You said if we randomly pick a number between 0 and 1, there is a 100% chance that we will not pick any rational number. Which is amazing, considering there are infinitely many. I guess that's a consequence of uncountably vs countably infinite. I guess if there's an infinite sequence of blue objects, and somebody inserts a red object at a random position within the sequence, there is also a 0% chance that anybody looking through the sequence for finite time could ever find the red one. Pretty amazing. Thanks!
You're right, if we pick a uniform random variable between 0 and 1 it has probability 0 of being rational! And you're right it's also true for any countably infinite set. And although we can pick a real number uniformly at random, there is no way to pick a rational number uniformly at random. Or even an integer uniformly at random. If they all have the same probability c, then we add up c+c+c+... and get either 0 if c=0 or infinity if c>0, but the probabilities should add up to 1. So if we want to pick an integer position at random, we have to either gives the integers different probabilities of being picked or restrict ourselves to a finite set of possibilities.
The normal numbers set reminds me of the Cantor set, which is uncountably infinite yet has measure 0. If we examine its other properties we notice It has a hausdorff dimension less than 1 .. which brings us to the next question: What is the Hausdorff dimension of the normal numbers set?
The answer is apparently 1 projecteuclid.org/journals/pacific-journal-of-mathematics/volume-95/issue-1/The-Hausdorff-dimension-of-a-set-of-normal-numbers/pjm/1102735539.pdf
I find it surprising that the Copeland-Erdős constant is normal - except for 2 and 5, all primes end in 1, 3, 7 or 9, so the first few hundred digits look far from normal. Yet it is.
@@RSchrE That's clearly the case - its's just that the beginning of the sequence doesn't look 'promising'. For example, taking a list of the primes below 100,000 (~9,500 primes), the count of single digits is still strongly skewed: 0s = 2725 1s = 6353 2s = 3906 3s = 6229 4s = 3772 5s = 3816 6s = 3741 7s = 6172 8s = 3690 9s = 6130 Zero is under-represented because I didn't pad the numbers below 10,000 with leading zeros, but you can see that while the frequency of 2, 4, 5, 6, 8 is more or less equal (which supports the hypothesis of normality), 1, 3, 7 and 9 are over-represented. Obviously, this is a very small sample, but it is still larger (~46,000 digits) than what a human brain can "eyeball" to detect patterns.
@@DrSeanGroathouse But doesn't the Lebesgue measure only output zero for sets that are finite or countably infinite? You said that there are an uncountable number of non-normal real numbers as well as an uncountable number normal reals. Was that a mistake and you meant to say a countable number of non-normal reals?
The fact that the probability of picking a rational number from the reals between 0 and 1 is 0 is so unintuitive to me. I understand how it follows from measure theory, but it also feels like a contradiction regarding our theory of uncountably infinite sets is just staring us in the face. You would think that if x is in set S, and you select s from S uniformly at random, then P(x = s) > 0, but that's not true when S is uncountably infinite (I think it's also not true if S is just countably infinite? In other words, is the probability of randomly selecting 0 from the set of natural numbers also zero? It seems like that is the case.).
It's definitely unintuitive, I think! And with countably infinite sets it's maybe even less intuitive. We can't pick an integer uniformly at random, since like you said the probabilities can't be positive, so they must be 0. But then if we add up countably many 0's, that's just 0. So there's no way to get the probabilities to sum to 1. So we also can't pick a rational number between 0 and 1 uniformly at random. But we can with real numbers!
@@DrSeanGroathouse this is a very helpful reply. After some head scratching it makes more sense now. The integers are sort of doing "the right thing": you have something infinite, so selecting uniformly at random doesn't make sense and is not defined. But you're telling me you can do it with the reals? Was there any abuse of terminology when you said you can pick uniformly from random from the reals? How much of this relies on the axiom of choice?
@@DrSeanGroathouse I'm thinking about again and recalling properties of continuous probability distributions. Isn't the probability of any particular number zero? Isn't the density around numbers that's the non-zero thing? In this case is it the density around the rationals is zero?
