal for open-architecture application integration. With Embedded Ch, C/C++ applications can be extended with all features of Ch. It allows execution of C/C++ scripts generated dynamically on-line. Embedded Ch has a small footprint. The pointer and time deterministic nature of the C language provides a perfect interface with hardware in real-time systems. * Third party software support Ch supports an increasing number of third party software applications. It includes Ch Control toolbox, Ch NAG Statistics Package, Ch CGI toolbox, Ch ODBC toolbox, Intel OpenCV for computer vision and image processing, National Instruments data acquisition toolkit NI-DAQ and motion control toolkit NI-Motion, Barret Technologys Barret Hand robotic manipulator control, and Perl Compatible Regular Expressions. Pricing and Availability Ch for Windows, Linux, Mac OS X, Solaris, and HP-UX is available. Ch Standard Edition is now free for both commercial and non-commercial uses in all platforms. Ch Professional Edition is free for academic use and costs $399 for personal or commercial use. More at http://www.softintegration.com === Subject: Re: maple vs mathematica > I just wonder what program to choose: maple 9 or mathematica 4.1 How about Ch? The feature comparison among Ch, MATLAB and Mathematica can be found at http://www.softintegration.com/products/features/ch_matlab_ mathematica.html and http://www.softintegration.com/products/features/ch_vs_ matlab.html. === Subject: Re: Why are irrational numbers not countable? === >Subject: Why are irrational numbers not countable? >If Im reading right, countable sets are defined as those that can be >mapped one to one to the natural numbers. Integers and rational >numbers are considered countable, even though there are more of them >(e.g. 2 integers for every natural number). First error is here. In the usual advanced undergraduate set theory one learns that the set of countable sets is not countable. This is called the power set. For a finite set of count n it would have 2^n elements. If the infinity for the naturals is aleph_0 the power set has cardinality aleph_1. If the continuum hypothesis is correct that is also the count of the number of reals. Correct the error and do the development again. The results are likely to be different so there is little point in pointing out the erors that result from the first error. > Sets of countable sets >are considered countable. So why not irrational numbers? >Consider any number with N decimal digits. The set of those numbers >can be mapped to 10^N or fewer natural numbers, and so is countable. >This will remain true even if you insert a decimal point somewhere in >the sequence of N digits. The set of all sets of numbers with N >decimal digits (N from 0 to infinity) must be countable as well. But >that necessarily includes the (infinite) sequences of decimal digits >which represent irrational numbers. >Or to put it really simplistically, put a 1 in front of any irrational >number (to handle cases like 0.0000ABCD... vs 0.ABCD... vs AB.CD...) >and remove the decimal point, and youre left with a unique natural >number - one which corresponds to no other irrational number. So all >irrational numbers map to a subset of the natural numbers, and any >subset of natural numbers can map one to one to the natural numbers - >i.e. countable. >So why arent irrational numbers considered countable? === Subject: Re: Why are irrational numbers not countable? === >>Subject: Why are irrational numbers not countable? >>If Im reading right, countable sets are defined as those that can be >>mapped one to one to the natural numbers. Integers and rational >>numbers are considered countable, even though there are more of them >>(e.g. 2 integers for every natural number). > First error is here. In the usual advanced undergraduate set theory > one learns that the set of countable sets is not countable. Its worse than that; there is not even such a thing as a set of all countable sets in the first place. In fact, there isnt even a set of all singleton sets; if there were, the union of that set would be the set of all sets, which doesnt exist in ZF. But I think you misread what he said, if you think the first error is here in the quoted paragraph. He didnt say anything about power sets or all countable sets; he merely observed (correctly) that its possible for one countably infinite set to be a proper subset of another. >This is > called the power set. For a finite set of count n it would have 2^n > elements. If the infinity for the naturals is aleph_0 the power > set has cardinality aleph_1. And your first error is here. The power set of a countably infinite set has cardinality 2^aleph_0. >If the continuum hypothesis is correct > that is also the count of the number of reals. It is provable without the continuum hypothesis that the cardinality of the reals is 2^aleph_0, which is the same as the cardinality of the power set of the naturals. The continuum hypothesis is the assertion that 2^aleph_0 = aleph_1, where aleph_1 is the cardinality of the first uncountable ordinal (= the cardinality of the set of all countable ordinals). Perhaps you meant to talk about the set of all countable ordinals in your first paragraph instead of the set of all countable sets, which doesnt exist. >Correct the error and > do the development again. The results are likely to be different so > there is little point in pointing out the erors that result from the > first error. Looks like good advice. -- Dave Seaman Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Why are irrational numbers not countable? [ deleted ] Some irrationals ARE countable. In fact, the roots of finite polynomial equations with integer coefficients p(x) = 0, ---a class that includes such irrational numbers as sqrt(2), sqrt(3), etc. as well as rational numbers, which are the roots of first-degree polynomials---are countable since we can establish a 1-1 correspondence between those roots and the positive integers. I leave the proof as a HW exercise for you. (Such numbers, BTW, are called algebraic.) However, as Cantor showed, the real numbers between 0 and 1 are not count- able, because if you propose a counting scheme for such numbers, then tabulate them in the order of enumeration, I can construct a number by taking the 1st decimal digit of the 1st item of the table and adding 1 (mod 10) to it. I put the result in the 1st place to the right of the decimal point. Then I take the 2nd digit of the 2nd number and add 1 (mod 10), placing it in the 2nd position to the right of the dp. By this process I construct a number that differs from EVERY number in your table, so your table did not include all the reals from 0 to 1, hence the assumption that you could map the reals 1-1 onto the integers was false. QED Of course, since one can map the real numbers from 0 to infinity 1-1 onto the reals from 0 to 1 by the invertible transformation y = x/(1-x), 0 < x < 1 [ inverse is x = y/(1+y) ] the cardinality of all positive reals is aleph_1 . The (invertible) transformation u = log(y), 0 < y < infinity [ inverse is y = exp(u) ] maps the positive reals onto the reals from -infinitey to +infinity, so ALL the reals have cardinality aleph_1 . The startling conclusion is that most real numbers are neither rational nor algebraic, but transcendental (like e = 2.71828... or pi = 3.14159...). Hope this helps. -- Julian V. Noble Professor Emeritus of Physics ^^^^^^^^^^^^^^^^^^ http://galileo.phys.virginia.edu/~jvn/ Science knows only one commandment: contribute to science. -- Bertolt Brecht, Galileo. === Subject: Re: Why are irrational numbers not countable? <3F50BE37.FF1DE4E3@virginia.edu> [...] > so ALL the reals have cardinality aleph_1 . [etc.] 2^aleph_0 is the actual answer. The hypothesis aleph_1 = 2^aleph_0 is undecidable. In some set theory models it holds, in others it does not hold. (Search keyphrase: Continuum Hypothesis). === Subject: Re: Why are irrational numbers not countable? > If Im reading right, countable sets are defined as those that can be > mapped one to one to the natural numbers. Integers and rational > numbers are considered countable, even though there are more of them > (e.g. 2 integers for every natural number). Sets of countable sets > are considered countable. So why not irrational numbers? [...] [erroneous attempts at construction explained in another reply] [...] > So why arent irrational numbers considered countable? (Wrong newsgroup: sci.math is more appropriate.) The word considered does not fit. It would be used fittingly if you, for example, talked about a politician whom some consider liberal, while others consider him conservative. When talking about countability in the simple situations you mentioned, there is no room for mere considering. Rationals are *provably* countable, and irrationals are *provably* uncountable. For proofs, there are dozens of books which present them at different levels of difficulty. There is actually a very explicit way of establishing one-to-one correspondence between the irrationals from (0,1) and *infinite* sequences of positive integers. This correspondence uses continued fractions, which are a lot of fun to study for their own sake. And the set of infinite sequences of positive integers can be proved uncountable, through a simple diagonal process. === Subject: Re: simultaneous diagonalizability > can anyone suggest me some thorough reference on the simultaneous > diagonalizability of two or more matrices? > I would realy appreciate it. > NB: Two are more matrices A1,A2,... are simultaneously diagonalizable if > there is a matrix P such that P^T(Ai)P is diagonal for all i. > I know an equivalent characterization but i would need some deeper > analysis, special cases, etc. For more on a very closely related subject see the book Simultaneous triangulization by Heydar Radjavi and Peter Rosenthal, published by springer in its Universitext series in 2000. HTH, Felix. === Subject: Re: simultaneous diagonalizability > Imre Polik a dit : > > can anyone suggest me some thorough reference on the simultaneous > diagonalizability of two or more matrices? > I would realy appreciate it. > If they commute with each other and they are diagonalizable, you can > diagonalize them simultaneously. > This result is powerful, and I dont know any other result on > simultaneous diagonalisation. > Exercise : > Let A and B two square matrices, with AB = BA. > Let P(X, Y) = det(AX - BY) > Prove that P(B, A) = 0. Since simultaneously diagonalizable matrices obviously commute, this condition is both necessary and sufficient. In fact it is the case that ANY set of diagonalizable matrices is simultaneously diagonalizable iff they all commute with each other. In one direction, the proof is obvious, while in the other what you have to do is show that any one of the matrices can only permute the eigenvalues of the other. === Subject: Re: Probability calculation > Hello - > Could someone tell me, in general terms, since Im not really a > mathematician -- > if: > 1) one were to be thinking about a specific topic, and > 2) then within seconds, randomly pick up a 1500 page book that > addresses this topic only once in a couple paragraphs only, > 3) open it to the exact page > 3) and then, to specifically look at the exact place within the page > that the topic exists > I know you probably dont have all the data to make any kind of > analysis, but what are the ballpark chances of something like that > happening? Or, what are the minimum chances of it happening? Well, if the topic occurs on 1 page of a 1500 page book, then the probability of you randomly opening the book to that page are 1 in 1500 or ~.0006667 John === Subject: Re: Probability calculation >> Hello - >> Could someone tell me, in general terms, since Im not really a >> mathematician -- >> if: >> 1) one were to be thinking about a specific topic, and 2) then within >> seconds, randomly pick up a 1500 page book that addresses this topic >> only once in a couple paragraphs only, 3) open it to the exact page >> 3) and then, to specifically look at the exact place within the page >> that the topic exists >> I know you probably dont have all the data to make any kind of >> analysis, but what are the ballpark chances of something like that >> happening? Or, what are the minimum chances of it happening? > Well, if the topic occurs on 1 page of a 1500 page book, then the > probability of you randomly opening the book to that page are 1 in 1500 > or ~.0006667 Yes, but is that really what he wants to know? People are often excessively impressed with that sort of coincidence because they ask a question like What are the odds of this exact improbable event happening at this time when they should be asking What are the odds of some events of this probability or less occurring to (me|anyone I know) in a given (year|lifetime)? --- Roy Stogner === Subject: Re: Looking for examples of bad data analysis <3F4E6089.47930C6F@SpamMeSenseless.us.ibm.com> In the darkness, Wood reached over and surreptitiously > _removed_the_prism_ from the spectrograph. An N-ray spectrum was > produced all the same. Sounds more like a hoax rather than bad analysis. Jeffery Winkler === Subject: Re: Looking for examples of bad data analysis >> If you happen to know of any illustrative example that mislead many >> researchers, please let me know. >The easiest mistake is in cosmology, when calculating the function >between a things distance from us and its speed. Normally its taken >that V = D*H where H is a constant (I think its called the hubble >constant.) is exact, but this violates relativity in that you can get >V = c. > There are several misconceptions here, among them: > 1) That the recessional velocity is a physical velocity constrained to > be less than c > 2) That V = D*H holds without modification over cosmological distances > 3) That H is large enough that the recessional velocity approaches c > for less than cosmological distances > 4) That cosmologists are not, in fact, aware of all this. > You may find it enlightening to temporarily set (2) aside, and given > the fact that the Hubble constant is roughly the inverse of the age of > the universe consider just what does happen at the distance where the > recessional velocity is equal to c. I see. I can understand if c is not a VELOCITY but rather the measure of the increase in the amount of space between two points divided by the time interval over which it happened... but then that would pose a problem. Which time interval are we using? And which space interval? If the universe is expanding, and if it IS possible to go past C, meaning we dont have to use any SR formulas, then this means that velocities add similar to the way they do in Newtonian physics. But I take it that GR is a more proper way to analyze this than Newtonian physics. Also, perhaps the reason there is Ôsuperluminal motion in far away galaxies is because the recessional velocity sets different speed boundaries, IE in another galaxy, relative to our galaxys reference frame, some speed less than c is the limit going toward our galaxy and some speed greater than c is the limit going away from our galaxy. Of course, in the other galaxys proper reference frames this isnt the case. Am I right? Or am I misconstruing the data again? This seems to be the only way to make sense of the hubble constant and recession velocity and such. (...Starblade Riven Darksquall...) === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, > 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to the >last line). >One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >--Mark But he assumes .999... = 1 in his equation before it is proven. Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 > 9(1) =/= 9(.999...) >> Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). >> He assumes 9(1) = 9 (going from the second to last line that you >> quoted, to the last line). >> One reason this proof is deficient is because of the assumption >> that 10(.9999999...) = 9.9999999... (which is true, but needs to be >> proven). >> --Mark > But he assumes .999... = 1 in his equation before it is proven. Would you point out where he makes this assumption? I repeat the entire proof, expanded a bit, with line numbers added for your convenience: [1] 9=9 [2] 9=9.9999999...-.9999999... [3] 9(1)=10(.9999999...)-.9999999 [4] Let x = .9999999... and substitute in [3] [5] 9(1) = 10x - x [6] 9(1) = 9(x) [7] Therefore x=1 In which line is the assumption .99999... = 1 used? --Mark === Subject: Re: PROOF that 0.99999... = 1 In sci.math, Mark Nudelman : > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, >> 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) > Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). > He assumes 9(1) = 9 (going from the second to last line that you > quoted, to the last line). > One reason this proof is deficient is because of the assumption > that 10(.9999999...) = 9.9999999... (which is true, but needs to be > proven). > --Mark >> But he assumes .999... = 1 in his equation before it is proven. > Would you point out where he makes this assumption? I repeat the entire > proof, expanded a bit, with line numbers added for your convenience: > [1] 9=9 > [2] 9=9.9999999...