Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ ¹ ² ³ °
 

You are not logged in. #1 20080202 16:20:27
The most difficult proofProving this is the most difficult, most frustrating to any mathematician. Character is who you are when no one is looking. #2 20080202 21:35:57
Re: The most difficult proofcounter example to corollary 1 Last edited by lucadeltodesco (20080202 21:36:26) The Beginning Of All Things To End. The End Of All Things To Come. #3 20080202 22:25:44
Re: The most difficult proofHas 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. #4 20080203 00:09:13
Re: The most difficult proof
Luca's counterexample works as long as he divides by 10^(1 million) instead, and that also disproves Corollary 2. Why did the vector cross the road? It wanted to be normal. #5 20080203 00:41:15
Re: The most difficult proofI concede mathsy is correct, however much I'd like to hate it! Character is who you are when no one is looking. #6 20080203 00:52:20
Re: The most difficult proof
silly me ^^ The Beginning Of All Things To End. The End Of All Things To Come. #7 20080203 01:06:58
Re: The most difficult proofCorollary 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. Character is who you are when no one is looking. #8 20080203 01:33:04
Re: The most difficult proofSorry, I'm not sure what Corollary 3 is saying. Why did the vector cross the road? It wanted to be normal. #9 20080204 12:24:50
Re: The most difficult proofganesh, 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..." #10 20080204 21:49:54
Re: The most difficult proofThanks for the info, Ricky. Character is who you are when no one is looking. #11 20080205 01:08:07
Re: The most difficult proofRicky, Character is who you are when no one is looking. #12 20080205 03:38:01
Re: The most difficult proofif 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.
Last edited by lucadeltodesco (20080205 03:39:08) The Beginning Of All Things To End. The End Of All Things To Come. #13 20080205 05:23:45
Re: The most difficult proof
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 nonrepeating, nonterminating, 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 counterintuitive 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. Last edited by iheartmaths (20080205 05:43:40) Don't quote me on that. #14 20080205 06:47:07
Re: The most difficult proofTrue, but a string of a million zeroes in a decimal expansion doesn't mean that it's recurring. Why did the vector cross the road? It wanted to be normal. #15 20080205 08:59:32
Re: The most difficult proof
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).
Just because you don't know what digit is next does not imply it is random. Pi is by all means not random.
Probability only makes sense when there is a concept of chance. Pi's digits are determined whether you know them or not.
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..." #16 20080205 10:52:50
Re: The most difficult proof
And just because you know what digit is next does not imply it is not random.
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.
Is that a conjecture? Why did the vector cross the road? It wanted to be normal. #17 20080206 00:40:16
Re: The most difficult proofDigits of Pi
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..." #18 20080206 02:57:53
Re: The most difficult proofThat 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.
Fine, I'll settle for pi's first 30 million digits exhibiting normality.
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. Why did the vector cross the road? It wanted to be normal. #19 20080206 03:16:42
Re: The most difficult proof
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...
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. "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..." #20 20080206 04:17:44
Re: The most difficult proof
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. Why did the vector cross the road? It wanted to be normal. #21 20080206 07:55:25
Re: The most difficult proofGreat discussion you guys are having guys. Although you lost me at some times I really enjoy all this mathematical talk. #22 20080206 10:25:15
Re: The most difficult proof
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 counterarguement, 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. #23 20080206 21:05:28
Re: The most difficult proofGuys, 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. #24 20080207 04:11:55
Re: The most difficult proofYou 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..." #25 20081030 05:37:48
Re: The most difficult proofI think pi/1 million will satisfy ganish's proposal. X'(yXβ)=0 