@@compositeboson123 Which is kind of the same thing, I think. Start from your usual ax² + bx + c = 0 multiply by 4a both sides 4a²x² + 4abx + 4ac = 0 add b² to both sides ("complete the square") 4a²x² + 4abx + 4ac + b² = b² subtract 4ac from both sides 4a²x² + 4abx + b² = b² - 4ac note that the LH is now (2ax + b)², so take the square root 2ax + b = ±√(b² - 4ac) now, when you solve the for "first degree" x, the two solutions are 2√(b² - 4ac) apart.
The probability that pi is not normal is 0 as you say as per measure theory. Doesn't this serve as proof? If it doesn't, when are probability-based proofs correct and when are they not?
it doesn't serve as a proof since a probability of 1 doesn't always mean certain and a probability of 0 does not always mean impossible. 3B1B made a video about this. The title is "Why probability of 0 does not mean impossible" Nevertheless, the fact that measure theory tells us that almost all reals are normal tells us that it is more likely that Pi is normal and this helps us in forming a proof of this result because we know what we should expect and we could use a proof by contradiction by starting from the most likely false premise that Pi is not normal and reach a contradiction to conclude that it is normal.
0:17 if pi is normal, that means both of these digit sequenses are digits from pi
2:27 thank you for confirming that being rational and being normal is mutually exclusive.
I don‘t know.
You said if we randomly pick a number between 0 and 1, there is a 100% chance that we will not pick any rational number. Which is amazing, considering there are infinitely many. I guess that's a consequence of uncountably vs countably infinite. I guess if there's an infinite sequence of blue objects, and somebody inserts a red object at a random position within the sequence, there is also a 0% chance that anybody looking through the sequence for finite time could ever find the red one. Pretty amazing. Thanks!
You're right, if we pick a uniform random variable between 0 and 1 it has probability 0 of being rational! And you're right it's also true for any countably infinite set.
And although we can pick a real number uniformly at random, there is no way to pick a rational number uniformly at random. Or even an integer uniformly at random. If they all have the same probability c, then we add up c+c+c+... and get either 0 if c=0 or infinity if c>0, but the probabilities should add up to 1. So if we want to pick an integer position at random, we have to either gives the integers different probabilities of being picked or restrict ourselves to a finite set of possibilities.
Ty for ur work Dr Sean
I'm glad you liked it!
The normal numbers set reminds me of the Cantor set, which is uncountably infinite yet has measure 0. If we examine its other properties we notice It has a hausdorff dimension less than 1 .. which brings us to the next question:
What is the Hausdorff dimension of the normal numbers set?
The answer is apparently 1 projecteuclid.org/journals/pacific-journal-of-mathematics/volume-95/issue-1/The-Hausdorff-dimension-of-a-set-of-normal-numbers/pjm/1102735539.pdf
I've just noticed now that I wrongly was referring to the set of normal numbers when actually I was referring to the set of non normal numbers.
Since you talked about pi, suggestion: please talk about other popular irrational/transcendental constants like e, Sqrt(2), golden ratio, etc
Thanks for the suggestions, I added them to my list!
Awesome video! Is there a name for numbers that are normal in all bases?
Glad you liked it! Those are called 'absolutely normal' or sometimes just 'normal'
i had some pie.. it was normal..
I find it surprising that the Copeland-Erdős constant is normal - except for 2 and 5, all primes end in 1, 3, 7 or 9, so the first few hundred digits look far from normal. Yet it is.
Maybe that, since the sequences of digits grow larger and larger, the single last digit doesn’t matter in the end.