-.9999999... > [3] 9(1)=10(.9999999...)-.9999999 > [4] Let x = .9999999... and substitute in [3] > [5] 9(1) = 10x - x > [6] 9(1) = 9(x) > [7] Therefore x=1 > In which line is the assumption .99999... = 1 used? [3]. The possibility of an infinite borrow generates headaches. > --Mark -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 > In sci.math, Mark Nudelman >> Would you point out where he makes this assumption? I repeat the >> entire proof, expanded a bit, with line numbers added for your >> convenience: >> [1] 9=9 >> [2] 9=9.9999999...-.9999999... >> [3] 9(1)=10(.9999999...)-.9999999 >> [4] Let x = .9999999... and substitute in [3] >> [5] 9(1) = 10x - x >> [6] 9(1) = 9(x) >> [7] Therefore x=1 >> In which line is the assumption .99999... = 1 used? > [3]. The possibility of an infinite borrow generates headaches. Going from [2] to [3] merely assumes that 10(.99999...) = 9.9999.... This is indeed problematic and needs to be proven, as does the assumption that 9 = 9.99999...- 0.999999 in going from [1] to [2], but I dont see that either of these steps uses the assumption that .99999... = 1. --Mark === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 > 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >>assumes 9(1) = 9 (going from the second to last line that you quoted, to >>the >>last line). >>One reason this proof is deficient is because of the assumption that >>10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ jesus christ! do you know anything about mathematical induction?????????????? let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = 9*(1.11111111...) = 9.999999.. then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - 1)/(1 - 1/10))| = |1/10^n| = 1/10^n < e for all n >= N > -log(e) that means, the difference between the infinitely repeating decimal with period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this work for any number, not just 9) if you dont believe that x_n = 9.9999999999999999999 then thats your fault, you need to learn some simple math.... just try to find me a number sticktly between .999999999999..... and 1! you can do this for all x if you want... x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) if x is terminating or repeating in its tail, then the sum has a simple solution and its easy to calculate the answer. if you put x = 1, the {x} = 0 x = .99999...... then sum is just over 9/10^k which is easily to compute again, the only thing that you can have any sorta problem with is how .9999999999 could be reprsented by the sum, but that is your problem... as any halfwit knows that. === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, >> 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to >the >last line). >One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >--Mark >> But he assumes .999... = 1 in his equation before it is proven. >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >jesus christ! >do you know anything about mathematical induction?????????????? >let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >9*(1.11111111...) = 9.999999.. >then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >1)/(1 - 1/10))| >= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >that means, the difference between the infinitely repeating decimal with >period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this work >for any number, not just 9) >if you dont believe that x_n = 9.9999999999999999999 then thats your fault, >you need to learn some simple math.... just try to find me a number sticktly >between .999999999999..... and 1! >you can do this for all x if you want... >x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) >if x is terminating or repeating in its tail, then the sum has a simple >solution and its easy to calculate the answer. >if you put x = 1, the {x} = 0 >x = .99999...... >then sum is just over 9/10^k which is easily to compute >again, the only thing that you can have any sorta problem with is how >.9999999999 could be reprsented by the sum, but that is your problem... as >any halfwit knows that. Hey .999... IS NOT A REAL NUMBER (PERIOD). See math link below: http://mathworld.wolfram.com/HyperrealNumber.html .999... is of the form of a hyper-real number because there is a space between the real numbers between .999... and 1. .999... | | 1 ^ | See space A Hyperreal number is of the form Where n is a real number, x < n x = .999... n = 1 .999... < 1 THEREFORE, .999... =/= 1 Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >>assumes 9(1) = 9 (going from the second to last line that you quoted, to >>the >>last line). >>One reason this proof is deficient is because of the assumption that >>10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ >>jesus christ! >>do you know anything about mathematical induction?????????????? >>let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >>9*(1.11111111...) = 9.999999.. >>then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >>1)/(1 - 1/10))| >>= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >>that means, the difference between the infinitely repeating decimal with >>period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this >>work >>for any number, not just 9) >>if you dont believe that x_n = 9.9999999999999999999 then thats your >>fault, >>you need to learn some simple math.... just try to find me a number >>sticktly >>between .999999999999..... and 1! >>you can do this for all x if you want... >>x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) >>if x is terminating or repeating in its tail, then the sum has a simple >>solution and its easy to calculate the answer. >>if you put x = 1, the {x} = 0 >>x = .99999...... >>then sum is just over 9/10^k which is easily to compute >>again, the only thing that you can have any sorta problem with is how >>.9999999999 could be reprsented by the sum, but that is your problem... as >>any halfwit knows that. > Hey .999... IS NOT A REAL NUMBER (PERIOD). > See math link below: > http://mathworld.wolfram.com/HyperrealNumber.html > .999... is of the form of a hyper-real number because there is a space > between > the real numbers between .999... and 1. > .999... | | 1 > ^ > | > See space > A Hyperreal number is of the form > Where n is a real number, > x < n > x = .999... > n = 1 > .999... < 1 > THEREFORE, > .999... =/= 1 > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! === Subject: Re: PROOF that 0.99999... = 1 In sci.math, Jon Slaughter <10sbusfei7k2lee@corp.supernews.com>: [snipped for sanity] >> .999... =/= 1 >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ > your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! FSVO genius. Most of us use an alternate word with one less letter. :-) -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, >> 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to >the >last line). >>One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >jesus christ! >do you know anything about mathematical induction?????????????? >let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >9*(1.11111111...) = 9.999999.. >then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >1)/(1 - 1/10))| >= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >that means, the difference between the infinitely repeating decimal with >period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this >work >for any number, not just 9) >if you dont believe that x_n = 9.9999999999999999999 then thats your >fault, >you need to learn some simple math.... just try to find me a number >sticktly >between .999999999999..... and 1! >you can do this for all x if you want... >x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) >if x is terminating or repeating in its tail, then the sum has a simple >solution and its easy to calculate the answer. >if you put x = 1, the {x} = 0 >x = .99999...... >then sum is just over 9/10^k which is easily to compute >again, the only thing that you can have any sorta problem with is how >.9999999999 could be reprsented by the sum, but that is your problem... as >any halfwit knows that. >> Hey .999... IS NOT A REAL NUMBER (PERIOD). >> See math link below: >> http://mathworld.wolfram.com/HyperrealNumber.html >> .999... is of the form of a hyper-real number because there is a space >> between >> the real numbers between .999... and 1. >> .999... | | 1 >> ^ >> | >> See space >> A Hyperreal number is of the form >> Where n is a real number, >> x < n >> x = .999... >> n = 1 >> .999... < 1 >> THEREFORE, >> .999... =/= 1 >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! youre not your Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > .999... | | 1 > ^ > | > See space Pure scribble. > A Hyperreal number is of the form You would not know a hyperreal if it bit you. You have not the foggiest notion of how the real number system R is extended to *R. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> .999... | | 1 >> ^ >> | >> See space >Pure scribble. >> A Hyperreal number is of the form >You would not know a hyperreal if it bit you. You have not the foggiest >notion of how the real number system R is extended to *R. >Bob Kolker Hey, I thought you said I didnt know what it was. You are wrong again, and again, again. http://mathworld.wolfram.com/HyperrealNumber.html Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 In sci.math, S. Enterprize Company > > .999... | | 1 > ^ > | > See space >>Pure scribble. > > A Hyperreal number is of the form >>You would not know a hyperreal if it bit you. You have not the foggiest >>notion of how the real number system R is extended to *R. >>Bob Kolker > Hey, I thought you said I didnt know what it was. You are wrong again, and > again, again. > http://mathworld.wolfram.com/HyperrealNumber.html Like that tells him *anything*. Heres a few Qs for you. [1] If d is such that 0 < d < 1/n for all n in N, what is d^2? d^3? sqrt(d)? [2] Why is 5/5 != 9/9? 5/5 = 1, of course; 0.2 * 5 = 1. 9/9, by contrast, is 0.111... * 9 = 0.999... = 1 - d. In base 12, 1/9 = 0.14(12) but 1/5 = .24972497...(12) ; therefore in this case 9/9 = 1 but 5/5 = 1-d. Does it matter what base one uses for arithmetic? [3] Explain how one computes D_10[.999..., w-1], where w (omega) is the first transfinite ordinal, and D_10[r,n] is rs nth digit to the right of the decimal point, if n is an integer, then evaluate D_10[(.999... + 9)/10, w-1] and D_10[.999... * 10 - 9, w-1]. (n can be negative but thats not all that important here.) [.sigsnip] -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 > jesus christ! > do you know anything about mathematical induction?????????????? Enterprise does not even know what end shit comes out of. He is a total mathematical incompetent. He makes JSH look intelligent by comparison. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> jesus christ! >> do you know anything about mathematical induction?????????????? >Enterprise does not even know what end shit comes out of. He is a total >mathematical incompetent. He makes JSH look intelligent by comparison. >Bob Kolker Whats a hyper-real number? Do you even know anything about math? Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > Whats a hyper-real number? Do you even know anything about math? No. But I do know how the hyperrals are constructed. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 In sci.math, robert j. kolker : >> Whats a hyper-real number? Do you even know anything about math? > No. But I do know how the hyperrals are constructed. > Bob Kolker http://mathworld.wolfram.com/HyperrealNumber.html is extremely bare-bones (is there one hyperreal? more than one? arithmetic operations? proofs?) but at least its a start. A reference link http://members.tripod.com/PhilipApps/line.html looks to be little more than my attempts at d-math, though there might be more than one d -- or H, its dual. No doubt one could claim at least three theories: [1] An infinite hierarchy of d < 1/n for all n in N: 0 < ... < d^4 < d^3 < d^2 < d < 1, with a more or less standard algebra (e.g., (1-d)^3 = 1 - 3d + 3d^2 - d^3). [2] d^k = d for some k in N. [3] Some other esoteric condition. I suppose one might even notate this as R[d] -- a standard polynomial group over R, with a slightly weird ordering. And again, I must complain that S. Enterprize is being extremely sloppy here. (Not that Im all that neat, but hopefully my notations clear at least.) -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 >> Whats a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are You dont even know what a hyper-real number is??? And you are name calling people here like you know everything?????? Why not admit you ARE WRONG! constructed. >Bob Kolker Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > Whats a hyper-real number? Do you even know anything about math? >>No. But I do know how the hyperrals are > You dont even know what a hyper-real number is??? And you are name calling > people here like you know everything?????? Why not admit you ARE WRONG! Quick. Define an ultra-filter. No, dont look it up. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> Whats a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are >> You dont even know what a hyper-real number is??? And you are name >calling >> people here like you know everything?????? Why not admit you ARE WRONG! >Quick. Define an ultra-filter. No, dont look it up. >Bob Kolker Oh this is so hard to understand, I might need to take an asprin for a headache. Ill define it with an example. Suppose you have alot of people here making noise here on this NG and they dont know what they are talking about with .999..., and then comes along an ultrafilter F_Smart1234 with the correct information. What we do is apply ultrafilter F_Smart1234 to the whole set S of noise on the NG, and then just the pure correct answer is shown. The ultrafilter is then said to be a success and has worked very well, and is therefore proven. Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > Whats a hyper-real number? Do you even know anything about math? >>No. But I do know how the hyperrals are > > You dont even know what a hyper-real number is??? And you are name >>calling > people here like you know everything?????? Why not admit you ARE WRONG! >>Quick. Define an ultra-filter. No, dont look it up. >>Bob Kolker > Oh this is so hard to understand, I might need to take an asprin for a >headache. > Ill define it with an example. Suppose you have alot of people here >making >noise here on this NG and they dont know what they are talking about with >.999..., and then comes along an ultrafilter F_Smart1234 with the correct >information. What we do is apply ultrafilter F_Smart1234 to the whole set S >noise on the NG, and then just the pure correct answer is shown. > The ultrafilter is then said to be a success and has worked very well, and >is therefore proven. Your turn. Perform a ANOVA statistical test between .999... and 1. And of course go into details explaining what the ANOVA test is. hurry hurry dont look... Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 >> Whats a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are >> You dont even know what a hyper-real number is??? And you are name >calling >> people here like you know everything?????? Why not admit you ARE WRONG! >Quick. Define an ultra-filter. No, dont look it up. Oh, but I do have the right to refresh my memory. I even gave you time to do this and you still dont know what a hyper-real number is. >Bob Kolker Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: A Quantum Poem for Xmas === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, > 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to the >last line). >One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >--Mark But he assumes .999... = 1 in his equation before it is proven. Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 > 9(1) =/= 9(.999...) >> Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). >> He assumes 9(1) = 9 (going from the second to last line that you >> quoted, to the last line). >> One reason this proof is deficient is because of the assumption >> that 10(.9999999...) = 9.9999999... (which is true, but needs to be >> proven). >> --Mark > But he assumes .999... = 1 in his equation before it is proven. Would you point out where he makes this assumption? I repeat the entire proof, expanded a bit, with line numbers added for your convenience: [1] 9=9 [2] 9=9.9999999...-.9999999... [3] 9(1)=10(.9999999...)-.9999999 [4] Let x = .9999999... and substitute in [3] [5] 9(1) = 10x - x [6] 9(1) = 9(x) [7] Therefore x=1 In which line is the assumption .99999... = 1 used? --Mark === Subject: Re: PROOF that 0.99999... = 1 In sci.math, Mark Nudelman : > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, >> 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) > Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). > He assumes 9(1) = 9 (going from the second to last line that you > quoted, to the last line). > One reason this proof is deficient is because of the assumption > that 10(.9999999...) = 9.9999999... (which is true, but needs to be > proven). > --Mark >> But he assumes .999... = 1 in his equation before it is proven. > Would you point out where he makes this assumption? I repeat the entire > proof, expanded a bit, with line numbers added for your convenience: > [1] 9=9 > [2] 9=9.9999999...