Is π Random? Exploring the Elusive Normal Numbers 

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 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'm glad you liked it!

Thanks for the suggestions, I added them to my list!

Glad you liked it! Those are called 'absolutely normal' or sometimes just 'normal'

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.

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!

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?

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