mathematical paradox?

ixfd64 · 2005-09-11, 19:50

Here is a problem I have been wondering about for quite some time. Is it possible for transcendental constants (like pi or e), or perhaps very large prime numbers (like those over 10^5,000,000 digits), to have long repeating strings of digits or other interesting patterns?

In other words, is it possible for pi to have, say, a string of five trillion ones?

Code:

11111.....(4,999,999,999,990 ones).....11111

The chances of this is one out of this is nearly zero. But then, pi goes on infinitely.

What do you think?

cyrix · 2005-09-11, 20:02

Hi ixfd64!

If $\pi$ is random (which is unproven, but most people think so), then you can find every finite string (of numbers) in the decimal expansion of $\pi$ .

cyrix

Citrix · 2005-09-11, 21:35

As for prime numbers you can generate a prime with a large number of 11111's.

Citrix

jinydu · 2005-09-11, 23:46

Quote:

Originally Posted by ixfd64

The chances of this is one out of this is nearly zero. But then, pi goes on infinitely.

If the decimal expansion of $\pi$ is random (obviously, $\pi$ itself is not random), then such a sequence of ones must occur. (Actual) infinity wins out over "nearly zero".

Peter Nelson · 2005-09-12, 01:04

Just out of interest, the BINARY expansion of EVERY MERSENNE PRIME is a long string of 1s without zeroes.

There is not yet a proof that the decimal expansion of pi (or certain other numbers) has its digits randomly arranged. If it is then it is possible to find within it any arbitrary string of digits. If not, you might find a certain string , or might not. For example there is a number whose decimal expansion contains only odd digits eg 1/3. You will not find any sequence in there looking like 88888 etc

ColdFury · 2005-09-12, 01:49

Quote:

If Pi is random (which is unproven, but most people think so),

Pi is a constant. You mean if the digits of Pi are normally distributed.

Mystwalker · 2005-09-12, 22:35

Is there a need for some form of distribution at all?

IANAM, but I'd think that it's enough that Pi is transcendent, which AFAIK means that the decimal representation

a) has an infinite amount of positions after the decimal point and
b) doesn't have a recurring decimal.

As a result, every finite pattern should be included, right?

cheesehead · 2005-09-13, 02:54

Quote:

Originally Posted by Mystwalker

IANAM, but I'd think that it's enough that Pi is transcendent, which AFAIK means that the decimal representation

a) has an infinite amount of positions after the decimal point and
b) doesn't have a recurring decimal.

As a result, every finite pattern should be included, right?

Not necessarily. Ya gotta be careful with those transcendentals! Just because a number is transcendental doesn't mean it has every "special" property. Your a) and b) are correct, but they are not sufficient to guarantee that every finite pattern is included.

The Thue Constant (http://mathworld.wolfram.com/ThueConstant.html) is transcendental, but you'll never find two or more consecutive 0s in its base-2 representation! Every 0 digit in its base-2 representation has a 1 digit on each side, due to its definition. Now, this just means that not every finite binary pattern appears in the binary representation of this particular transcendental constant, but I'm fairly sure that for any base n, a constant can be constructed that is transcendental (proving transcendence is the hard part) but does not include certain finite base-n digit strings in the constant's base-n representation.

- - - -

Check out "We are in Digits of Pi and Live Forever" at http://sprott.physics.wisc.edu/pickover/pimatrix.html !!!

jinydu · 2005-09-13, 03:42

I checked the "We are in Digits of Pi" link and, being lazy, read only the first two posts. But it seems that at least the first two posters there understand neither the definition of an irrational number nor cheesehead's point.

In any case, if you want to find a number that certainly does contain every finite sequence of numbers, there's a much easier way to do so. Just consider the number:

0.123456789101112131415161718...

Here's another transcendental number that doesn't contain every possible combination of (binary) bits:

http://mathworld.wolfram.com/LiouvillesConstant.html

For instance, the sequence 1001 never appears in that number.

ixfd64 · 2005-09-13, 07:38

Yes, it's possible for transcendental numbers to have patterns.