-.9999999... > [3] 9(1)=10(.9999999...)-.9999999 > [4] Let x = .9999999... and substitute in [3] > [5] 9(1) = 10x - x > [6] 9(1) = 9(x) > [7] Therefore x=1 > In which line is the assumption .99999... = 1 used? [3]. The possibility of an infinite borrow generates headaches. > --Mark -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 > In sci.math, Mark Nudelman >> Would you point out where he makes this assumption? I repeat the >> entire proof, expanded a bit, with line numbers added for your >> convenience: >> [1] 9=9 >> [2] 9=9.9999999...-.9999999... >> [3] 9(1)=10(.9999999...)-.9999999 >> [4] Let x = .9999999... and substitute in [3] >> [5] 9(1) = 10x - x >> [6] 9(1) = 9(x) >> [7] Therefore x=1 >> In which line is the assumption .99999... = 1 used? > [3]. The possibility of an infinite borrow generates headaches. Going from [2] to [3] merely assumes that 10(.99999...) = 9.9999.... This is indeed problematic and needs to be proven, as does the assumption that 9 = 9.99999...- 0.999999 in going from [1] to [2], but I dont see that either of these steps uses the assumption that .99999... = 1. --Mark === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 > 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >>assumes 9(1) = 9 (going from the second to last line that you quoted, to >>the >>last line). >>One reason this proof is deficient is because of the assumption that >>10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ jesus christ! do you know anything about mathematical induction?????????????? let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = 9*(1.11111111...) = 9.999999.. then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - 1)/(1 - 1/10))| = |1/10^n| = 1/10^n < e for all n >= N > -log(e) that means, the difference between the infinitely repeating decimal with period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this work for any number, not just 9) if you dont believe that x_n = 9.9999999999999999999 then thats your fault, you need to learn some simple math.... just try to find me a number sticktly between .999999999999..... and 1! you can do this for all x if you want... x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) if x is terminating or repeating in its tail, then the sum has a simple solution and its easy to calculate the answer. if you put x = 1, the {x} = 0 x = .99999...... then sum is just over 9/10^k which is easily to compute again, the only thing that you can have any sorta problem with is how .9999999999 could be reprsented by the sum, but that is your problem... as any halfwit knows that. === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, >> 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to >the >last line). >One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >--Mark >> But he assumes .999... = 1 in his equation before it is proven. >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >jesus christ! >do you know anything about mathematical induction?????????????? >let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >9*(1.11111111...) = 9.999999.. >then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >1)/(1 - 1/10))| >= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >that means, the difference between the infinitely repeating decimal with >period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this work >for any number, not just 9) >if you dont believe that x_n = 9.9999999999999999999 then thats your fault, >you need to learn some simple math.... just try to find me a number sticktly >between .999999999999..... and 1! >you can do this for all x if you want... >x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) >if x is terminating or repeating in its tail, then the sum has a simple >solution and its easy to calculate the answer. >if you put x = 1, the {x} = 0 >x = .99999...... >then sum is just over 9/10^k which is easily to compute >again, the only thing that you can have any sorta problem with is how >.9999999999 could be reprsented by the sum, but that is your problem... as >any halfwit knows that. Hey .999... IS NOT A REAL NUMBER (PERIOD). See math link below: http://mathworld.wolfram.com/HyperrealNumber.html .999... is of the form of a hyper-real number because there is a space between the real numbers between .999... and 1. .999... | | 1 ^ | See space A Hyperreal number is of the form Where n is a real number, x < n x = .999... n = 1 .999... < 1 THEREFORE, .999... =/= 1 Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >>assumes 9(1) = 9 (going from the second to last line that you quoted, to >>the >>last line). >>One reason this proof is deficient is because of the assumption that >>10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ >>jesus christ! >>do you know anything about mathematical induction?????????????? >>let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >>9*(1.11111111...) = 9.999999.. >>then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >>1)/(1 - 1/10))| >>= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >>that means, the difference between the infinitely repeating decimal with >>period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this >>work >>for any number, not just 9) >>if you dont believe that x_n = 9.9999999999999999999 then thats your >>fault, >>you need to learn some simple math.... just try to find me a number >>sticktly >>between .999999999999..... and 1! >>you can do this for all x if you want... >>x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) >>if x is terminating or repeating in its tail, then the sum has a simple >>solution and its easy to calculate the answer. >>if you put x = 1, the {x} = 0 >>x = .99999...... >>then sum is just over 9/10^k which is easily to compute >>again, the only thing that you can have any sorta problem with is how >>.9999999999 could be reprsented by the sum, but that is your problem... as >>any halfwit knows that. > Hey .999... IS NOT A REAL NUMBER (PERIOD). > See math link below: > http://mathworld.wolfram.com/HyperrealNumber.html > .999... is of the form of a hyper-real number because there is a space > between > the real numbers between .999... and 1. > .999... | | 1 > ^ > | > See space > A Hyperreal number is of the form > Where n is a real number, > x < n > x = .999... > n = 1 > .999... < 1 > THEREFORE, > .999... =/= 1 > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! === Subject: Re: PROOF that 0.99999... = 1 In sci.math, Jon Slaughter <10sbusfei7k2lee@corp.supernews.com>: [snipped for sanity] >> .999... =/= 1 >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ > your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! FSVO genius. Most of us use an alternate word with one less letter. :-) -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, >> 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to >the >last line). >>One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >jesus christ! >do you know anything about mathematical induction?????????????? >let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >9*(1.11111111...) = 9.999999.. >then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >1)/(1 - 1/10))| >= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >that means, the difference between the infinitely repeating decimal with >period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this >work >for any number, not just 9) >if you dont believe that x_n = 9.9999999999999999999 then thats your >fault, >you need to learn some simple math.... just try to find me a number >sticktly >between .999999999999..... and 1! >you can do this for all x if you want... >x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) >if x is terminating or repeating in its tail, then the sum has a simple >solution and its easy to calculate the answer. >if you put x = 1, the {x} = 0 >x = .99999...... >then sum is just over 9/10^k which is easily to compute >again, the only thing that you can have any sorta problem with is how >.9999999999 could be reprsented by the sum, but that is your problem... as >any halfwit knows that. >> Hey .999... IS NOT A REAL NUMBER (PERIOD). >> See math link below: >> http://mathworld.wolfram.com/HyperrealNumber.html >> .999... is of the form of a hyper-real number because there is a space >> between >> the real numbers between .999... and 1. >> .999... | | 1 >> ^ >> | >> See space >> A Hyperreal number is of the form >> Where n is a real number, >> x < n >> x = .999... >> n = 1 >> .999... < 1 >> THEREFORE, >> .999... =/= 1 >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! youre not your Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > .999... | | 1 > ^ > | > See space Pure scribble. > A Hyperreal number is of the form You would not know a hyperreal if it bit you. You have not the foggiest notion of how the real number system R is extended to *R. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> .999... | | 1 >> ^ >> | >> See space >Pure scribble. >> A Hyperreal number is of the form >You would not know a hyperreal if it bit you. You have not the foggiest >notion of how the real number system R is extended to *R. >Bob Kolker Hey, I thought you said I didnt know what it was. You are wrong again, and again, again. http://mathworld.wolfram.com/HyperrealNumber.html Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 In sci.math, S. Enterprize Company > > .999... | | 1 > ^ > | > See space >>Pure scribble. > > A Hyperreal number is of the form >>You would not know a hyperreal if it bit you. You have not the foggiest >>notion of how the real number system R is extended to *R. >>Bob Kolker > Hey, I thought you said I didnt know what it was. You are wrong again, and > again, again. > http://mathworld.wolfram.com/HyperrealNumber.html Like that tells him *anything*. Heres a few Qs for you. [1] If d is such that 0 < d < 1/n for all n in N, what is d^2? d^3? sqrt(d)? [2] Why is 5/5 != 9/9? 5/5 = 1, of course; 0.2 * 5 = 1. 9/9, by contrast, is 0.111... * 9 = 0.999... = 1 - d. In base 12, 1/9 = 0.14(12) but 1/5 = .24972497...(12) ; therefore in this case 9/9 = 1 but 5/5 = 1-d. Does it matter what base one uses for arithmetic? [3] Explain how one computes D_10[.999..., w-1], where w (omega) is the first transfinite ordinal, and D_10[r,n] is rs nth digit to the right of the decimal point, if n is an integer, then evaluate D_10[(.999... + 9)/10, w-1] and D_10[.999... * 10 - 9, w-1]. (n can be negative but thats not all that important here.) [.sigsnip] -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 > jesus christ! > do you know anything about mathematical induction?????????????? Enterprise does not even know what end shit comes out of. He is a total mathematical incompetent. He makes JSH look intelligent by comparison. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> jesus christ! >> do you know anything about mathematical induction?????????????? >Enterprise does not even know what end shit comes out of. He is a total >mathematical incompetent. He makes JSH look intelligent by comparison. >Bob Kolker Whats a hyper-real number? Do you even know anything about math? Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > Whats a hyper-real number? Do you even know anything about math? No. But I do know how the hyperrals are constructed. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 In sci.math, robert j. kolker : >> Whats a hyper-real number? Do you even know anything about math? > No. But I do know how the hyperrals are constructed. > Bob Kolker http://mathworld.wolfram.com/HyperrealNumber.html is extremely bare-bones (is there one hyperreal? more than one? arithmetic operations? proofs?) but at least its a start. A reference link http://members.tripod.com/PhilipApps/line.html looks to be little more than my attempts at d-math, though there might be more than one d -- or H, its dual. No doubt one could claim at least three theories: [1] An infinite hierarchy of d < 1/n for all n in N: 0 < ... < d^4 < d^3 < d^2 < d < 1, with a more or less standard algebra (e.g., (1-d)^3 = 1 - 3d + 3d^2 - d^3). [2] d^k = d for some k in N. [3] Some other esoteric condition. I suppose one might even notate this as R[d] -- a standard polynomial group over R, with a slightly weird ordering. And again, I must complain that S. Enterprize is being extremely sloppy here. (Not that Im all that neat, but hopefully my notations clear at least.) -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 >> Whats a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are You dont even know what a hyper-real number is??? And you are name calling people here like you know everything?????? Why not admit you ARE WRONG! constructed. >Bob Kolker Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > Whats a hyper-real number? Do you even know anything about math? >>No. But I do know how the hyperrals are > You dont even know what a hyper-real number is??? And you are name calling > people here like you know everything?????? Why not admit you ARE WRONG! Quick. Define an ultra-filter. No, dont look it up. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> Whats a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are >> You dont even know what a hyper-real number is??? And you are name >calling >> people here like you know everything?????? Why not admit you ARE WRONG! >Quick. Define an ultra-filter. No, dont look it up. >Bob Kolker Oh this is so hard to understand, I might need to take an asprin for a headache. Ill define it with an example. Suppose you have alot of people here making noise here on this NG and they dont know what they are talking about with .999..., and then comes along an ultrafilter F_Smart1234 with the correct information. What we do is apply ultrafilter F_Smart1234 to the whole set S of noise on the NG, and then just the pure correct answer is shown. The ultrafilter is then said to be a success and has worked very well, and is therefore proven. Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > Whats a hyper-real number? Do you even know anything about math? >>No. But I do know how the hyperrals are > > You dont even know what a hyper-real number is??? And you are name >>calling > people here like you know everything?????? Why not admit you ARE WRONG! >>Quick. Define an ultra-filter. No, dont look it up. >>Bob Kolker > Oh this is so hard to understand, I might need to take an asprin for a >headache. > Ill define it with an example. Suppose you have alot of people here >making >noise here on this NG and they dont know what they are talking about with >.999..., and then comes along an ultrafilter F_Smart1234 with the correct >information. What we do is apply ultrafilter F_Smart1234 to the whole set S >noise on the NG, and then just the pure correct answer is shown. > The ultrafilter is then said to be a success and has worked very well, and >is therefore proven. Your turn. Perform a ANOVA statistical test between .999... and 1. And of course go into details explaining what the ANOVA test is. hurry hurry dont look... Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 >> Whats a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are >> You dont even know what a hyper-real number is??? And you are name >calling >> people here like you know everything?????? Why not admit you ARE WRONG! >Quick. Define an ultra-filter. No, dont look it up. Oh, but I do have the right to refresh my memory. I even gave you time to do this and you still dont know what a hyper-real number is. >Bob Kolker Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: A Quantum Poem for Xmas === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, > 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to the >last line). >One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >--Mark But he assumes .999... = 1 in his equation before it is proven. Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 > 9(1) =/= 9(.999...) >> Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). >> He assumes 9(1) = 9 (going from the second to last line that you >> quoted, to the last line). >> One reason this proof is deficient is because of the assumption >> that 10(.9999999...) = 9.9999999... (which is true, but needs to be >> proven). >> --Mark > But he assumes .999... = 1 in his equation before it is proven. Would you point out where he makes this assumption? I repeat the entire proof, expanded a bit, with line numbers added for your convenience: [1] 9=9 [2] 9=9.9999999...-.9999999... [3] 9(1)=10(.9999999...)