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## #1 2008-02-01 17:20:27

ganesh
Registered: 2005-06-28
Posts: 23,792

### The most difficult proof

Proving this is the most difficult, most frustrating to any mathematician.

Theorem :- The number pi does not ever contain a string of 1 million continuos zeros after the decimal.

Corollary 1 :- No irrational number ever contains a string of 1 million continuos zeros after the decimal.

Corollary 2 :- No irrational number ever contains a string of 1 million continuos any of the numbers from 0 to 9 after the decimal.

Attempt proving or disproving this, by counter-example or otherwise!

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #2 2008-02-01 22:35:57

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

### Re: The most difficult proof

counter example to corollary 1

divide pi by 1 million? 0.00...1million zeros... 314159...

Last edited by luca-deltodesco (2008-02-01 22:36:26)

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## #3 2008-02-01 23:25:44

Identity
Member
Registered: 2007-04-18
Posts: 934

### Re: The most difficult proof

Has this theorem been proven yet? It seems to me like a guess in the dark, given how much we know about the digit sequences in irrational numbers.

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## #4 2008-02-02 01:09:13

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: The most difficult proof

luca-deltodesco wrote:

counter example to corollary 1

divide pi by 1 million? 0.00...1million zeros... 314159...

Luca's counter-example works as long as he divides by 10^(1 million) instead, and that also disproves Corollary 2.

I would guess that the theorem itself is false also.
For any particular digit of pi, the chance that it and the next 999999 are "0" is 1/10^(1 million).
Extremely minimal, but since pi's decimal expansion has infinite digits, a sequence like that is bound to pop up eventually (assuming pi's digits are random).

Why did the vector cross the road?
It wanted to be normal.

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## #5 2008-02-02 01:41:15

ganesh
Registered: 2005-06-28
Posts: 23,792

### Re: The most difficult proof

I concede mathsy is correct, however much I'd like to hate it!
However minimal the possibility, it isn't NIL by any yardstick!
Hence, we'd have to assume it's somehwere possible down the track,
disproving it is almost next to impossible!

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #6 2008-02-02 01:52:20

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

### Re: The most difficult proof

mathsyperson wrote:

Luca's counter-example works as long as he divides by 10^(1 million) instead

silly me ^^

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## #7 2008-02-02 02:06:58

ganesh
Registered: 2005-06-28
Posts: 23,792

### Re: The most difficult proof

Corollary 3:-  No irrational number ever contains a string of 1 million continuos any of the numbers from 0 to 1,000,000 after the decimal.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #8 2008-02-02 02:33:04

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: The most difficult proof

Sorry, I'm not sure what Corollary 3 is saying.

Also, I've decided that for luca's disproof, we'd actually have to divide π by 10^(1 million +1), since there's a 3 before the point that needs to be considered.

Why did the vector cross the road?
It wanted to be normal.

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## #9 2008-02-03 13:24:50

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

### Re: The most difficult proof

ganesh, I think you have a misunderstanding what exactly a real number is.  Every sequence of digits corresponds to a unique real number (9 repeating is typically barred by definition so that this holds).   Thus, every finite sequence of digits occurs in some irrational number.  To see this, we construct a number with the sequence in question, and tack on the digits of pi.  This must be an irrational number, otherwise pi*(length of that sequence) would be rational.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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## #10 2008-02-03 22:49:54

ganesh
Registered: 2005-06-28
Posts: 23,792

### Re: The most difficult proof

Thanks for the info, Ricky.
I was pondering over this for a while.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #11 2008-02-04 02:08:07

ganesh
Registered: 2005-06-28
Posts: 23,792

### Re: The most difficult proof

Ricky,
Inspite of all what you have stated, I find the concept of finding a million zeros on the trot quite difficult, in the places of decimal of pi after the dot. I shall put it this way. Since because an indieent can occur does not make it occur. Just because the possibility cannot be eleminated, it never means it should happen at some point of time. This brings us back to square one. Infinity cannot be intrpreted to one's convenience. It is, exactly as it is. We have no business to tamper with it. My question was obviously too good one for a computer professional. I am a mathematician. I don't care the way the computer scientists work. We mathematicians show the way to the rest of the world, including the Computer sicientists. You know, who teahces whom. The symbols are the concept of the mathematicians. We mathematicians express things in the shortest possible way. The rest of the world listens to us, learns from us. How on earth do you think, as a mathematician I can accept the fact that there are a million consecutive zeros in the value of pi aftert the deciaml simply because the list is endless? ISN'T THAT ABSURDITY?

