Irrational Numbers

CraigD · 04-04-2008

Quote:

Originally Posted by ughaibu

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 Distribution of Digits in Irrational Numbers.

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.

Quote:

Originally Posted by ughaibu

3) an infinite normal non-repeating string will contain all possible strings of digits

Ughaibu, can you give a proof, or a link to a proof, of this?

A cool line of thought

ughaibu · 04-05-2008

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.

ughaibu · 04-06-2008

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

CraigD · 04-06-2008

Quote:

Originally Posted by ughaibu

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

The 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 $b^{-m}$ , a special case of a disjunctive sequence. Simply normal sequences need hold only for $m=1$ , 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 $\frac{a}{10^{10n}-1}$ , where $a$ is a simply normal positive $10n$ -digit integer. For example, $1234567890/9999999999 = .1234567890 \, 1234567890 \, \dots$ . 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.

Moontanman · 04-06-2008

Is the square root of negative one irrational?

CraigD · 04-06-2008

Quote:

Originally Posted by Moontanman

Is the square root of negative one irrational?

No.

It’s an imaginary number and a complex number, but can be represented as fraction ( $\frac11 i$ ), so isn’t an irrational number.

Here’s a compact summary of the common number systems:

Natural (n) : 1, 2, 3, ...
Integer (z) = $n_1 - n_2$ : ..., -2, -1, 0, 1, 2, ...
Rational (r)= $\frac{z_1}{z_2}$ : ... -2 ... -3/2 ... -1 ... -1/5 ... 0 ... 1 ... 123/45 ...
Real (r) = $r_1^{r_2} - r_3^{r_4} + r_5^{r_6} \dots$ , where $r_1, \, r_3, \, r_4$ are non-negative: $2^{1/2} = \sqrt{2},\, \pi,$ etc.
Complex (c) = $r_1 +r_2i$ , where $i^2 = -1$ : 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).

Moontanman · 04-06-2008

Quote:

Originally Posted by CraigD

No.

It’s an imaginary number and a complex number, but can be represented as fraction ( $frac11 i$ ), so isn’t an irrational number.

Here’s a compact summary of the common number systems:

Natural (n) : 1, 2, 3, ...
Integer (z) = $n_1 - n_2$ : ..., -2, -1, 0, 1, 2, ...
Rational (r)= $frac{z_1}{z_2}$ : ... -2 ... -3/2 ... -1 ... -1/5 ... 0 ... 1 ... 123/45 ...
Real (r) = $r_1^{r_2} - r_3^{r_4} + r_5^{r_6} dots$ , where $r_1, , r_3, , r_4$ are non-negative: $2^{1/2} = sqrt{2},, pi,$ etc.
Complex (c) = $r_1 +r_2i$ , where $i^2 = -1$ : 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.

Nootropic · 05-24-2008

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.

ughaibu · 11-11-2008

Quote:

Originally Posted by CraigD

Step 2, however, remains a major unproven.

Does this article meet the requirement? Borel Normality and Algorithmic Randomness # - CiteSeerX

CraigD · 11-11-2008

Quote:

Originally Posted by ughaibu

Quote:

Originally Posted by CraigD

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),

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:

$\sum_{n=0}^{\infty} 10^{-(2^n)} = 0.11010001000000010000000000000001000000000000000000000000000001 \dots$

is both irrational and nonnormal in base 10.

To the best of my knowledge, irrational numbers such as $\pi$ and $\sqrt{2}$ 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.

Irrational Numbers

Hypography?

Share the love!

Sponsors

Friends