-.9999999 [4] Let x = .9999999... and substitute in [3] [5] 9(1) = 10x - x [6] 9(1) = 9(x) [7] Therefore x=1 In which line is the assumption .99999... = 1 used? --Mark === Subject: Re: PROOF that 0.99999... = 1 In sci.math, Mark Nudelman : > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, >> 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) > Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). > He assumes 9(1) = 9 (going from the second to last line that you > quoted, to the last line). > One reason this proof is deficient is because of the assumption > that 10(.9999999...) = 9.9999999... (which is true, but needs to be > proven). > --Mark >> But he assumes .999... = 1 in his equation before it is proven. > Would you point out where he makes this assumption? I repeat the entire > proof, expanded a bit, with line numbers added for your convenience: > [1] 9=9 > [2] 9=9.9999999...-.9999999... > [3] 9(1)=10(.9999999...)-.9999999 > [4] Let x = .9999999... and substitute in [3] > [5] 9(1) = 10x - x > [6] 9(1) = 9(x) > [7] Therefore x=1 > In which line is the assumption .99999... = 1 used? [3]. The possibility of an infinite borrow generates headaches. > --Mark -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 > In sci.math, Mark Nudelman >> Would you point out where he makes this assumption? I repeat the >> entire proof, expanded a bit, with line numbers added for your >> convenience: >> [1] 9=9 >> [2] 9=9.9999999...-.9999999... >> [3] 9(1)=10(.9999999...)-.9999999 >> [4] Let x = .9999999... and substitute in [3] >> [5] 9(1) = 10x - x >> [6] 9(1) = 9(x) >> [7] Therefore x=1 >> In which line is the assumption .99999... = 1 used? > [3]. The possibility of an infinite borrow generates headaches. Going from [2] to [3] merely assumes that 10(.99999...) = 9.9999.... This is indeed problematic and needs to be proven, as does the assumption that 9 = 9.99999...- 0.999999 in going from [1] to [2], but I dont see that either of these steps uses the assumption that .99999... = 1. --Mark === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 > 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >>assumes 9(1) = 9 (going from the second to last line that you quoted, to >>the >>last line). >>One reason this proof is deficient is because of the assumption that >>10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ jesus christ! do you know anything about mathematical induction?????????????? let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = 9*(1.11111111...) = 9.999999.. then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - 1)/(1 - 1/10))| = |1/10^n| = 1/10^n < e for all n >= N > -log(e) that means, the difference between the infinitely repeating decimal with period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this work for any number, not just 9) if you dont believe that x_n = 9.9999999999999999999 then thats your fault, you need to learn some simple math.... just try to find me a number sticktly between .999999999999..... and 1! you can do this for all x if you want... x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) if x is terminating or repeating in its tail, then the sum has a simple solution and its easy to calculate the answer. if you put x = 1, the {x} = 0 x = .99999...... then sum is just over 9/10^k which is easily to compute again, the only thing that you can have any sorta problem with is how .9999999999 could be reprsented by the sum, but that is your problem... as any halfwit knows that. === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, >> 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to >the >last line). >One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >--Mark >> But he assumes .999... = 1 in his equation before it is proven. >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >jesus christ! >do you know anything about mathematical induction?????????????? >let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >9*(1.11111111...) = 9.999999.. >then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >1)/(1 - 1/10))| >= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >that means, the difference between the infinitely repeating decimal with >period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this work >for any number, not just 9) >if you dont believe that x_n = 9.9999999999999999999 then thats your fault, >you need to learn some simple math.... just try to find me a number sticktly >between .999999999999..... and 1! >you can do this for all x if you want... >x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) >if x is terminating or repeating in its tail, then the sum has a simple >solution and its easy to calculate the answer. >if you put x = 1, the {x} = 0 >x = .99999...... >then sum is just over 9/10^k which is easily to compute >again, the only thing that you can have any sorta problem with is how >.9999999999 could be reprsented by the sum, but that is your problem... as >any halfwit knows that. Hey .999... IS NOT A REAL NUMBER (PERIOD). See math link below: http://mathworld.wolfram.com/HyperrealNumber.html .999... is of the form of a hyper-real number because there is a space between the real numbers between .999... and 1. .999... | | 1 ^ | See space A Hyperreal number is of the form Where n is a real number, x < n x = .999... n = 1 .999... < 1 THEREFORE, .999... =/= 1 Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >>assumes 9(1) = 9 (going from the second to last line that you quoted, to >>the >>last line). >>One reason this proof is deficient is because of the assumption that >>10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ >>jesus christ! >>do you know anything about mathematical induction?????????????? >>let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >>9*(1.11111111...) = 9.999999.. >>then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >>1)/(1 - 1/10))| >>= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >>that means, the difference between the infinitely repeating decimal with >>period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this >>work >>for any number, not just 9) >>if you dont believe that x_n = 9.9999999999999999999 then thats your >>fault, >>you need to learn some simple math.... just try to find me a number >>sticktly >>between .999999999999..... and 1! >>you can do this for all x if you want... >>x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) >>if x is terminating or repeating in its tail, then the sum has a simple >>solution and its easy to calculate the answer. >>if you put x = 1, the {x} = 0 >>x = .99999...... >>then sum is just over 9/10^k which is easily to compute >>again, the only thing that you can have any sorta problem with is how >>.9999999999 could be reprsented by the sum, but that is your problem... as >>any halfwit knows that. > Hey .999... IS NOT A REAL NUMBER (PERIOD). > See math link below: > http://mathworld.wolfram.com/HyperrealNumber.html > .999... is of the form of a hyper-real number because there is a space > between > the real numbers between .999... and 1. > .999... | | 1 > ^ > | > See space > A Hyperreal number is of the form > Where n is a real number, > x < n > x = .999... > n = 1 > .999... < 1 > THEREFORE, > .999... =/= 1 > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! === Subject: Re: PROOF that 0.99999... = 1 In sci.math, Jon Slaughter <10sbusfei7k2lee@corp.supernews.com>: [snipped for sanity] >> .999... =/= 1 >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ > your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! FSVO genius. Most of us use an alternate word with one less letter. :-) -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, >> 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to >the >last line). >>One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >jesus christ! >do you know anything about mathematical induction?????????????? >let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >9*(1.11111111...) = 9.999999.. >then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >1)/(1 - 1/10))| >= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >that means, the difference between the infinitely repeating decimal with >period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this >work >for any number, not just 9) >if you dont believe that x_n = 9.9999999999999999999 then thats your >fault, >you need to learn some simple math.... just try to find me a number >sticktly >between .999999999999..... and 1! >you can do this for all x if you want... >x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) >if x is terminating or repeating in its tail, then the sum has a simple >solution and its easy to calculate the answer. >if you put x = 1, the {x} = 0 >x = .99999...... >then sum is just over 9/10^k which is easily to compute >again, the only thing that you can have any sorta problem with is how >.9999999999 could be reprsented by the sum, but that is your problem... as >any halfwit knows that. >> Hey .999... IS NOT A REAL NUMBER (PERIOD). >> See math link below: >> http://mathworld.wolfram.com/HyperrealNumber.html >> .999... is of the form of a hyper-real number because there is a space >> between >> the real numbers between .999... and 1. >> .999... | | 1 >> ^ >> | >> See space >> A Hyperreal number is of the form >> Where n is a real number, >> x < n >> x = .999... >> n = 1 >> .999... < 1 >> THEREFORE, >> .999... =/= 1 >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! youre not your Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > .999... | | 1 > ^ > | > See space Pure scribble. > A Hyperreal number is of the form You would not know a hyperreal if it bit you. You have not the foggiest notion of how the real number system R is extended to *R. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> .999... | | 1 >> ^ >> | >> See space >Pure scribble. >> A Hyperreal number is of the form >You would not know a hyperreal if it bit you. You have not the foggiest >notion of how the real number system R is extended to *R. >Bob Kolker Hey, I thought you said I didnt know what it was. You are wrong again, and again, again. http://mathworld.wolfram.com/HyperrealNumber.html Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 In sci.math, S. Enterprize Company > > .999... | | 1 > ^ > | > See space >>Pure scribble. > > A Hyperreal number is of the form >>You would not know a hyperreal if it bit you. You have not the foggiest >>notion of how the real number system R is extended to *R. >>Bob Kolker > Hey, I thought you said I didnt know what it was. You are wrong again, and > again, again. > http://mathworld.wolfram.com/HyperrealNumber.html Like that tells him *anything*. Heres a few Qs for you. [1] If d is such that 0 < d < 1/n for all n in N, what is d^2? d^3? sqrt(d)? [2] Why is 5/5 != 9/9? 5/5 = 1, of course; 0.2 * 5 = 1. 9/9, by contrast, is 0.111... * 9 = 0.999... = 1 - d. In base 12, 1/9 = 0.14(12) but 1/5 = .24972497...(12) ; therefore in this case 9/9 = 1 but 5/5 = 1-d. Does it matter what base one uses for arithmetic? [3] Explain how one computes D_10[.999..., w-1], where w (omega) is the first transfinite ordinal, and D_10[r,n] is rs nth digit to the right of the decimal point, if n is an integer, then evaluate D_10[(.999... + 9)/10, w-1] and D_10[.999... * 10 - 9, w-1]. (n can be negative but thats not all that important here.) [.sigsnip] -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 > jesus christ! > do you know anything about mathematical induction?????????????? Enterprise does not even know what end shit comes out of. He is a total mathematical incompetent. He makes JSH look intelligent by comparison. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> jesus christ! >> do you know anything about mathematical induction?????????????? >Enterprise does not even know what end shit comes out of. He is a total >mathematical incompetent. He makes JSH look intelligent by comparison. >Bob Kolker Whats a hyper-real number? Do you even know anything about math? Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > Whats a hyper-real number? Do you even know anything about math? No. But I do know how the hyperrals are constructed. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 In sci.math, robert j. kolker : >> Whats a hyper-real number? Do you even know anything about math? > No. But I do know how the hyperrals are constructed. > Bob Kolker http://mathworld.wolfram.com/HyperrealNumber.html is extremely bare-bones (is there one hyperreal? more than one? arithmetic operations? proofs?) but at least its a start. A reference link http://members.tripod.com/PhilipApps/line.html looks to be little more than my attempts at d-math, though there might be more than one d -- or H, its dual. No doubt one could claim at least three theories: [1] An infinite hierarchy of d < 1/n for all n in N: 0 < ... < d^4 < d^3 < d^2 < d < 1, with a more or less standard algebra (e.g., (1-d)^3 = 1 - 3d + 3d^2 - d^3). [2] d^k = d for some k in N. [3] Some other esoteric condition. I suppose one might even notate this as R[d] -- a standard polynomial group over R, with a slightly weird ordering. And again, I must complain that S. Enterprize is being extremely sloppy here. (Not that Im all that neat, but hopefully my notations clear at least.) -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 >> Whats a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are You dont even know what a hyper-real number is??? And you are name calling people here like you know everything?????? Why not admit you ARE WRONG! constructed. >Bob Kolker Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > Whats a hyper-real number? Do you even know anything about math? >>No. But I do know how the hyperrals are > You dont even know what a hyper-real number is??? And you are name calling > people here like you know everything?????? Why not admit you ARE WRONG! Quick. Define an ultra-filter. No, dont look it up. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> Whats a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are >> You dont even know what a hyper-real number is??? And you are name >calling >> people here like you know everything?????? Why not admit you ARE WRONG! >Quick. Define an ultra-filter. No, dont look it up. >Bob Kolker Oh this is so hard to understand, I might need to take an asprin for a headache. Ill define it with an example. Suppose you have alot of people here making noise here on this NG and they dont know what they are talking about with .999..., and then comes along an ultrafilter F_Smart1234 with the correct information. What we do is apply ultrafilter F_Smart1234 to the whole set S of noise on the NG, and then just the pure correct answer is shown. The ultrafilter is then said to be a success and has worked very well, and is therefore proven. Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > Whats a hyper-real number? Do you even know anything about math? >>No. But I do know how the hyperrals are > > You dont even know what a hyper-real number is??? And you are name >>calling > people here like you know everything?????? Why not admit you ARE WRONG! >>Quick. Define an ultra-filter. No, dont look it up. >>Bob Kolker > Oh this is so hard to understand, I might need to take an asprin for a >headache. > Ill define it with an example. Suppose you have alot of people here >making >noise here on this NG and they dont know what they are talking about with >.999..., and then comes along an ultrafilter F_Smart1234 with the correct >information. What we do is apply ultrafilter F_Smart1234 to the whole set S >noise on the NG, and then just the pure correct answer is shown. > The ultrafilter is then said to be a success and has worked very well, and >is therefore proven. Your turn. Perform a ANOVA statistical test between .999... and 1. And of course go into details explaining what the ANOVA test is. hurry hurry dont look... Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 >> Whats a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are >> You dont even know what a hyper-real number is??? And you are name >calling >> people here like you know everything?????? Why not admit you ARE WRONG! >Quick. Define an ultra-filter. No, dont look it up. Oh, but I do have the right to refresh my memory. I even gave you time to do this and you still dont know what a hyper-real number is. >Bob Kolker Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: A Quantum Poem for Xmas === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, > 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to the >last line). >One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >--Mark But he assumes .999... = 1 in his equation before it is proven. Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 > 9(1) =/= 9(.999...) >> Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). >> He assumes 9(1) = 9 (going from the second to last line that you >> quoted, to the last line). >> One reason this proof is deficient is because of the assumption >> that 10(.9999999...) = 9.9999999... (which is true, but needs to be >> proven). >> --Mark > But he assumes .999... = 1 in his equation before it is proven. Would you point out where he makes this assumption? I repeat the entire proof, expanded a bit, with line numbers added for your convenience: [1] 9=9 [2] 9=9.9999999...-.9999999... [3] 9(1)=10(.9999999...)