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #12 2008-02-04 04:38:01

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

### Re: The most difficult proof

if you treat digits of pi as being random, then it is perfectly possible for it to happen, if you flip a coin, its perfectly possible for you to flip heads 1 million times in a row, as is it possible for the coin to land on its side 1 million times in a row. its very unlikely, but not impossible.

but at the same time, pi is irrational, it has no pattern to it, so for all intensive purposes, it is random.

i find it quite absurd that you could think pi would contain the sequence of digits 1415926535897...etc

Last edited by luca-deltodesco (2008-02-04 04:39:08)

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## #13 2008-02-04 06:23:45

iheartmaths
Member
Registered: 2008-01-08
Posts: 31

### Re: The most difficult proof

ganesh wrote:

Proving this is the most difficult, most frustrating to any mathematician.

Theorem :- The number pi does not ever contain a string of 1 million continuos zeros after the decimal.

Corollary 1 :- No irrational number ever contains a string of 1 million continuos zeros after the decimal.

Corollary 2 :- No irrational number ever contains a string of 1 million continuos any of the numbers from 0 to 9 after the decimal.

Attempt proving or disproving this, by counter-example or otherwise!

Now, I'm quite the amateur, so correct me when I veer off the mathematician's path here, as I most likely will do. If pi is non-repeating, non-terminating, then as many digits of pi that you can calculate is only equal to ∞-n/∞ of its digits, right? So, if we looked at pi and said, "Well, it cannot contain one million zeros. And look, we've calculated it one billion places, and no zero or string of numbers 0 through 9." Then aren't we haunted by the fact that this is only ∞-n/∞ of its digits, and the fact that there are infinite numbers left. So, even if we calculated it to some astronomically high digit place and found nothing, we'd have to calculate even more, and more, and more, forever. Well, interestingly enough, I would think that the probability of 14159265358979 is equal to 111..., or 222.... However, if we accept the fact that it is infinite, then however counter-intuitive it seems, doesn't the same probability go for a million zeroes? This is only my unprofessional opinion. But it would be interesting to discuss.

However, my friend, another armchair mathematician, brought up an interesting point. He said that if pi is a division problem, such as circumference divided by diameter, then in order for it to contain one million continuous zeros, pi would have to be terminating, or at least repeating, in that to contain one million zeros, circumference/diameter = 3.14 ...000..., which means it couldn't possibly jump back to 1, right?

Either way, I think it's another interesting point.

Last edited by iheartmaths (2008-02-04 06:43:40)

Don't quote me on that.

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## #14 2008-02-04 07:47:07

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: The most difficult proof

True, but a string of a million zeroes in a decimal expansion doesn't mean that it's recurring.
By definition, recurring digits have to go on forever, and so if there are only a million in the string then it doesn't.

Not sure if this is mathematically sound, but consider the following table:

n  |  s
----------
0  |  0
1  |  0
2  |  1
3  |  3
4  |  3
5  |  5
6  |  5

This table shows that in the first 10^n digits of π's decimal expansion, the longest string of zeroes has s of them. I only went up to the first million digits because it becomes hard to find them after that, but it looks like the pattern is roughly linear. Therefore, we could reasonably expect there to be a string of 1 million zeroes somewhere within the first 10^million digits, or at least closish to that order of magnitude.

Why did the vector cross the road?
It wanted to be normal.

