ughaibu Posted August 2, 2006 Report Share Posted August 2, 2006 CraigD: Thanks for the reply. I understand that isolated sequences dont constitute repetition, if they did repetition would be established by a sequence of more than ten. Between any pair of repeating sequences there will exist a finite sequence, this intervening sequence will also inevitably repeat between the same terminal sequences and thus we will have a repeating sequence composed of our initial three sequences. This pair of repeating conglomerate sequences in turn will be seperated by a finite sequence, a finite sequence which will also repeat in the same situation. Quote Link to comment Share on other sites More sharing options...

ughaibu Posted August 2, 2006 Report Share Posted August 2, 2006 My above post is badly worded. An initial intervening sequence won't inevitably repeat in the same situation but to avoid it doing so the number of available intervening sequences will be reduced, thus increasing the frequency of repetition of an alternative intervening sequence. Quote Link to comment Share on other sites More sharing options...

ughaibu Posted April 4, 2008 Report Share Posted April 4, 2008 I think that what I was trying to say can be expressed as follows: 1) any expansion of an irrational number consists of an infinite string of non-repeating digits, thus the cardinality of this string is aleph-zero 2) if such strings do not repeat, it seems plausible that they will be normal 3) an infinite normal non-repeating string will contain all possible strings of digits 4) it follows that such a string will be the power set of aleph-zero, ie aleph-one 5) a set of elements, such as an infinite string of non-repeating digits, cant have both the cardinalities of aleph-zero and aleph-one. Quote Link to comment Share on other sites More sharing options...

CraigD Posted April 4, 2008 Report Share Posted April 4, 2008 I think that what I was trying to say can be expressed as follows: 1) any expansion of an irrational number consists of an infinite string of non-repeating digits, thus the cardinality of this string is aleph-zero 2) if such strings do not repeat, it seems plausible that they will be normal ...It’s not, to my knowledge, proven that irrational numbers in general, or specific ones (notably pi) have a normal distribution of their digits (that is, if carried out to enough digits, have nearly the same number of 0s, 1s, 2s, 3s, etc.) We discussed the subject a few months ago in 13369. Without this plausible but unproven theorem, ughaibu’s proof fails at step 2 – not a bad thing, as it hints that this might be a way to get at a RAA proof of the normalcy or non-normalcy of the digits of irrational numbers in general.3) an infinite normal non-repeating string will contain all possible strings of digitsUghaibu, can you give a proof, or a link to a proof, of this? :) A cool line of thought ;) Quote Link to comment Share on other sites More sharing options...

ughaibu Posted April 5, 2008 Report Share Posted April 5, 2008 CraigD: thanks for the reply. I'd assumed my (3) was an established result, but I guess, after reading posts by Qfwfq and yourself on the thread you linked to, that I was mistaken. It's interesting, I'll think about it. Quote Link to comment Share on other sites More sharing options...

ughaibu Posted April 6, 2008 Report Share Posted April 6, 2008 At the following link, I find "any normal sequence (a sequence in which each string of equal length appears with equal frequency) is disjunctive" and "a disjunctive sequence is an infinite sequence (over a finite alphabet of characters) in which every finite string appears as a substring" Disjunctive sequence - Wikipedia, the free encyclopedia Quote Link to comment Share on other sites More sharing options...

CraigD Posted April 6, 2008 Report Share Posted April 6, 2008 At the following link, I find "any normal sequence (a sequence in which each string of equal length appears with equal frequency) is disjunctive" and "a disjunctive sequence is an infinite sequence (over a finite alphabet of characters) in which every finite string appears as a substring" Disjunctive sequence - Wikipedia, the free encyclopediaThe wikipedia article “normal number” also has a lot of relevant discussion on the subject. It’s important to note that “normal sequence” in this context is defined differently than in Distribution of Digits in Irrational Numbers. The latter definition is of “simply normality”, while the stronger definition of normality requires that the frequency of occurance of any subsequence composed of m members of an alphabet of b symbols approaches precisely [math]b^{-m}[/math], a special case of a disjunctive sequence. Simply normal sequences need hold only for [math]m=1[/math], so all normal sequences are simply normal, but not vice versa. Disjunctive sequences are more general cases of normal sequences, where the frequency of every subsequence need not approach a specific value, only be non-zero. It’s easy to give examples of simply normal sequences that are not disjunctive, with the family of sequences of the decimal digits of rational numbers of the form [math]\frac{a}{10^{10n}-1}[/math], where [math]a[/math] is a simply normal positive [math]10n[/math]-digit integer. For example, [math]1234567890/9999999999 = .1234567890 \, 1234567890 \, \dots[/math]. Although the sequence 1,2,3,4,5,6,7,8,9,0, 1,2,3,4,5,6,7,8,9,0 ... is simply normal, it is not disjunctive, because it does not contain many subsequences, such as “1,0”. So, provided the strict definition of normality is used, step 3 in post #20 is true by definition. Step 2, however, remains a major unproven. Quote Link to comment Share on other sites More sharing options...

Moontanman Posted April 7, 2008 Report Share Posted April 7, 2008 Is the square root of negative one irrational? Quote Link to comment Share on other sites More sharing options...