-.9999999 [4] Let x = .9999999... and substitute in [3] [5] 9(1) = 10x - x [6] 9(1) = 9(x) [7] Therefore x=1 In which line is the assumption .99999... = 1 used? --Mark === Subject: Re: PROOF that 0.99999... = 1 In sci.math, Mark Nudelman : > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, >> 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) > Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). > He assumes 9(1) = 9 (going from the second to last line that you > quoted, to the last line). > One reason this proof is deficient is because of the assumption > that 10(.9999999...) = 9.9999999... (which is true, but needs to be > proven). > --Mark >> But he assumes .999... = 1 in his equation before it is proven. > Would you point out where he makes this assumption? I repeat the entire > proof, expanded a bit, with line numbers added for your convenience: > [1] 9=9 > [2] 9=9.9999999...-.9999999... > [3] 9(1)=10(.9999999...)-.9999999 > [4] Let x = .9999999... and substitute in [3] > [5] 9(1) = 10x - x > [6] 9(1) = 9(x) > [7] Therefore x=1 > In which line is the assumption .99999... = 1 used? [3]. The possibility of an infinite borrow generates headaches. > --Mark -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 > In sci.math, Mark Nudelman >> Would you point out where he makes this assumption? I repeat the >> entire proof, expanded a bit, with line numbers added for your >> convenience: >> [1] 9=9 >> [2] 9=9.9999999...-.9999999... >> [3] 9(1)=10(.9999999...)-.9999999 >> [4] Let x = .9999999... and substitute in [3] >> [5] 9(1) = 10x - x >> [6] 9(1) = 9(x) >> [7] Therefore x=1 >> In which line is the assumption .99999... = 1 used? > [3]. The possibility of an infinite borrow generates headaches. Going from [2] to [3] merely assumes that 10(.99999...) = 9.9999.... This is indeed problematic and needs to be proven, as does the assumption that 9 = 9.99999...- 0.999999 in going from [1] to [2], but I dont see that either of these steps uses the assumption that .99999... = 1. --Mark === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 > 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >>assumes 9(1) = 9 (going from the second to last line that you quoted, to >>the >>last line). >>One reason this proof is deficient is because of the assumption that >>10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ jesus christ! do you know anything about mathematical induction?????????????? let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = 9*(1.11111111...) = 9.999999.. then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - 1)/(1 - 1/10))| = |1/10^n| = 1/10^n < e for all n >= N > -log(e) that means, the difference between the infinitely repeating decimal with period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this work for any number, not just 9) if you dont believe that x_n = 9.9999999999999999999 then thats your fault, you need to learn some simple math.... just try to find me a number sticktly between .999999999999..... and 1! you can do this for all x if you want... x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) if x is terminating or repeating in its tail, then the sum has a simple solution and its easy to calculate the answer. if you put x = 1, the {x} = 0 x = .99999...... then sum is just over 9/10^k which is easily to compute again, the only thing that you can have any sorta problem with is how .9999999999 could be reprsented by the sum, but that is your problem... as any halfwit knows that. === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, >> 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to >the >last line). >One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >--Mark >> But he assumes .999... = 1 in his equation before it is proven. >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >jesus christ! >do you know anything about mathematical induction?????????????? >let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >9*(1.11111111...) = 9.999999.. >then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >1)/(1 - 1/10))| >= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >that means, the difference between the infinitely repeating decimal with >period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this work >for any number, not just 9) >if you dont believe that x_n = 9.9999999999999999999 then thats your fault, >you need to learn some simple math.... just try to find me a number sticktly >between .999999999999..... and 1! >you can do this for all x if you want... >x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) >if x is terminating or repeating in its tail, then the sum has a simple >solution and its easy to calculate the answer. >if you put x = 1, the {x} = 0 >x = .99999...... >then sum is just over 9/10^k which is easily to compute >again, the only thing that you can have any sorta problem with is how >.9999999999 could be reprsented by the sum, but that is your problem... as >any halfwit knows that. Hey .999... IS NOT A REAL NUMBER (PERIOD). See math link below: http://mathworld.wolfram.com/HyperrealNumber.html .999... is of the form of a hyper-real number because there is a space between the real numbers between .999... and 1. .999... | | 1 ^ | See space A Hyperreal number is of the form Where n is a real number, x < n x = .999... n = 1 .999... < 1 THEREFORE, .999... =/= 1 Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >>assumes 9(1) = 9 (going from the second to last line that you quoted, to >>the >>last line). >>One reason this proof is deficient is because of the assumption that >>10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ >>jesus christ! >>do you know anything about mathematical induction?????????????? >>let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >>9*(1.11111111...) = 9.999999.. >>then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >>1)/(1 - 1/10))| >>= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >>that means, the difference between the infinitely repeating decimal with >>period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this >>work >>for any number, not just 9) >>if you dont believe that x_n = 9.9999999999999999999 then thats your >>fault, >>you need to learn some simple math.... just try to find me a number >>sticktly >>between .999999999999..... and 1! >>you can do this for all x if you want... >>x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) >>if x is terminating or repeating in its tail, then the sum has a simple >>solution and its easy to calculate the answer. >>if you put x = 1, the {x} = 0 >>x = .99999...... >>then sum is just over 9/10^k which is easily to compute >>again, the only thing that you can have any sorta problem with is how >>.9999999999 could be reprsented by the sum, but that is your problem... as >>any halfwit knows that. > Hey .999... IS NOT A REAL NUMBER (PERIOD). > See math link below: > http://mathworld.wolfram.com/HyperrealNumber.html > .999... is of the form of a hyper-real number because there is a space > between > the real numbers between .999... and 1. > .999... | | 1 > ^ > | > See space > A Hyperreal number is of the form > Where n is a real number, > x < n > x = .999... > n = 1 > .999... < 1 > THEREFORE, > .999... =/= 1 > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! === Subject: Re: PROOF that 0.99999... = 1 In sci.math, Jon Slaughter <10sbusfei7k2lee@corp.supernews.com>: [snipped for sanity] >> .999... =/= 1 >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ > your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! FSVO genius. Most of us use an alternate word with one less letter. :-) -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, >> 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to >the >last line). >>One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >jesus christ! >do you know anything about mathematical induction?????????????? >let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >9*(1.11111111...) = 9.999999.. >then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >1)/(1 - 1/10))| >= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >that means, the difference between the infinitely repeating decimal with >period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this >work >for any number, not just 9) >if you dont believe that x_n = 9.9999999999999999999 then thats your >fault, >you need to learn some simple math.... just try to find me a number >sticktly >between .999999999999..... and 1! >you can do this for all x if you want... >x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) >if x is terminating or repeating in its tail, then the sum has a simple >solution and its easy to calculate the answer. >if you put x = 1, the {x} = 0 >x = .99999...... >then sum is just over 9/10^k which is easily to compute >again, the only thing that you can have any sorta problem with is how >.9999999999 could be reprsented by the sum, but that is your problem... as >any halfwit knows that. >> Hey .999... IS NOT A REAL NUMBER (PERIOD). >> See math link below: >> http://mathworld.wolfram.com/HyperrealNumber.html >> .999... is of the form of a hyper-real number because there is a space >> between >> the real numbers between .999... and 1. >> .999... | | 1 >> ^ >> | >> See space >> A Hyperreal number is of the form >> Where n is a real number, >> x < n >> x = .999... >> n = 1 >> .999... < 1 >> THEREFORE, >> .999... =/= 1 >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! youre not your Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > .999... | | 1 > ^ > | > See space Pure scribble. > A Hyperreal number is of the form You would not know a hyperreal if it bit you. You have not the foggiest notion of how the real number system R is extended to *R. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> .999... | | 1 >> ^ >> | >> See space >Pure scribble. >> A Hyperreal number is of the form >You would not know a hyperreal if it bit you. You have not the foggiest >notion of how the real number system R is extended to *R. >Bob Kolker Hey, I thought you said I didnt know what it was. You are wrong again, and again, again. http://mathworld.wolfram.com/HyperrealNumber.html Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 In sci.math, S. Enterprize Company > > .999... | | 1 > ^ > | > See space >>Pure scribble. > > A Hyperreal number is of the form >>You would not know a hyperreal if it bit you. You have not the foggiest >>notion of how the real number system R is extended to *R. >>Bob Kolker > Hey, I thought you said I didnt know what it was. You are wrong again, and > again, again. > http://mathworld.wolfram.com/HyperrealNumber.html Like that tells him *anything*. Heres a few Qs for you. [1] If d is such that 0 < d < 1/n for all n in N, what is d^2? d^3? sqrt(d)? [2] Why is 5/5 != 9/9? 5/5 = 1, of course; 0.2 * 5 = 1. 9/9, by contrast, is 0.111... * 9 = 0.999... = 1 - d. In base 12, 1/9 = 0.14(12) but 1/5 = .24972497...(12) ; therefore in this case 9/9 = 1 but 5/5 = 1-d. Does it matter what base one uses for arithmetic? [3] Explain how one computes D_10[.999..., w-1], where w (omega) is the first transfinite ordinal, and D_10[r,n] is rs nth digit to the right of the decimal point, if n is an integer, then evaluate D_10[(.999... + 9)/10, w-1] and D_10[.999... * 10 - 9, w-1]. (n can be negative but thats not all that important here.) [.sigsnip] -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 > jesus christ! > do you know anything about mathematical induction?????????????? Enterprise does not even know what end shit comes out of. He is a total mathematical incompetent. He makes JSH look intelligent by comparison. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> jesus christ! >> do you know anything about mathematical induction?????????????? >Enterprise does not even know what end shit comes out of. He is a total >mathematical incompetent. He makes JSH look intelligent by comparison. >Bob Kolker Whats a hyper-real number? Do you even know anything about math? Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > Whats a hyper-real number? Do you even know anything about math? No. But I do know how the hyperrals are constructed. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 In sci.math, robert j. kolker : >> Whats a hyper-real number? Do you even know anything about math? > No. But I do know how the hyperrals are constructed. > Bob Kolker http://mathworld.wolfram.com/HyperrealNumber.html is extremely bare-bones (is there one hyperreal? more than one? arithmetic operations? proofs?) but at least its a start. A reference link http://members.tripod.com/PhilipApps/line.html looks to be little more than my attempts at d-math, though there might be more than one d -- or H, its dual. No doubt one could claim at least three theories: [1] An infinite hierarchy of d < 1/n for all n in N: 0 < ... < d^4 < d^3 < d^2 < d < 1, with a more or less standard algebra (e.g., (1-d)^3 = 1 - 3d + 3d^2 - d^3). [2] d^k = d for some k in N. [3] Some other esoteric condition. I suppose one might even notate this as R[d] -- a standard polynomial group over R, with a slightly weird ordering. And again, I must complain that S. Enterprize is being extremely sloppy here. (Not that Im all that neat, but hopefully my notations clear at least.) -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 >> Whats a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are You dont even know what a hyper-real number is??? And you are name calling people here like you know everything?????? Why not admit you ARE WRONG! constructed. >Bob Kolker Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > Whats a hyper-real number? Do you even know anything about math? >>No. But I do know how the hyperrals are > You dont even know what a hyper-real number is??? And you are name calling > people here like you know everything?????? Why not admit you ARE WRONG! Quick. Define an ultra-filter. No, dont look it up. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> Whats a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are >> You dont even know what a hyper-real number is??? And you are name >calling >> people here like you know everything?????? Why not admit you ARE WRONG! >Quick. Define an ultra-filter. No, dont look it up. >Bob Kolker Oh this is so hard to understand, I might need to take an asprin for a headache. Ill define it with an example. Suppose you have alot of people here making noise here on this NG and they dont know what they are talking about with .999..., and then comes along an ultrafilter F_Smart1234 with the correct information. What we do is apply ultrafilter F_Smart1234 to the whole set S of noise on the NG, and then just the pure correct answer is shown. The ultrafilter is then said to be a success and has worked very well, and is therefore proven. Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > Whats a hyper-real number? Do you even know anything about math? >>No. But I do know how the hyperrals are > > You dont even know what a hyper-real number is??? And you are name >>calling > people here like you know everything?????? Why not admit you ARE WRONG! >>Quick. Define an ultra-filter. No, dont look it up. >>Bob Kolker > Oh this is so hard to understand, I might need to take an asprin for a >headache. > Ill define it with an example. Suppose you have alot of people here >making >noise here on this NG and they dont know what they are talking about with >.999..., and then comes along an ultrafilter F_Smart1234 with the correct >information. What we do is apply ultrafilter F_Smart1234 to the whole set S >noise on the NG, and then just the pure correct answer is shown. > The ultrafilter is then said to be a success and has worked very well, and >is therefore proven. Your turn. Perform a ANOVA statistical test between .999... and 1. And of course go into details explaining what the ANOVA test is. hurry hurry dont look... Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 >> Whats a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are >> You dont even know what a hyper-real number is??? And you are name >calling >> people here like you know everything?????? Why not admit you ARE WRONG! >Quick. Define an ultra-filter. No, dont look it up. Oh, but I do have the right to refresh my memory. I even gave you time to do this and you still dont know what a hyper-real number is. >Bob Kolker Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: A Quantum Poem for Xmas === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, > 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to the >last line). >One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >--Mark But he assumes .999... = 1 in his equation before it is proven. Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 > 9(1) =/= 9(.999...) >> Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). >> He assumes 9(1) = 9 (going from the second to last line that you >> quoted, to the last line). >> One reason this proof is deficient is because of the assumption >> that 10(.9999999...) = 9.9999999... (which is true, but needs to be >> proven). >> --Mark > But he assumes .999... = 1 in his equation before it is proven. Would you point out where he makes this assumption? I repeat the entire proof, expanded a bit, with line numbers added for your convenience: [1] 9=9 [2] 9=9.9999999...