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## #15 2008-02-04 09:59:32

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

### Re: The most difficult proof

Inspite of all what you have stated, I find the concept of finding a million zeros on the trot quite difficult, in the places of decimal of pi after the dot. I shall put it this way. Since because an indieent can occur does not make it occur. Just because the possibility cannot be eleminated, it never means it should happen at some point of time. This brings us back to square one. Infinity cannot be intrpreted to one's convenience. It is, exactly as it is. We have no business to tamper with it. My question was obviously too good one for a computer professional. I am a mathematician. I don't care the way the computer scientists work. We mathematicians show the way to the rest of the world, including the Computer sicientists. You know, who teahces whom. The symbols are the concept of the mathematicians. We mathematicians express things in the shortest possible way. The rest of the world listens to us, learns from us. How on earth do you think, as a mathematician I can accept the fact that there are a million consecutive zeros in the value of pi aftert the deciaml simply because the list is endless? ISN'T THAT ABSURDITY?

I would never ask you to.  I have no idea whether or not there are a million zeros after the decimal point consecutively.  My intuition says no, but I don't wish to research such a question (mostly out of lack of significance and interest).

but at the same time, pi is irrational, it has no pattern to it, so for all intensive purposes, it is random.

Just because you don't know what digit is next does not imply it is random.  Pi is by all means not random.

Well, interestingly enough, I would think that the probability of 14159265358979 is equal to 111..., or 222.... However, if we accept the fact that it is infinite, then however counter-intuitive it seems, doesn't the same probability go for a million zeroes? This is only my unprofessional opinion. But it would be interesting to discuss.

Probability only makes sense when there is a concept of chance.  Pi's digits are determined whether you know them or not.

but it looks like the pattern is roughly linear.

You mean logarithmic.  Any argument which involves the phrase "looks like" is simply a conjecture.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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## #16 2008-02-04 11:52:50

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: The most difficult proof

Just because you don't know what digit is next does not imply it is random.  Pi is by all means not random.

And just because you know what digit is next does not imply it is not random.

Here's a list of (pseudo-)random numbers that I'm getting off my calculator:
87, 21, 5, 37, 96, 26, 9, 41, 84, 50.

What comes after the 37? Why, 96 of course. I knew that, and yet these numbers are still random.

Besides, the point is that pi's digits have the properties of randomness: unpredictability and uniform distribution. Therefore even if they actually aren't random, it's not unreasonable to analyse them as if they are.

You mean logarithmic.  Any argument which involves the phrase "looks like" is simply a conjecture.

I was comparing n to s, so linear is right. And of course it's conjecture, the only way to completely prove it one way or another would be to actually find a string of a million digits somewhere (which is next to impossible). You say conjecture dismissively, but using evidence to suggest an answer is better than just using intuition to guess one.

Maybe off-topic, but:

Any argument which involves the phrase "looks like" is simply a conjecture.

Is that a conjecture?

Why did the vector cross the road?
It wanted to be normal.

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## #17 2008-02-05 01:40:16

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

### Re: The most difficult proof

Digits of Pi

Specifically, look for BBP formula.

Besides, the point is that pi's digits have the properties of randomness: unpredictability and uniform distribution. Therefore even if they actually aren't random, it's not unreasonable to analyse them as if they are.

This is also not known to be true.  It is still an open question whether or not pi is normal.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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## #18 2008-02-05 03:57:53

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: The most difficult proof

That BBP formula finds the nth digit of π in base 16, which makes it largely irrelevant to this discussion since there's no easy way of converting infinite (or very large) strings of digits between bases. Not that I know or can think of, anyway.

Ricky wrote:

It is still an open question whether or not pi is normal.

Fine, I'll settle for pi's first 30 million digits exhibiting normality.

It is not known if π is normal (Wagon 1985, Bailey and Crandall 2001), although the first 30 million digits are very uniformly distributed (Bailey 1988).

So I suppose I'm conjecturing again. I'd say it's almost certainly true though, and the only reason it's not definite is that it's (virtually) impossible to prove.

(Incidentally, I like how those two people get credited for saying they don't know something.)

Why did the vector cross the road?
It wanted to be normal.