@@RSchrE That's clearly the case - its's just that the beginning of the sequence doesn't look 'promising'. For example, taking a list of the primes below 100,000 (~9,500 primes), the count of single digits is still strongly skewed:
0s = 2725
1s = 6353
2s = 3906
3s = 6229
4s = 3772
5s = 3816
6s = 3741
7s = 6172
8s = 3690
9s = 6130
Zero is under-represented because I didn't pad the numbers below 10,000 with leading zeros, but you can see that while the frequency of 2, 4, 5, 6, 8 is more or less equal (which supports the hypothesis of normality), 1, 3, 7 and 9 are over-represented. Obviously, this is a very small sample, but it is still larger (~46,000 digits) than what a human brain can "eyeball" to detect patterns.
4:08 WHICH measure are we talking about?
Lebesgue measure :)
@@DrSeanGroathouse But doesn't the Lebesgue measure only output zero for sets that are finite or countably infinite? You said that there are an uncountable number of non-normal real numbers as well as an uncountable number normal reals. Was that a mistake and you meant to say a countable number of non-normal reals?
@@Lucky10279 Not necessarily, as a counterexample, there is the Cantor set. Look it up!
Not necessarily, as a counterexample, there is the Cantor set. Look it up!
My beliefs: Pi is normal and we will never be able to prove it.
The fact that the probability of picking a rational number from the reals between 0 and 1 is 0 is so unintuitive to me. I understand how it follows from measure theory, but it also feels like a contradiction regarding our theory of uncountably infinite sets is just staring us in the face. You would think that if x is in set S, and you select s from S uniformly at random, then P(x = s) > 0, but that's not true when S is uncountably infinite (I think it's also not true if S is just countably infinite? In other words, is the probability of randomly selecting 0 from the set of natural numbers also zero? It seems like that is the case.).
It's definitely unintuitive, I think! And with countably infinite sets it's maybe even less intuitive. We can't pick an integer uniformly at random, since like you said the probabilities can't be positive, so they must be 0. But then if we add up countably many 0's, that's just 0. So there's no way to get the probabilities to sum to 1. So we also can't pick a rational number between 0 and 1 uniformly at random. But we can with real numbers!
@@DrSeanGroathouse this is a very helpful reply. After some head scratching it makes more sense now. The integers are sort of doing "the right thing": you have something infinite, so selecting uniformly at random doesn't make sense and is not defined. But you're telling me you can do it with the reals? Was there any abuse of terminology when you said you can pick uniformly from random from the reals? How much of this relies on the axiom of choice?
@@DrSeanGroathouse I'm thinking about again and recalling properties of continuous probability distributions. Isn't the probability of any particular number zero? Isn't the density around numbers that's the non-zero thing? In this case is it the density around the rationals is zero?
edit: pi is not normal, like not at all buttt its also normal because that would be cool
why the discriminant *of the quadratic polynomial* = b^2-4ac
edit: corrected the sentence
Are you asking how to derive the quadratic formula?
@@dlevi67 no the discriminant
edit: *I think
@@compositeboson123 Which is kind of the same thing, I think.
Start from your usual
ax² + bx + c = 0
multiply by 4a both sides
4a²x² + 4abx + 4ac = 0
add b² to both sides ("complete the square")
4a²x² + 4abx + 4ac + b² = b²
subtract 4ac from both sides
4a²x² + 4abx + b² = b² - 4ac
note that the LH is now (2ax + b)², so take the square root
2ax + b = ±√(b² - 4ac)
now, when you solve the for "first degree" x, the two solutions are 2√(b² - 4ac) apart.
@@dlevi67 thanks
The probability that pi is not normal is 0 as you say as per measure theory. Doesn't this serve as proof?
If it doesn't, when are probability-based proofs correct and when are they not?
it doesn't serve as a proof since a probability of 1 doesn't always mean certain and a probability of 0 does not always mean impossible. 3B1B made a video about this. The title is "Why probability of 0 does not mean impossible"
Nevertheless, the fact that measure theory tells us that almost all reals are normal tells us that it is more likely that Pi is normal and this helps us in forming a proof of this result because we know what we should expect and we could use a proof by contradiction by starting from the most likely false premise that Pi is not normal and reach a contradiction to conclude that it is normal.
🦦