For example, 0.1010010001000010000010000001... is one.

cheesehead · 2005-09-13, 11:10

Quote:

Originally Posted by jinydu

I checked the "We are in Digits of Pi" link and, being lazy, read only the first two posts. But it seems that at least the first two posters there understand neither the definition of an irrational number nor cheesehead's point.

It gets better as it goes along, like many of our threads. Please read at least down to the Britney Spears photo. And the second page is better (not counting photo).

( Why Britney Spears? Well, it may be because of the "Britney Spears guide to Semiconductor Physics: semiconductor physics, Edge Emitting Lasers and VCSELs" page at http://britneyspears.ac/lasers.htm Or not.

But be sure to read how actress Hedy Lamarr (a genuine electrical engineer) co-invented spread-spectrum radio transmission after she escaped to the U.S. from pre-WWII Austria. U.S. Patent 2,292,387 for the "Secret Communication System" was granted on August 11, 1942. The patent is actually under her married name at the time - Hedy Kiesler Markey.

http://britneyspears.ac/physics/intro/hedy.htm)

2005-09-11, 19:50	#1
ixfd64 Bemusing Prompter "Danny" Dec 2002 California 100100111101₂ Posts	mathematical paradox? Here is a problem I have been wondering about for quite some time. Is it possible for transcendental constants (like pi or e), or perhaps very large prime numbers (like those over 10^5,000,000 digits), to have long repeating strings of digits or other interesting patterns? In other words, is it possible for pi to have, say, a string of five trillion ones? Code: 11111.....(4,999,999,999,990 ones).....11111 The chances of this is one out of this is nearly zero. But then, pi goes on infinitely. What do you think?

2005-09-13, 03:42	#9
jinydu Dec 2003 Hopefully Near M48 2·3·293 Posts	I checked the "We are in Digits of Pi" link and, being lazy, read only the first two posts. But it seems that at least the first two posters there understand neither the definition of an irrational number nor cheesehead's point. In any case, if you want to find a number that certainly does contain every finite sequence of numbers, there's a much easier way to do so. Just consider the number: 0.123456789101112131415161718... Here's another transcendental number that doesn't contain every possible combination of (binary) bits: http://mathworld.wolfram.com/LiouvillesConstant.html For instance, the sequence 1001 never appears in that number. Last fiddled with by jinydu on 2005-09-13 at 03:47

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2005-09-11, 20:02	#2
cyrix Jul 2003 Thuringia; Germany 2·29 Posts	Hi ixfd64! If $\pi$ is random (which is unproven, but most people think so), then you can find every finite string (of numbers) in the decimal expansion of $\pi$ . cyrix

2005-09-11, 21:35	#3
Citrix Jun 2003 1,579 Posts	As for prime numbers you can generate a prime with a large number of 11111's. Citrix

2005-09-12, 01:04	#5
Peter Nelson Oct 2004 23² Posts	Just out of interest, the BINARY expansion of EVERY MERSENNE PRIME is a long string of 1s without zeroes. There is not yet a proof that the decimal expansion of pi (or certain other numbers) has its digits randomly arranged. If it is then it is possible to find within it any arbitrary string of digits. If not, you might find a certain string , or might not. For example there is a number whose decimal expansion contains only odd digits eg 1/3. You will not find any sequence in there looking like 88888 etc

2005-09-12, 22:35	#7
Mystwalker Jul 2004 Potsdam, Germany 831₁₀ Posts	Is there a need for some form of distribution at all? IANAM, but I'd think that it's enough that Pi is transcendent, which AFAIK means that the decimal representation a) has an infinite amount of positions after the decimal point and b) doesn't have a recurring decimal. As a result, every finite pattern should be included, right?

2005-09-13, 07:38	#10
ixfd64 Bemusing Prompter "Danny" Dec 2002 California 5·11·43 Posts	Yes, it's possible for transcendental numbers to have patterns. For example, 0.1010010001000010000010000001... is one.