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## #19 2008-02-05 04:16:42

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

### Re: The most difficult proof

That BBP formula finds the nth digit of π in base 16, which makes it largely irrelevant to this discussion since there's no easy way of converting infinite (or very large) strings of digits between bases. Not that I know or can think of, anyway.

However, they do determine the digits of base 10.  And if the base 16 digits are not random and determine the base 10 digits, then...

So I suppose I'm conjecturing again. I'd say it's almost certainly true though, and the only reason it's not definite is that it's (virtually) impossible to prove.

(Incidentally, I like how those two people get credited for saying they don't know something.)

There are so many weird things in mathematics, I don't know how you could say that.  I do agree that if I was a betting man, all my money would go in on pi's normality.  However, mathematics isn't about betting.

Typically, people get credit for saying they don't know something only once they have shown that there really is no easy way of approaching it.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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## #20 2008-02-05 05:17:44

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: The most difficult proof

There are so many weird things in mathematics, I don't know how you could say that.

True, but there are even more elegant things. Maths works out nicely so often, and π is crucial to so many of its branches, so for π's digits to be so very uniform for the first 30 million and then stop later on is inconceivable to me.

Still, I fully realise that that's not a rigorous argument (or anywhere close), and we both agree that if we had to guess then we'd go with pi being normal, so I don't think there's much more to discuss.

Why did the vector cross the road?
It wanted to be normal.

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## #21 2008-02-05 08:55:25

LuisRodg
Real Member
Registered: 2007-10-23
Posts: 322

### Re: The most difficult proof

Great discussion you guys are having guys. Although you lost me at some times I really enjoy all this mathematical talk.

Im just a noob math major, hopefully I can become knowledgeable as all you guys.

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## #22 2008-02-05 11:25:15

iheartmaths
Member
Registered: 2008-01-08
Posts: 31

### Re: The most difficult proof

Ricky wrote:

Probability only makes sense when there is a concept of chance.  Pi's digits are determined whether you know them or not.

Determined? yes, I agree; known? Not at all. At least not after n number of decimal places. I'm not, however, saying that this is a valid counter-arguement, because I don't subscribe to the thinking of if we can't disprove it, it must exist. I will however still believe that in the higher decimal places if pi, its digits become a faith, in a sense. What I mean is that after n digits, we have to accept the fact that there is still an inifinite number of digits left. So, with that said, probability could very well be involved in pi, in that the division could take a turn for one million continuous zeros.

Don't quote me on that.

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## #23 2008-02-05 22:05:28

quantoholic
Member
Registered: 2008-02-05
Posts: 1

### Re: The most difficult proof

Guys, I m sorry to interrupt your beautiful discussion which i just bumped in to while surfing. I m a newbie to this site and dont know the rules n regulations but just got registered to post to this discussion. Please forgive me as i m not a mathematician either.

If this discussion is to take a larger form about Irrational numbers then i think you should check something which is called as the Liouville's number. It is irrational as far as i know and it does contain a million zeroes back to back. it is defined as the summation of 10^(-n!) for n = 1 to infinity.

Simply put 0.1100010000000000000000010000000000000.................

Giving 1's only for the digits which are factorial numbers 1 2 6 24 120 and so on.
so there is bound to be a value of some factorial greater than 1.
It is irrational because it neither terminates nor repeats.
Dont have any idea about the pi part though. And some guy called Mahler proved that pi is not a Liouville number. I could not understand this part though.

please rectify me if i am wrong. Thanks

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## #24 2008-02-06 05:11:55

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

### Re: The most difficult proof

You are certainly right on all counts, but I don't understand your comment about pi not being a Liouville number.  Certainly it is not equal to the number you described.  However, pi is irrational.  All ways I know to prove this involve advanced knowledge of mathematics.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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## #25 2008-10-29 06:37:48

George,Y
Member
Registered: 2006-03-12
Posts: 1,315

### Re: The most difficult proof

I think pi/1 million will satisfy ganish's proposal.

X'(y-Xβ)=0

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