"NO BOX CONTAINS THE BOX NUMBERS THAT DON'T CONTAIN THEIR OWN BOX NUMBER" ~ XEN
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From: Daryl McCullough on 9 Jun 2010 13:03 |-|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. -- Daryl McCullough Ithaca, NY
From: |-|ercules on 9 Jun 2010 13:15 "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. Herc
From: Daryl McCullough on 9 Jun 2010 15:15 |-|ercules says... > >"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. As I said, your questions are completely incoherent nonsense. I'm trying to give coherent responses, but you have to meet my half-way, and not continue to spout gobbledy-goop. >ALL (INFINITELY MANY) digits of ALL digit sequences are computable! Okay, that's false, but you seem to believe it. Why? >Are you saying you can find a new_sequence_of_digits using >diagonalization on a computable reals list? I already answered. Absolutely I believe that. I gave you an example. >It's MEANINGLESS. Different digit to what? You've already been told, many, many times. If L is a list of reals, then antidiagonal(L) is a real that whose first digit is different from that of the first real in list L, whose second digit is different from that of the second real in list L, whose third digit is different from the third real in L, etc. The antidiagonal is different from every single real on the list. >IF there are NUMEROUS different digits then there must be some FINITE >substring >between them (inclusive) that is not a computable digit sequence. Okay, that's false. But you seem to believe it. Why? -- Daryl McCullough Ithaca, NY
From: |-|ercules on 9 Jun 2010 17:13 "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote ... > |-|ercules says... >> >>"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. > > As I said, your questions are completely incoherent nonsense. I'm > trying to give coherent responses, but you have to meet my half-way, > and not continue to spout gobbledy-goop. > >>ALL (INFINITELY MANY) digits of ALL digit sequences are computable! > > Okay, that's false, but you seem to believe it. Why? What about ALL digit sequences are computable to ALL finite lengths. If you agree, you just took your first steps into a larger world. You should contemplate for a moment that I do follow your diagonal argument. I just disagree that specifying "it's different at digit N to the Nth real and it's different..." literally gives any new sequence of digits that are not computable. Examples don't prove it for the entire domain. Herc
From: |-|ercules on 9 Jun 2010 17:17 "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. Herc
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