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From: David Bernier on 9 Jun 2010 07:40 Daryl McCullough wrote: > |-|ercules says... >> >> (from the "Xenides dies" thread) >> >>> For *all* N, the sequence differs from the Nth entry in the list at >>> the Nth digit (and possibly other positions as well). It is new >>> because for *every* sequence in the list, the question "is it the same >>> as this sequence" is answered "no". >>> >> >> So you think the antidiagonal comes up with an actual NEW SEQUENCE OF DIGITS >> and this does not contradict that ALL sequences of digits are on the computable >> list of reals up to all (an infinite amount of) digit positions? > > You start with a completely crystal clear statement: > The antidiagonal number is not equal to any number on the list. > > Then you paraphrase this clear statement to get a completely > muddled statement: > >> the antidiagonal comes up with an actual NEW SEQUENCE OF DIGITS >> and this does not contradict that ALL sequences of digits are on >> the computable list of reals up to all (an infinite amount of) >> digit positions > > Why do you prefer to use muddled, incoherent statements instead of > clear ones? > > The antidiagonal is not equal to any of the numbers on the list. > What is unclear about that? I have an idea. Herc could practice with constructing the anti-diagonal for an effectively computable list containing every rational number r in [0, 1) which can be represented in base 10 by a terminating decimal expansion. Then, if one disallows using 0.99999... as the anti-diagonal, I think the anti-diagonal must be a number v such that 0<v<1 and v is irrational. David Bernier
From: David Bernier on 9 Jun 2010 14:34 |-|ercules wrote: > "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote ... >> |-|ercules says... >> >>> Are you saying you find a new_sequence_of_digits using >>> diagonalization on >>> infinite lists? >>> >>> Like 260? >>> >>> How so when ALL (INFINITELY MANY) digits of ALL digit sequences are >>> computable? >> >> Yes, all finite sequences of digits are computable. So let's >> go ahead and assume that we have a list that includes every >> finite sequence of digits. For definiteness, let's use the following >> rule: Let real number n be the decimal representation of n, preceded >> by a decimal point. So, we have: >> >> L_0 = 0.0 >> L_1 = 0.1 >> L_2 = 0.2 >> ... >> L_10 = 0.10 >> L_11 = 0.11 >> ... >> L_100 = 0.100 >> ... >> >> The way I've enumerated them, some numbers appear more than once. For >> example, L_1 is equal (as a real number) to L_10. But that doesn't >> matter. >> >> With this list, every finite decimal expansion occurs somewhere on the >> list. >> But not every *infinite* decimal expansion occurs. So let's apply our >> diagonalization procedure: we get >> >> r = 0.5555555555... >> >> This number is not on the list. It's not equal to L_0, it's not equal >> to L_1, it's not equal to L_2, etc. >> >> The finite approximations are on the list, however. 0.5 is equal to >> L_5. 0.55 is equal to L_55. 0.555 is equal to L_555. etc. But the >> full number 0.555... (which happens to be the decimal representation >> of the fraction 5/9) is not anywhere on the list. > > > You split my 2 related questions and gave trivial solutions to each. > > Try again! > > ALL (INFINITELY MANY) digits of ALL digit sequences are computable! > Are you saying you can find a new_sequence_of_digits using > diagonalization on a computable reals list? > Like 260 in the first example? > > > ________________________________________ > > > I can't find the post, but someone posted yesterday that there is a > DIFFERENT DIGIT at N=something etc. etc. > > And there can be numerous (infinite) different digits along the > expansion, for some real not on the computable reals list. > > If someone can find that post great. > > If not, it's utter BS! > > It's MEANINGLESS. Different digit to what? > > IF there are NUMEROUS different digits then there must be some FINITE > substring between them (inclusive) > that is not a computable digit sequence. To the digit d, one can associate anti(d) = d+3 (for d = 0, 1, 2, 3 or 4) = d-3 (for d = 5, 6, 7, 8 or 9). ($$$) anti(0) = 3 anti(1) = 4 anti(2) = 5 anti(3) = 6 anti(4) = 7 anti(5) = 2 (***) anti(6) = 3 anti(7) = 4 anti(8) = 5 anti(9) = 6 Let's say we have a list where the 10th number is 55/123. The decimal notation for 55/123 is: 0.4471544715 4471544715 4471544715 44715................ The tenth decimal after the point in the number above is '5'. anti(5) = 2 by (***) above or by ($$$). In the absence of the first nine numbers on the list, we can just put a question mark '?' for that position of ANTI, the anti-diagonal (or an anti-diagonal) for some list (whether finite or infinite). So as of now, we have: ANTI = 0.?????????2 ?????????? ................... ANTI can't be the same as 55/123 because two numbers that differ by 3 units in the tenth decimal position can't possibly be the same. Another thing is that blanks in terminating decimals on the list should be interpreted as zeros. David Bernier
From: David Bernier on 9 Jun 2010 18:31 |-|ercules wrote: > "David Bernier" <david250(a)videotron.ca> wrote... >> |-|ercules wrote: >>> "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote ... >>>> |-|ercules says... >>>> >>>>> Are you saying you find a new_sequence_of_digits using >>>>> diagonalization on >>>>> infinite lists? >>>>> >>>>> Like 260? >>>>> >>>>> How so when ALL (INFINITELY MANY) digits of ALL digit sequences are >>>>> computable? >>>> >>>> Yes, all finite sequences of digits are computable. So let's >>>> go ahead and assume that we have a list that includes every >>>> finite sequence of digits. For definiteness, let's use the following >>>> rule: Let real number n be the decimal representation of n, preceded >>>> by a decimal point. So, we have: >>>> >>>> L_0 = 0.0 >>>> L_1 = 0.1 >>>> L_2 = 0.2 >>>> ... >>>> L_10 = 0.10 >>>> L_11 = 0.11 >>>> ... >>>> L_100 = 0.100 >>>> ... >>>> >>>> The way I've enumerated them, some numbers appear more than once. For >>>> example, L_1 is equal (as a real number) to L_10. But that doesn't >>>> matter. >>>> >>>> With this list, every finite decimal expansion occurs somewhere on the >>>> list. >>>> But not every *infinite* decimal expansion occurs. So let's apply our >>>> diagonalization procedure: we get >>>> >>>> r = 0.5555555555... >>>> >>>> This number is not on the list. It's not equal to L_0, it's not equal >>>> to L_1, it's not equal to L_2, etc. >>>> >>>> The finite approximations are on the list, however. 0.5 is equal to >>>> L_5. 0.55 is equal to L_55. 0.555 is equal to L_555. etc. But the >>>> full number 0.555... (which happens to be the decimal representation >>>> of the fraction 5/9) is not anywhere on the list. >>> >>> >>> You split my 2 related questions and gave trivial solutions to each. >>> >>> Try again! >>> >>> ALL (INFINITELY MANY) digits of ALL digit sequences are computable! >>> Are you saying you can find a new_sequence_of_digits using >>> diagonalization on a computable reals list? >>> Like 260 in the first example? >>> >>> >>> ________________________________________ >>> >>> >>> I can't find the post, but someone posted yesterday that there is a >>> DIFFERENT DIGIT at N=something etc. etc. >>> >>> And there can be numerous (infinite) different digits along the >>> expansion, for some real not on the computable reals list. >>> >>> If someone can find that post great. >>> >>> If not, it's utter BS! >>> >>> It's MEANINGLESS. Different digit to what? >>> >>> IF there are NUMEROUS different digits then there must be some FINITE >>> substring between them (inclusive) >>> that is not a computable digit sequence. >> >> To the digit d, one can associate >> anti(d) = d+3 (for d = 0, 1, 2, 3 or 4) >> = d-3 (for d = 5, 6, 7, 8 or 9). ($$$) >> >> anti(0) = 3 >> anti(1) = 4 >> anti(2) = 5 >> anti(3) = 6 >> anti(4) = 7 >> anti(5) = 2 (***) >> anti(6) = 3 >> anti(7) = 4 >> anti(8) = 5 >> anti(9) = 6 >> >> Let's say we have a list where the 10th number is 55/123. >> The decimal notation for 55/123 is: >> >> 0.4471544715 4471544715 4471544715 44715................ >> >> The tenth decimal after the point in the number above is '5'. >> >> anti(5) = 2 by (***) above or by ($$$). >> >> In the absence of the first nine numbers on the list, >> we can just put a question mark '?' for that position >> of ANTI, the anti-diagonal (or an anti-diagonal) >> for some list (whether finite or infinite). >> >> So as of now, we have: >> ANTI = 0.?????????2 ?????????? ................... >> >> ANTI can't be the same as 55/123 because two >> numbers that differ by 3 units in the tenth decimal >> position can't possibly be the same. >> >> Another thing is that blanks in terminating decimals >> on the list should be interpreted as zeros. >> >> David Bernier > > > > Some time ago you said unless I come up with solid proof disputing > Cantor's proof > nobody would accept it. > > If you can help me find the post from yesterday that stated there is a > different > digit at such and such, and possibly numerous other different digits. > > THIS is provably false, and it should sway some of you, I'll go through > all the posts > again I missed it last time I searched. Concerning computable reals, there's one subtle point that I think is worth thinking about. We can simplify to cases where a computable real is >= 0 by adding a minus sign in front if needed. Also, we can forget about the integer part for now and just look at the fractional part, a number b such that 0 <= b < 1. There's a class of recursive (programmable) functions that, given an integer k >=1, return a digit in {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. f: N\{0} -> {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, with f recursive. Within recursive functions, there are the primitive recursive functions. These are functions that can be written in such a way that syntax guarantees they will halt. See for example this section in Wikipedia: < http://en.wikipedia.org/wiki/Primitive_recursive_function#Computer_language_definition > Computer programs that use while( condition Holds) { do: A ... B .. C ... } _may_ produce functions that are recursive (return a value for every k in N\{0}) but _not_ primitive recursive. One is based on the Ackermann function A(m, n): < http://en.wikipedia.org/wiki/Ackermann_function#Definition_and_properties > Let f(k) := A(k, k) . This f is not primitive recursive. It doesn't yield single digit outputs, but that's a side issue. The trouble with having while(.) loops is that it can be hard to see that one always exits the while(.) loop in finite time. Some computable reals can only be computed in the sense above by recursive non primitive recursive functions, for example if g(k) = 0.1000010000000001 ..... 1 ..... & k'th '1' at decimal position A(k, k), k>=1. Many (primitive recursive) functions will give digits of pi - 3. I don't think it matters in the diagonal argument (computability-bound) that several programs give pi-3. For primitive recursive functions, a program with the right syntax is proof that it always returns a value. With recursive but not primitive recursive functions, proving they always terminate on every input can be much harder. David Bernier
From: David Bernier on 10 Jun 2010 00:42
|-|ercules wrote: > "David Bernier" <david250(a)videotron.ca> wrote... >> |-|ercules wrote: >>> "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote ... >>>> |-|ercules says... >>>> >>>>> Are you saying you find a new_sequence_of_digits using >>>>> diagonalization on >>>>> infinite lists? >>>>> >>>>> Like 260? >>>>> >>>>> How so when ALL (INFINITELY MANY) digits of ALL digit sequences are >>>>> computable? >>>> >>>> Yes, all finite sequences of digits are computable. So let's >>>> go ahead and assume that we have a list that includes every >>>> finite sequence of digits. For definiteness, let's use the following >>>> rule: Let real number n be the decimal representation of n, preceded >>>> by a decimal point. So, we have: >>>> >>>> L_0 = 0.0 >>>> L_1 = 0.1 >>>> L_2 = 0.2 >>>> ... >>>> L_10 = 0.10 >>>> L_11 = 0.11 >>>> ... >>>> L_100 = 0.100 >>>> ... >>>> >>>> The way I've enumerated them, some numbers appear more than once. For >>>> example, L_1 is equal (as a real number) to L_10. But that doesn't >>>> matter. >>>> >>>> With this list, every finite decimal expansion occurs somewhere on the >>>> list. >>>> But not every *infinite* decimal expansion occurs. So let's apply our >>>> diagonalization procedure: we get >>>> >>>> r = 0.5555555555... >>>> >>>> This number is not on the list. It's not equal to L_0, it's not equal >>>> to L_1, it's not equal to L_2, etc. >>>> >>>> The finite approximations are on the list, however. 0.5 is equal to >>>> L_5. 0.55 is equal to L_55. 0.555 is equal to L_555. etc. But the >>>> full number 0.555... (which happens to be the decimal representation >>>> of the fraction 5/9) is not anywhere on the list. >>> >>> >>> You split my 2 related questions and gave trivial solutions to each. >>> >>> Try again! >>> >>> ALL (INFINITELY MANY) digits of ALL digit sequences are computable! >>> Are you saying you can find a new_sequence_of_digits using >>> diagonalization on a computable reals list? >>> Like 260 in the first example? >>> >>> >>> ________________________________________ >>> >>> >>> I can't find the post, but someone posted yesterday that there is a >>> DIFFERENT DIGIT at N=something etc. etc. >>> >>> And there can be numerous (infinite) different digits along the >>> expansion, for some real not on the computable reals list. >>> >>> If someone can find that post great. >>> >>> If not, it's utter BS! >>> >>> It's MEANINGLESS. Different digit to what? >>> >>> IF there are NUMEROUS different digits then there must be some FINITE >>> substring between them (inclusive) >>> that is not a computable digit sequence. >> >> To the digit d, one can associate >> anti(d) = d+3 (for d = 0, 1, 2, 3 or 4) >> = d-3 (for d = 5, 6, 7, 8 or 9). ($$$) >> >> anti(0) = 3 >> anti(1) = 4 >> anti(2) = 5 >> anti(3) = 6 >> anti(4) = 7 >> anti(5) = 2 (***) >> anti(6) = 3 >> anti(7) = 4 >> anti(8) = 5 >> anti(9) = 6 >> >> Let's say we have a list where the 10th number is 55/123. >> The decimal notation for 55/123 is: >> >> 0.4471544715 4471544715 4471544715 44715................ >> >> The tenth decimal after the point in the number above is '5'. >> >> anti(5) = 2 by (***) above or by ($$$). >> >> In the absence of the first nine numbers on the list, >> we can just put a question mark '?' for that position >> of ANTI, the anti-diagonal (or an anti-diagonal) >> for some list (whether finite or infinite). >> >> So as of now, we have: >> ANTI = 0.?????????2 ?????????? ................... >> >> ANTI can't be the same as 55/123 because two >> numbers that differ by 3 units in the tenth decimal >> position can't possibly be the same. >> >> Another thing is that blanks in terminating decimals >> on the list should be interpreted as zeros. >> >> David Bernier > > > > Some time ago you said unless I come up with solid proof disputing > Cantor's proof > nobody would accept it. To convince a Platonic-minded/realist set theorist that Cantor was wrong, one way would be to show that ZFC is an inconsistent theory, for example by showing that there exist sets X and Y such that X is an element of Y, and X is not an element of Y. > If you can help me find the post from yesterday that stated there is a > different > digit at such and such, and possibly numerous other different digits. I don't know which post you're referring to. If it was Thursday for you when you wrote the above, yesterday would have been _Wednesday_ ... >> Date: Thu, 10 Jun 2010 07:17:28 +1000 link: < http://groups.google.ca/group/alt.sci.physics/msg/81350943e9a30a20?hl=en&dmode=source > So, you posted the message I'm replying to here around 7:15 am, June 10 your time. David Bernier > THIS is provably false, and it should sway some of you, I'll go through > all the posts > again I missed it last time I searched. > > Herc > |