-.9999999... [3] 9(1)=10(.9999999...)-.9999999 [4] Let x = .9999999... and substitute in [3] [5] 9(1) = 10x - x [6] 9(1) = 9(x) [7] Therefore x=1 In which line is the assumption .99999... = 1 used? --Mark === Subject: Re: PROOF that 0.99999... = 1 In sci.math, Mark Nudelman : > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, >> 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) > Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). > He assumes 9(1) = 9 (going from the second to last line that you > quoted, to the last line). > One reason this proof is deficient is because of the assumption > that 10(.9999999...) = 9.9999999... (which is true, but needs to be > proven). > --Mark >> But he assumes .999... = 1 in his equation before it is proven. > Would you point out where he makes this assumption? I repeat the entire > proof, expanded a bit, with line numbers added for your convenience: > [1] 9=9 > [2] 9=9.9999999...-.9999999... > [3] 9(1)=10(.9999999...)-.9999999 > [4] Let x = .9999999... and substitute in [3] > [5] 9(1) = 10x - x > [6] 9(1) = 9(x) > [7] Therefore x=1 > In which line is the assumption .99999... = 1 used? [3]. The possibility of an infinite borrow generates headaches. > --Mark -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 > In sci.math, Mark Nudelman >> Would you point out where he makes this assumption? I repeat the >> entire proof, expanded a bit, with line numbers added for your >> convenience: >> [1] 9=9 >> [2] 9=9.9999999...-.9999999... >> [3] 9(1)=10(.9999999...)-.9999999 >> [4] Let x = .9999999... and substitute in [3] >> [5] 9(1) = 10x - x >> [6] 9(1) = 9(x) >> [7] Therefore x=1 >> In which line is the assumption .99999... = 1 used? > [3]. The possibility of an infinite borrow generates headaches. Going from [2] to [3] merely assumes that 10(.99999...) = 9.9999.... This is indeed problematic and needs to be proven, as does the assumption that 9 = 9.99999...- 0.999999 in going from [1] to [2], but I dont see that either of these steps uses the assumption that .99999... = 1. --Mark === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 > 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >>assumes 9(1) = 9 (going from the second to last line that you quoted, to >>the >>last line). >>One reason this proof is deficient is because of the assumption that >>10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ jesus christ! do you know anything about mathematical induction?????????????? let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = 9*(1.11111111...) = 9.999999.. then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - 1)/(1 - 1/10))| = |1/10^n| = 1/10^n < e for all n >= N > -log(e) that means, the difference between the infinitely repeating decimal with period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this work for any number, not just 9) if you dont believe that x_n = 9.9999999999999999999 then thats your fault, you need to learn some simple math.... just try to find me a number sticktly between .999999999999..... and 1! you can do this for all x if you want... x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) if x is terminating or repeating in its tail, then the sum has a simple solution and its easy to calculate the answer. if you put x = 1, the {x} = 0 x = .99999...... then sum is just over 9/10^k which is easily to compute again, the only thing that you can have any sorta problem with is how .9999999999 could be reprsented by the sum, but that is your problem... as any halfwit knows that. === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, >> 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to >the >last line). >One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >--Mark >> But he assumes .999... = 1 in his equation before it is proven. >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >jesus christ! >do you know anything about mathematical induction?????????????? >let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >9*(1.11111111...) = 9.999999.. >then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >1)/(1 - 1/10))| >= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >that means, the difference between the infinitely repeating decimal with >period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this work >for any number, not just 9) >if you dont believe that x_n = 9.9999999999999999999 then thats your fault, >you need to learn some simple math.... just try to find me a number sticktly >between .999999999999..... and 1! >you can do this for all x if you want... >x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) >if x is terminating or repeating in its tail, then the sum has a simple >solution and its easy to calculate the answer. >if you put x = 1, the {x} = 0 >x = .99999...... >then sum is just over 9/10^k which is easily to compute >again, the only thing that you can have any sorta problem with is how >.9999999999 could be reprsented by the sum, but that is your problem... as >any halfwit knows that. Hey .999... IS NOT A REAL NUMBER (PERIOD). See math link below: http://mathworld.wolfram.com/HyperrealNumber.html .999... is of the form of a hyper-real number because there is a space between the real numbers between .999... and 1. .999... | | 1 ^ | See space A Hyperreal number is of the form Where n is a real number, x < n x = .999... n = 1 .999... < 1 THEREFORE, .999... =/= 1 Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >>assumes 9(1) = 9 (going from the second to last line that you quoted, to >>the >>last line). >>One reason this proof is deficient is because of the assumption that >>10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ >>jesus christ! >>do you know anything about mathematical induction?????????????? >>let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >>9*(1.11111111...) = 9.999999.. >>then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >>1)/(1 - 1/10))| >>= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >>that means, the difference between the infinitely repeating decimal with >>period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this >>work >>for any number, not just 9) >>if you dont believe that x_n = 9.9999999999999999999 then thats your >>fault, >>you need to learn some simple math.... just try to find me a number >>sticktly >>between .999999999999..... and 1! >>you can do this for all x if you want... >>x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) >>if x is terminating or repeating in its tail, then the sum has a simple >>solution and its easy to calculate the answer. >>if you put x = 1, the {x} = 0 >>x = .99999...... >>then sum is just over 9/10^k which is easily to compute >>again, the only thing that you can have any sorta problem with is how >>.9999999999 could be reprsented by the sum, but that is your problem... as >>any halfwit knows that. > Hey .999... IS NOT A REAL NUMBER (PERIOD). > See math link below: > http://mathworld.wolfram.com/HyperrealNumber.html > .999... is of the form of a hyper-real number because there is a space > between > the real numbers between .999... and 1. > .999... | | 1 > ^ > | > See space > A Hyperreal number is of the form > Where n is a real number, > x < n > x = .999... > n = 1 > .999... < 1 > THEREFORE, > .999... =/= 1 > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! === Subject: Re: PROOF that 0.99999... = 1 In sci.math, Jon Slaughter <10sbusfei7k2lee@corp.supernews.com>: [snipped for sanity] >> .999... =/= 1 >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ > your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! FSVO genius. Most of us use an alternate word with one less letter. :-) -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, >> 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to >the >last line). >>One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >jesus christ! >do you know anything about mathematical induction?????????????? >let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >9*(1.11111111...) = 9.999999.. >then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >1)/(1 - 1/10))| >= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >that means, the difference between the infinitely repeating decimal with >period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this >work >for any number, not just 9) >if you dont believe that x_n = 9.9999999999999999999 then thats your >fault, >you need to learn some simple math.... just try to find me a number >sticktly >between .999999999999..... and 1! >you can do this for all x if you want... >x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) >if x is terminating or repeating in its tail, then the sum has a simple >solution and its easy to calculate the answer. >if you put x = 1, the {x} = 0 >x = .99999...... >then sum is just over 9/10^k which is easily to compute >again, the only thing that you can have any sorta problem with is how >.9999999999 could be reprsented by the sum, but that is your problem... as >any halfwit knows that. >> Hey .999... IS NOT A REAL NUMBER (PERIOD). >> See math link below: >> http://mathworld.wolfram.com/HyperrealNumber.html >> .999... is of the form of a hyper-real number because there is a space >> between >> the real numbers between .999... and 1. >> .999... | | 1 >> ^ >> | >> See space >> A Hyperreal number is of the form >> Where n is a real number, >> x < n >> x = .999... >> n = 1 >> .999... < 1 >> THEREFORE, >> .999... =/= 1 >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! youre not your Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > .999... | | 1 > ^ > | > See space Pure scribble. > A Hyperreal number is of the form You would not know a hyperreal if it bit you. You have not the foggiest notion of how the real number system R is extended to *R. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> .999... | | 1 >> ^ >> | >> See space >Pure scribble. >> A Hyperreal number is of the form >You would not know a hyperreal if it bit you. You have not the foggiest >notion of how the real number system R is extended to *R. >Bob Kolker Hey, I thought you said I didnt know what it was. You are wrong again, and again, again. http://mathworld.wolfram.com/HyperrealNumber.html Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 In sci.math, S. Enterprize Company > > .999... | | 1 > ^ > | > See space >>Pure scribble. > > A Hyperreal number is of the form >>You would not know a hyperreal if it bit you. You have not the foggiest >>notion of how the real number system R is extended to *R. >>Bob Kolker > Hey, I thought you said I didnt know what it was. You are wrong again, and > again, again. > http://mathworld.wolfram.com/HyperrealNumber.html Like that tells him *anything*. Heres a few Qs for you. [1] If d is such that 0 < d < 1/n for all n in N, what is d^2? d^3? sqrt(d)? [2] Why is 5/5 != 9/9? 5/5 = 1, of course; 0.2 * 5 = 1. 9/9, by contrast, is 0.111... * 9 = 0.999... = 1 - d. In base 12, 1/9 = 0.14(12) but 1/5 = .24972497...(12) ; therefore in this case 9/9 = 1 but 5/5 = 1-d. Does it matter what base one uses for arithmetic? [3] Explain how one computes D_10[.999..., w-1], where w (omega) is the first transfinite ordinal, and D_10[r,n] is rs nth digit to the right of the decimal point, if n is an integer, then evaluate D_10[(.999... + 9)/10, w-1] and D_10[.999... * 10 - 9, w-1]. (n can be negative but thats not all that important here.) [.sigsnip] -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 > jesus christ! > do you know anything about mathematical induction?????????????? Enterprise does not even know what end shit comes out of. He is a total mathematical incompetent. He makes JSH look intelligent by comparison. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> jesus christ! >> do you know anything about mathematical induction?????????????? >Enterprise does not even know what end shit comes out of. He is a total >mathematical incompetent. He makes JSH look intelligent by comparison. >Bob Kolker Whats a hyper-real number? Do you even know anything about math? Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > Whats a hyper-real number? Do you even know anything about math? No. But I do know how the hyperrals are constructed. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 In sci.math, robert j. kolker : >> Whats a hyper-real number? Do you even know anything about math? > No. But I do know how the hyperrals are constructed. > Bob Kolker http://mathworld.wolfram.com/HyperrealNumber.html is extremely bare-bones (is there one hyperreal? more than one? arithmetic operations? proofs?) but at least its a start. A reference link http://members.tripod.com/PhilipApps/line.html looks to be little more than my attempts at d-math, though there might be more than one d -- or H, its dual. No doubt one could claim at least three theories: [1] An infinite hierarchy of d < 1/n for all n in N: 0 < ... < d^4 < d^3 < d^2 < d < 1, with a more or less standard algebra (e.g., (1-d)^3 = 1 - 3d + 3d^2 - d^3). [2] d^k = d for some k in N. [3] Some other esoteric condition. I suppose one might even notate this as R[d] -- a standard polynomial group over R, with a slightly weird ordering. And again, I must complain that S. Enterprize is being extremely sloppy here. (Not that Im all that neat, but hopefully my notations clear at least.) -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 >> Whats a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are You dont even know what a hyper-real number is??? And you are name calling people here like you know everything?????? Why not admit you ARE WRONG! constructed. >Bob Kolker Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > Whats a hyper-real number? Do you even know anything about math? >>No. But I do know how the hyperrals are > You dont even know what a hyper-real number is??? And you are name calling > people here like you know everything?????? Why not admit you ARE WRONG! Quick. Define an ultra-filter. No, dont look it up. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> Whats a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are >> You dont even know what a hyper-real number is??? And you are name >calling >> people here like you know everything?????? Why not admit you ARE WRONG! >Quick. Define an ultra-filter. No, dont look it up. >Bob Kolker Oh this is so hard to understand, I might need to take an asprin for a headache. Ill define it with an example. Suppose you have alot of people here making noise here on this NG and they dont know what they are talking about with .999..., and then comes along an ultrafilter F_Smart1234 with the correct information. What we do is apply ultrafilter F_Smart1234 to the whole set S of noise on the NG, and then just the pure correct answer is shown. The ultrafilter is then said to be a success and has worked very well, and is therefore proven. Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > Whats a hyper-real number? Do you even know anything about math? >>No. But I do know how the hyperrals are > > You dont even know what a hyper-real number is??? And you are name >>calling > people here like you know everything?????? Why not admit you ARE WRONG! >>Quick. Define an ultra-filter. No, dont look it up. >>Bob Kolker > Oh this is so hard to understand, I might need to take an asprin for a >headache. > Ill define it with an example. Suppose you have alot of people here >making >noise here on this NG and they dont know what they are talking about with >.999..., and then comes along an ultrafilter F_Smart1234 with the correct >information. What we do is apply ultrafilter F_Smart1234 to the whole set S >noise on the NG, and then just the pure correct answer is shown. > The ultrafilter is then said to be a success and has worked very well, and >is therefore proven. Your turn. Perform a ANOVA statistical test between .999... and 1. And of course go into details explaining what the ANOVA test is. hurry hurry dont look... Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 >> Whats a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are >> You dont even know what a hyper-real number is??? And you are name >calling >> people here like you know everything?????? Why not admit you ARE WRONG! >Quick. Define an ultra-filter. No, dont look it up. Oh, but I do have the right to refresh my memory. I even gave you time to do this and you still dont know what a hyper-real number is. >Bob Kolker Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: A Quantum Poem for Xmas === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, > 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to the >last line). >One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >--Mark But he assumes .999... = 1 in his equation before it is proven. Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 > 9(1) =/= 9(.999...) >> Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). >> He assumes 9(1) = 9 (going from the second to last line that you >> quoted, to the last line). >> One reason this proof is deficient is because of the assumption >> that 10(.9999999...) = 9.9999999... (which is true, but needs to be >> proven). >> --Mark > But he assumes .999... = 1 in his equation before it is proven. Would you point out where he makes this assumption? I repeat the entire proof, expanded a bit, with line numbers added for your convenience: [1] 9=9 [2] 9=9.9999999...-.9999999... [3] 9(1)=10(.9999999...)-.9999999 [4] Let x = .9999999... and substitute in [3] [5] 9(1) = 10x - x [6] 9(1) = 9(x) [7] Therefore x=1 In which line is the assumption .99999... = 1 used? --Mark === Subject: Re: PROOF that 0.99999... = 1 In sci.math, Mark Nudelman : > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, >> 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) > Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). > He assumes 9(1) = 9 (going from the second to last line that you > quoted, to the last line). > One reason this proof is deficient is because of the assumption > that 10(.9999999...) = 9.9999999... (which is true, but needs to be > proven). > --Mark >> But he assumes .999... = 1 in his equation before it is proven. > Would you point out where he makes this assumption? I repeat the entire > proof, expanded a bit, with line numbers added for your convenience: > [1] 9=9 > [2] 9=9.9999999...-.9999999... > [3] 9(1)=10(.9999999...)-.9999999 > [4] Let x = .9999999... and substitute in [3] > [5] 9(1) = 10x - x > [6] 9(1) = 9(x) > [7] Therefore x=1 > In which line is the assumption .99999... = 1 used? [3]. The possibility of an infinite borrow generates headaches. > --Mark -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 > In sci.math, Mark Nudelman >> Would you point out where he makes this assumption? I repeat the >> entire proof, expanded a bit, with line numbers added for your >> convenience: >> [1] 9=9 >> [2] 9=9.9999999...-.9999999... >> [3] 9(1)=10(.9999999...)-.9999999 >> [4] Let x = .9999999... and substitute in [3] >> [5] 9(1) = 10x - x >> [6] 9(1) = 9(x) >> [7] Therefore x=1 >> In which line is the assumption .99999... = 1 used? > [3]. The possibility of an infinite borrow generates headaches. Going from [2] to [3] merely assumes that 10(.99999...) = 9.9999.... This is indeed problematic and needs to be proven, as does the assumption that 9 = 9.99999...- 0.999999 in going from [1] to [2], but I dont see that either of these steps uses the assumption that .99999... = 1. --Mark === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 > 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >>assumes 9(1) = 9 (going from the second to last line that you quoted, to >>the >>last line). >>One reason this proof is deficient is because of the assumption that >>10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ jesus christ! do you know anything about mathematical induction?????????????? let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = 9*(1.11111111...) = 9.999999.. then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - 1)/(1 - 1/10))| = |1/10^n| = 1/10^n < e for all n >= N > -log(e) that means, the difference between the infinitely repeating decimal with period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this work for any number, not just 9) if you dont believe that x_n = 9.9999999999999999999 then thats your fault, you need to learn some simple math.... just try to find me a number sticktly between .999999999999..... and 1! you can do this for all x if you want... x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) if x is terminating or repeating in its tail, then the sum has a simple solution and its easy to calculate the answer. if you put x = 1, the {x} = 0 x = .99999...... then sum is just over 9/10^k which is easily to compute again, the only thing that you can have any sorta problem with is how .9999999999 could be reprsented by the sum, but that is your problem... as any halfwit knows that. === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, >> 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to >the >last line). >One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >--Mark >> But he assumes .999... = 1 in his equation before it is proven. >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >jesus christ! >do you know anything about mathematical induction?????????????? >let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >9*(1.11111111...) = 9.999999.. >then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >1)/(1 - 1/10))| >= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >that means, the difference between the infinitely repeating decimal with >period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this work >for any number, not just 9) >if you dont believe that x_n = 9.9999999999999999999 then thats your fault, >you need to learn some simple math.... just try to find me a number sticktly >between .999999999999..... and 1! >you can do this for all x if you want... >x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) >if x is terminating or repeating in its tail, then the sum has a simple >solution and its easy to calculate the answer. >if you put x = 1, the {x} = 0 >x = .99999...... >then sum is just over 9/10^k which is easily to compute >again, the only thing that you can have any sorta problem with is how >.9999999999 could be reprsented by the sum, but that is your problem... as >any halfwit knows that. Hey .999... IS NOT A REAL NUMBER (PERIOD). See math link below: http://mathworld.wolfram.com/HyperrealNumber.html .999... is of the form of a hyper-real number because there is a space between the real numbers between .999... and 1. .999... | | 1 ^ | See space A Hyperreal number is of the form Where n is a real number, x < n x = .999... n = 1 .999... < 1 THEREFORE, .999... =/= 1 Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >>assumes 9(1) = 9 (going from the second to last line that you quoted, to >>the >>last line). >>One reason this proof is deficient is because of the assumption that >>10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ >>jesus christ! >>do you know anything about mathematical induction?????????????? >>let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >>9*(1.11111111...) = 9.999999.. >>then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >>1)/(1 - 1/10))| >>= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >>that means, the difference between the infinitely repeating decimal with >>period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this >>work >>for any number, not just 9) >>if you dont believe that x_n = 9.9999999999999999999 then thats your >>fault, >>you need to learn some simple math.... just try to find me a number >>sticktly >>between .999999999999..... and 1! >>you can do this for all x if you want... >>x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) >>if x is terminating or repeating in its tail, then the sum has a simple >>solution and its easy to calculate the answer. >>if you put x = 1, the {x} = 0 >>x = .99999...... >>then sum is just over 9/10^k which is easily to compute >>again, the only thing that you can have any sorta problem with is how >>.9999999999 could be reprsented by the sum, but that is your problem... as >>any halfwit knows that. > Hey .999... IS NOT A REAL NUMBER (PERIOD). > See math link below: > http://mathworld.wolfram.com/HyperrealNumber.html > .999... is of the form of a hyper-real number because there is a space > between > the real numbers between .999... and 1. > .999... | | 1 > ^ > | > See space > A Hyperreal number is of the form > Where n is a real number, > x < n > x = .999... > n = 1 > .999... < 1 > THEREFORE, > .999... =/= 1 > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! === Subject: Re: PROOF that 0.99999... = 1 In sci.math, Jon Slaughter <10sbusfei7k2lee@corp.supernews.com>: [snipped for sanity] >> .999... =/= 1 >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ > your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! FSVO genius. Most of us use an alternate word with one less letter. :-) -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, >> 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to >the >last line). >>One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >jesus christ! >do you know anything about mathematical induction?????????????? >let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >9*(1.11111111...) = 9.999999.. >then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >1)/(1 - 1/10))| >= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >that means, the difference between the infinitely repeating decimal with >period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this >work >for any number, not just 9) >if you dont believe that x_n = 9.9999999999999999999 then thats your >fault, >you need to learn some simple math.... just try to find me a number >sticktly >between .999999999999..... and 1! >you can do this for all x if you want... >x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) >if x is terminating or repeating in its tail, then the sum has a simple >solution and its easy to calculate the answer. >if you put x = 1, the {x} = 0 >x = .99999...... >then sum is just over 9/10^k which is easily to compute >again, the only thing that you can have any sorta problem with is how >.9999999999 could be reprsented by the sum, but that is your problem... as >any halfwit knows that. >> Hey .999... IS NOT A REAL NUMBER (PERIOD). >> See math link below: >> http://mathworld.wolfram.com/HyperrealNumber.html >> .999... is of the form of a hyper-real number because there is a space >> between >> the real numbers between .999... and 1. >> .999... | | 1 >> ^ >> | >> See space >> A Hyperreal number is of the form >> Where n is a real number, >> x < n >> x = .999... >> n = 1 >> .999... < 1 >> THEREFORE, >> .999... =/= 1 >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! youre not your Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > .999... | | 1 > ^ > | > See space Pure scribble. > A Hyperreal number is of the form You would not know a hyperreal if it bit you. You have not the foggiest notion of how the real number system R is extended to *R. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> .999... | | 1 >> ^ >> | >> See space >Pure scribble. >> A Hyperreal number is of the form >You would not know a hyperreal if it bit you. You have not the foggiest >notion of how the real number system R is extended to *R. >Bob Kolker Hey, I thought you said I didnt know what it was. You are wrong again, and again, again. http://mathworld.wolfram.com/HyperrealNumber.html Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 In sci.math, S. Enterprize Company > > .999... | | 1 > ^ > | > See space >>Pure scribble. > > A Hyperreal number is of the form >>You would not know a hyperreal if it bit you. You have not the foggiest >>notion of how the real number system R is extended to *R. >>Bob Kolker > Hey, I thought you said I didnt know what it was. You are wrong again, and > again, again. > http://mathworld.wolfram.com/HyperrealNumber.html Like that tells him *anything*. Heres a few Qs for you. [1] If d is such that 0 < d < 1/n for all n in N, what is d^2? d^3? sqrt(d)? [2] Why is 5/5 != 9/9? 5/5 = 1, of course; 0.2 * 5 = 1. 9/9, by contrast, is 0.111... * 9 = 0.999... = 1 - d. In base 12, 1/9 = 0.14(12) but 1/5 = .24972497...(12) ; therefore in this case 9/9 = 1 but 5/5 = 1-d. Does it matter what base one uses for arithmetic? [3] Explain how one computes D_10[.999..., w-1], where w (omega) is the first transfinite ordinal, and D_10[r,n] is rs nth digit to the right of the decimal point, if n is an integer, then evaluate D_10[(.999... + 9)/10, w-1] and D_10[.999... * 10 - 9, w-1]. (n can be negative but thats not all that important here.) [.sigsnip] -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 > jesus christ! > do you know anything about mathematical induction?????????????? Enterprise does not even know what end shit comes out of. He is a total mathematical incompetent. He makes JSH look intelligent by comparison. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> jesus christ! >> do you know anything about mathematical induction?????????????? >Enterprise does not even know what end shit comes out of. He is a total >mathematical incompetent. He makes JSH look intelligent by comparison. >Bob Kolker Whats a hyper-real number? Do you even know anything about math? Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > Whats a hyper-real number? Do you even know anything about math? No. But I do know how the hyperrals are constructed. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 In sci.math, robert j. kolker : >> Whats a hyper-real number? Do you even know anything about math? > No. But I do know how the hyperrals are constructed. > Bob Kolker http://mathworld.wolfram.com/HyperrealNumber.html is extremely bare-bones (is there one hyperreal? more than one? arithmetic operations? proofs?) but at least its a start. A reference link http://members.tripod.com/PhilipApps/line.html looks to be little more than my attempts at d-math, though there might be more than one d -- or H, its dual. No doubt one could claim at least three theories: [1] An infinite hierarchy of d < 1/n for all n in N: 0 < ... < d^4 < d^3 < d^2 < d < 1, with a more or less standard algebra (e.g., (1-d)^3 = 1 - 3d + 3d^2 - d^3). [2] d^k = d for some k in N. [3] Some other esoteric condition. I suppose one might even notate this as R[d] -- a standard polynomial group over R, with a slightly weird ordering. And again, I must complain that S. Enterprize is being extremely sloppy here. (Not that Im all that neat, but hopefully my notations clear at least.) -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 >> Whats a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are You dont even know what a hyper-real number is??? And you are name calling people here like you know everything?????? Why not admit you ARE WRONG! constructed. >Bob Kolker Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > Whats a hyper-real number? Do you even know anything about math? >>No. But I do know how the hyperrals are > You dont even know what a hyper-real number is??? And you are name calling > people here like you know everything?????? Why not admit you ARE WRONG! Quick. Define an ultra-filter. No, dont look it up. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> Whats a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are >> You dont even know what a hyper-real number is??? And you are name >calling >> people here like you know everything?????? Why not admit you ARE WRONG! >Quick. Define an ultra-filter. No, dont look it up. >Bob Kolker Oh this is so hard to understand, I might need to take an asprin for a headache. Ill define it with an example. Suppose you have alot of people here making noise here on this NG and they dont know what they are talking about with .999..., and then comes along an ultrafilter F_Smart1234 with the correct information. What we do is apply ultrafilter F_Smart1234 to the whole set S of noise on the NG, and then just the pure correct answer is shown. The ultrafilter is then said to be a success and has worked very well, and is therefore proven. Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > Whats a hyper-real number? Do you even know anything about math? >>No. But I do know how the hyperrals are > > You dont even know what a hyper-real number is??? And you are name >>calling > people here like you know everything?????? Why not admit you ARE WRONG! >>Quick. Define an ultra-filter. No, dont look it up. >>Bob Kolker > Oh this is so hard to understand, I might need to take an asprin for a >headache. > Ill define it with an example. Suppose you have alot of people here >making >noise here on this NG and they dont know what they are talking about with >.999..., and then comes along an ultrafilter F_Smart1234 with the correct >information. What we do is apply ultrafilter F_Smart1234 to the whole set S >noise on the NG, and then just the pure correct answer is shown. > The ultrafilter is then said to be a success and has worked very well, and >is therefore proven. Your turn. Perform a ANOVA statistical test between .999... and 1. And of course go into details explaining what the ANOVA test is. hurry hurry dont look... Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 >> Whats a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are >> You dont even know what a hyper-real number is??? And you are name >calling >> people here like you know everything?????? Why not admit you ARE WRONG! >Quick. Define an ultra-filter. No, dont look it up. Oh, but I do have the right to refresh my memory. I even gave you time to do this and you still dont know what a hyper-real number is. >Bob Kolker Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: A Quantum Poem for Xmas === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, > 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to the >last line). >One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >--Mark But he assumes .999... = 1 in his equation before it is proven. Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 > 9(1) =/= 9(.999...) >> Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). >> He assumes 9(1) = 9 (going from the second to last line that you >> quoted, to the last line). >> One reason this proof is deficient is because of the assumption >> that 10(.9999999...) = 9.9999999... (which is true, but needs to be >> proven). >> --Mark > But he assumes .999... = 1 in his equation before it is proven. Would you point out where he makes this assumption? I repeat the entire proof, expanded a bit, with line numbers added for your convenience: [1] 9=9 [2] 9=9.9999999...-.9999999... [3] 9(1)=10(.9999999...)-.9999999 [4] Let x = .9999999... and substitute in [3] [5] 9(1) = 10x - x [6] 9(1) = 9(x) [7] Therefore x=1 In which line is the assumption .99999... = 1 used? --Mark === Subject: Re: PROOF that 0.99999... = 1 In sci.math, Mark Nudelman : > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, >> 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) > Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). > He assumes 9(1) = 9 (going from the second to last line that you > quoted, to the last line). > One reason this proof is deficient is because of the assumption > that 10(.9999999...) = 9.9999999... (which is true, but needs to be > proven). > --Mark >> But he assumes .999... = 1 in his equation before it is proven. > Would you point out where he makes this assumption? I repeat the entire > proof, expanded a bit, with line numbers added for your convenience: > [1] 9=9 > [2] 9=9.9999999...-.9999999... > [3] 9(1)=10(.9999999...)-.9999999 > [4] Let x = .9999999... and substitute in [3] > [5] 9(1) = 10x - x > [6] 9(1) = 9(x) > [7] Therefore x=1 > In which line is the assumption .99999... = 1 used? [3]. The possibility of an infinite borrow generates headaches. > --Mark -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 > In sci.math, Mark Nudelman >> Would you point out where he makes this assumption? I repeat the >> entire proof, expanded a bit, with line numbers added for your >> convenience: >> [1] 9=9 >> [2] 9=9.9999999...-.9999999... >> [3] 9(1)=10(.9999999...)-.9999999 >> [4] Let x = .9999999... and substitute in [3] >> [5] 9(1) = 10x - x >> [6] 9(1) = 9(x) >> [7] Therefore x=1 >> In which line is the assumption .99999... = 1 used? > [3]. The possibility of an infinite borrow generates headaches. Going from [2] to [3] merely assumes that 10(.99999...) = 9.9999.... This is indeed problematic and needs to be proven, as does the assumption that 9 = 9.99999...- 0.999999 in going from [1] to [2], but I dont see that either of these steps uses the assumption that .99999... = 1. --Mark === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 > 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >>assumes 9(1) = 9 (going from the second to last line that you quoted, to >>the >>last line). >>One reason this proof is deficient is because of the assumption that >>10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ jesus christ! do you know anything about mathematical induction?????????????? let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = 9*(1.11111111...) = 9.999999.. then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - 1)/(1 - 1/10))| = |1/10^n| = 1/10^n < e for all n >= N > -log(e) that means, the difference between the infinitely repeating decimal with period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this work for any number, not just 9) if you dont believe that x_n = 9.9999999999999999999 then thats your fault, you need to learn some simple math.... just try to find me a number sticktly between .999999999999..... and 1! you can do this for all x if you want... x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) if x is terminating or repeating in its tail, then the sum has a simple solution and its easy to calculate the answer. if you put x = 1, the {x} = 0 x = .99999...... then sum is just over 9/10^k which is easily to compute again, the only thing that you can have any sorta problem with is how .9999999999 could be reprsented by the sum, but that is your problem... as any halfwit knows that. === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, >> 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to >the >last line). >One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >--Mark >> But he assumes .999... = 1 in his equation before it is proven. >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >jesus christ! >do you know anything about mathematical induction?????????????? >let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >9*(1.11111111...) = 9.999999.. >then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >1)/(1 - 1/10))| >= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >that means, the difference between the infinitely repeating decimal with >period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this work >for any number, not just 9) >if you dont believe that x_n = 9.9999999999999999999 then thats your fault, >you need to learn some simple math.... just try to find me a number sticktly >between .999999999999..... and 1! >you can do this for all x if you want... >x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) >if x is terminating or repeating in its tail, then the sum has a simple >solution and its easy to calculate the answer. >if you put x = 1, the {x} = 0 >x = .99999...... >then sum is just over 9/10^k which is easily to compute >again, the only thing that you can have any sorta problem with is how >.9999999999 could be reprsented by the sum, but that is your problem... as >any halfwit knows that. Hey .999... IS NOT A REAL NUMBER (PERIOD). See math link below: http://mathworld.wolfram.com/HyperrealNumber.html .999... is of the form of a hyper-real number because there is a space between the real numbers between .999... and 1. .999... | | 1 ^ | See space A Hyperreal number is of the form Where n is a real number, x < n x = .999... n = 1 .999... < 1 THEREFORE, .999... =/= 1 Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >>assumes 9(1) = 9 (going from the second to last line that you quoted, to >>the >>last line). >>One reason this proof is deficient is because of the assumption that >>10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ >>jesus christ! >>do you know anything about mathematical induction?????????????? >>let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >>9*(1.11111111...) = 9.999999.. >>then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >>1)/(1 - 1/10))| >>= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >>that means, the difference between the infinitely repeating decimal with >>period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this >>work >>for any number, not just 9) >>if you dont believe that x_n = 9.9999999999999999999 then thats your >>fault, >>you need to learn some simple math.... just try to find me a number >>sticktly >>between .999999999999..... and 1! >>you can do this for all x if you want... >>x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) >>if x is terminating or repeating in its tail, then the sum has a simple >>solution and its easy to calculate the answer. >>if you put x = 1, the {x} = 0 >>x = .99999...... >>then sum is just over 9/10^k which is easily to compute >>again, the only thing that you can have any sorta problem with is how >>.9999999999 could be reprsented by the sum, but that is your problem... as >>any halfwit knows that. > Hey .999... IS NOT A REAL NUMBER (PERIOD). > See math link below: > http://mathworld.wolfram.com/HyperrealNumber.html > .999... is of the form of a hyper-real number because there is a space > between > the real numbers between .999... and 1. > .999... | | 1 > ^ > | > See space > A Hyperreal number is of the form > Where n is a real number, > x < n > x = .999... > n = 1 > .999... < 1 > THEREFORE, > .999... =/= 1 > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! === Subject: Re: PROOF that 0.99999... = 1 In sci.math, Jon Slaughter <10sbusfei7k2lee@corp.supernews.com>: [snipped for sanity] >> .999... =/= 1 >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ > your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! FSVO genius. Most of us use an alternate word with one less letter. :-) -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, >> 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to >the >last line). >>One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >jesus christ! >do you know anything about mathematical induction?????????????? >let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >9*(1.11111111...) = 9.999999.. >then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >1)/(1 - 1/10))| >= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >that means, the difference between the infinitely repeating decimal with >period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this >work >for any number, not just 9) >if you dont believe that x_n = 9.9999999999999999999 then thats your >fault, >you need to learn some simple math.... just try to find me a number >sticktly >between .999999999999..... and 1! >you can do this for all x if you want... >x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) >if x is terminating or repeating in its tail, then the sum has a simple >solution and its easy to calculate the answer. >if you put x = 1, the {x} = 0 >x = .99999...... >then sum is just over 9/10^k which is easily to compute >again, the only thing that you can have any sorta problem with is how >.9999999999 could be reprsented by the sum, but that is your problem... as >any halfwit knows that. >> Hey .999... IS NOT A REAL NUMBER (PERIOD). >> See math link below: >> http://mathworld.wolfram.com/HyperrealNumber.html >> .999... is of the form of a hyper-real number because there is a space >> between >> the real numbers between .999... and 1. >> .999... | | 1 >> ^ >> | >> See space >> A Hyperreal number is of the form >> Where n is a real number, >> x < n >> x = .999... >> n = 1 >> .999... < 1 >> THEREFORE, >> .999... =/= 1 >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! youre not your Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > .999... | | 1 > ^ > | > See space Pure scribble. > A Hyperreal number is of the form You would not know a hyperreal if it bit you. You have not the foggiest notion of how the real number system R is extended to *R. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> .999... | | 1 >> ^ >> | >> See space >Pure scribble. >> A Hyperreal number is of the form >You would not know a hyperreal if it bit you. You have not the foggiest >notion of how the real number system R is extended to *R. >Bob Kolker Hey, I thought you said I didnt know what it was. You are wrong again, and again, again. http://mathworld.wolfram.com/HyperrealNumber.html Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 In sci.math, S. Enterprize Company > > .999... | | 1 > ^ > | > See space >>Pure scribble. > > A Hyperreal number is of the form >>You would not know a hyperreal if it bit you. You have not the foggiest >>notion of how the real number system R is extended to *R. >>Bob Kolker > Hey, I thought you said I didnt know what it was. You are wrong again, and > again, again. > http://mathworld.wolfram.com/HyperrealNumber.html Like that tells him *anything*. Heres a few Qs for you. [1] If d is such that 0 < d < 1/n for all n in N, what is d^2? d^3? sqrt(d)? [2] Why is 5/5 != 9/9? 5/5 = 1, of course; 0.2 * 5 = 1. 9/9, by contrast, is 0.111... * 9 = 0.999... = 1 - d. In base 12, 1/9 = 0.14(12) but 1/5 = .24972497...(12) ; therefore in this case 9/9 = 1 but 5/5 = 1-d. Does it matter what base one uses for arithmetic? [3] Explain how one computes D_10[.999..., w-1], where w (omega) is the first transfinite ordinal, and D_10[r,n] is rs nth digit to the right of the decimal point, if n is an integer, then evaluate D_10[(.999... + 9)/10, w-1] and D_10[.999... * 10 - 9, w-1]. (n can be negative but thats not all that important here.) [.sigsnip] -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 > jesus christ! > do you know anything about mathematical induction?????????????? Enterprise does not even know what end shit comes out of. He is a total mathematical incompetent. He makes JSH look intelligent by comparison. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> jesus christ! >> do you know anything about mathematical induction?????????????? >Enterprise does not even know what end shit comes out of. He is a total >mathematical incompetent. He makes JSH look intelligent by comparison. >Bob Kolker Whats a hyper-real number? Do you even know anything about math? Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > Whats a hyper-real number? Do you even know anything about math? No. But I do know how the hyperrals are constructed. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 In sci.math, robert j. kolker : >> Whats a hyper-real number? Do you even know anything about math? > No. But I do know how the hyperrals are constructed. > Bob Kolker http://mathworld.wolfram.com/HyperrealNumber.html is extremely bare-bones (is there one hyperreal? more than one? arithmetic operations? proofs?) but at least its a start. A reference link http://members.tripod.com/PhilipApps/line.html looks to be little more than my attempts at d-math, though there might be more than one d -- or H, its dual. No doubt one could claim at least three theories: [1] An infinite hierarchy of d < 1/n for all n in N: 0 < ... < d^4 < d^3 < d^2 < d < 1, with a more or less standard algebra (e.g., (1-d)^3 = 1 - 3d + 3d^2 - d^3). [2] d^k = d for some k in N. [3] Some other esoteric condition. I suppose one might even notate this as R[d] -- a standard polynomial group over R, with a slightly weird ordering. And again, I must complain that S. Enterprize is being extremely sloppy here. (Not that Im all that neat, but hopefully my notations clear at least.) -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 >> Whats a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are You dont even know what a hyper-real number is??? And you are name calling people here like you know everything?????? Why not admit you ARE WRONG! constructed. >Bob Kolker Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > Whats a hyper-real number? Do you even know anything about math? >>No. But I do know how the hyperrals are > You dont even know what a hyper-real number is??? And you are name calling > people here like you know everything?????? Why not admit you ARE WRONG! Quick. Define an ultra-filter. No, dont look it up. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> Whats a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are >> You dont even know what a hyper-real number is??? And you are name >calling >> people here like you know everything?????? Why not admit you ARE WRONG! >Quick. Define an ultra-filter. No, dont look it up. >Bob Kolker Oh this is so hard to understand, I might need to take an asprin for a headache. Ill define it with an example. Suppose you have alot of people here making noise here on this NG and they dont know what they are talking about with .999..., and then comes along an ultrafilter F_Smart1234 with the correct information. What we do is apply ultrafilter F_Smart1234 to the whole set S of noise on the NG, and then just the pure correct answer is shown. The ultrafilter is then said to be a success and has worked very well, and is therefore proven. Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > Whats a hyper-real number? Do you even know anything about math? >>No. But I do know how the hyperrals are > > You dont even know what a hyper-real number is??? And you are name >>calling > people here like you know everything?????? Why not admit you ARE WRONG! >>Quick. Define an ultra-filter. No, dont look it up. >>Bob Kolker > Oh this is so hard to understand, I might need to take an asprin for a >headache. > Ill define it with an example. Suppose you have alot of people here >making >noise here on this NG and they dont know what they are talking about with >.999..., and then comes along an ultrafilter F_Smart1234 with the correct >information. What we do is apply ultrafilter F_Smart1234 to the whole set S >noise on the NG, and then just the pure correct answer is shown. > The ultrafilter is then said to be a success and has worked very well, and >is therefore proven. Your turn. Perform a ANOVA statistical test between .999... and 1. And of course go into details explaining what the ANOVA test is. hurry hurry dont look... Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 >> Whats a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are >> You dont even know what a hyper-real number is??? And you are name >calling >> people here like you know everything?????? Why not admit you ARE WRONG! >Quick. Define an ultra-filter. No, dont look it up. Oh, but I do have the right to refresh my memory. I even gave you time to do this and you still dont know what a hyper-real number is. >Bob Kolker Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: A Quantum Poem for Xmas