CraigD Posted April 7, 2008 Report Share Posted April 7, 2008 Is the square root of negative one irrational?No. It’s an imaginary number and a complex number, but can be represented as fraction ([math]\frac11 i[/math]), so isn’t an irrational number. Here’s a compact summary of the common number systems:Natural (n) : 1, 2, 3, ... Integer (z) = [math]n_1 - n_2[/math] : ..., -2, -1, 0, 1, 2, ...Rational ®= [math]\frac{z_1}{z_2}[/math] : ... -2 ... -3/2 ... -1 ... -1/5 ... 0 ... 1 ... 123/45 ... Real ® = [math]r_1^{r_2} - r_3^{r_4} + r_5^{r_6} \dots[/math], where [math]r_1, \, r_3, \, r_4[/math] are non-negative: [math]2^{1/2} = \sqrt{2},\, \pi,[/math] etc.Complex © = [math]r_1 +r_2i[/math], where [math]i^2 = -1[/math] : 2 -3i, etc. (these can’t be written more neatly than this)There are several mathematically interesting number systems “outside and between” these common ones, but these 5 are the most important ones. Imaginary numbers are a subset of the complex numbers, while irrational numbers are a subset of the reals. Note that each system contains the preceding one, and can be made with an arithmetic expression of two to an infinite number of the preceding numbers (though a math purist might have issues with the rational to real step). Quote Link to comment Share on other sites More sharing options...

Moontanman Posted April 7, 2008 Report Share Posted April 7, 2008 No. It’s an imaginary number and a complex number, but can be represented as fraction ([math]frac11 i[/math]), so isn’t an irrational number. Here’s a compact summary of the common number systems:Natural (n) : 1, 2, 3, ... Integer (z) = [math]n_1 - n_2[/math] : ..., -2, -1, 0, 1, 2, ...Rational ®= [math]frac{z_1}{z_2}[/math] : ... -2 ... -3/2 ... -1 ... -1/5 ... 0 ... 1 ... 123/45 ... Real ® = [math]r_1^{r_2} - r_3^{r_4} + r_5^{r_6} dots[/math], where [math]r_1, , r_3, , r_4[/math] are non-negative: [math]2^{1/2} = sqrt{2},, pi,[/math] etc.Complex © = [math]r_1 +r_2i[/math], where [math]i^2 = -1[/math] : 2 -3i, etc. (these can’t be written more neatly than this)There are several mathematically interesting number systems “outside and between” these common ones, but these 5 are the most important ones. Imaginary numbers are a subset of the complex numbers, while irrational numbers are a subset of the reals. Note that each system contains the preceding one, and can be made with an arithmetic expression of two to an infinite number of the preceding numbers (though a math purist might have issues with the rational to real step). Thank you, I wasn't sure, it's been a long time since I had to know anything beyond real numbers. I used to work in statistical analysis but that was almost 15 years ago now. Quote Link to comment Share on other sites More sharing options...

Nootropic Posted May 25, 2008 Report Share Posted May 25, 2008 Constructing the reals from the rationals is interesting. Baby Rudin's first chapter utilizes Dedekind Cuts for this purpose and it's truly fascinating. Probably one of the more interesting facts about R is the fact that it is an infinite-dimensional extension of Q and C is a finite-dimensional extension of R. It really is much greater of a logical leap from Q to R than it is from R to C. CraigD 1 Quote Link to comment Share on other sites More sharing options...

ughaibu Posted November 12, 2008 Report Share Posted November 12, 2008 Step 2, however, remains a major unproven.Does this article meet the requirement? Borel Normality and Algorithmic Randomness # - CiteSeerX Quote Link to comment Share on other sites More sharing options...

CraigD Posted November 12, 2008 Report Share Posted November 12, 2008 Step 2, however, remains a major unproven.Does this article meet the requirement? Borel Normality and Algorithmic Randomness # - CiteSeerX No, it doesn’t. It seems a good and interesting proof (I’ve not yet read it in depth), :thumbs_up but it applies only to a particular class of random real numbers, not to all irrational numbers. We can easily construct irrational numbers that are not normal to a particular base, disproving the assertion that all irrational numbers are normal. For example: [math]\sum_{n=0}^{\infty} 10^{-(2^n)} = 0.11010001000000010000000000000001000000000000000000000000000001 \dots [/math] is both irrational and nonnormal in base 10. To the best of my knowledge, irrational numbers such as [imath]\pi[/imath] and [imath]\sqrt{2}[/imath] are believed to be normal, but have never been proved so or not so. Such a proof would, I believe, be very major, and likely of consequence in many other areas of number theory. Quote Link to comment Share on other sites More sharing options...

engineerdude Posted November 12, 2008 Report Share Posted November 12, 2008 hehe Quote Link to comment Share on other sites More sharing options...

ughaibu Posted November 12, 2008 Report Share Posted November 12, 2008 it applies only to a particular class of random real numbers, not to all irrational numbersYes, I realise that, nevertheless, if my argument in post 20 is sound, I think normality would be a problem for some non-repeating, non-terminating numbers. I suspect the problem with my argument is premise 4, that only finite sequences appear in the expansion, whereas the power set of aleph zero includes infinite sequences. However, I dont see why the expansion can not contain sequences of length (cardinality) aleph zero(?) Anyway, thanks for your reply. Quote Link to comment Share on other sites More sharing options...

## Recommended Posts

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.