I can’t disagree about what you want but I myself don’t really see the point in using the word emergent for a straightforward property of irrational numbers. I wouldn’t go so far as to say the term is useless but whatever use it could have would need to describe something more complex properties that are caused by simpler rules.
This isn’t a general property of irrational numbers, although with probability 1 any irrational number will have this property. In fact, any random real number will have this property with probability 1 (rational numbers have measure 0 since they form a countable set). This is pretty easy to prove if one is familiar with Lebesque measure.
There are irrational numbers which do not share this property. For example,
.101001000100001000001… is irrational and does not share this property.
This isn’t a general property of irrational numbers, although with probability 1 any irrational number will have this property.
Any irrational number drawn from what distribution? There are plenty of distributions that you could draw irrational numbers from which do not have this property, and which contain the same number of numbers in them. For example, the set of all irrational numbers in which every other digit is zero has the same cardinality as the set of all irrational numbers.
Yes, although generally when asking these sorts of questions one looks at the standard Lebesque measure on [0,1] or [0,1) since that’s easier to normalize. I’ve been told that this result also holds for any bell-curve distribution centered at 0 but I haven’t seen a proof of that and it isn’t at all obvious to me how to construct one.
Well, the quick way is to note that the bell-curve measure is absolutely continuous with respect to Lebesgue measure, as is any other measure given by an integrable distribution function on the real line. (If you want, you can do this by hand as well, comparing the probability of a small bounded open set in the bell curve distribution with its Lebesgue measure, taking limits, and then removing the condition of boundedness.)
I can’t disagree about what you want but I myself don’t really see the point in using the word emergent for a straightforward property of irrational numbers. I wouldn’t go so far as to say the term is useless but whatever use it could have would need to describe something more complex properties that are caused by simpler rules.
This isn’t a general property of irrational numbers, although with probability 1 any irrational number will have this property. In fact, any random real number will have this property with probability 1 (rational numbers have measure 0 since they form a countable set). This is pretty easy to prove if one is familiar with Lebesque measure.
There are irrational numbers which do not share this property. For example, .101001000100001000001… is irrational and does not share this property.
True enough. it would seem that irrational number is not the correct term for the set I refer to.
The property you are looking for is normalness to base 10. See normal number.
ETA: Actually, you want simple normalness to base 10 which is slighly weaker.
Any irrational number drawn from what distribution? There are plenty of distributions that you could draw irrational numbers from which do not have this property, and which contain the same number of numbers in them. For example, the set of all irrational numbers in which every other digit is zero has the same cardinality as the set of all irrational numbers.
I’m presuming he’s talking about measure, using the standard Lebesgue measure on R
Yes, although generally when asking these sorts of questions one looks at the standard Lebesque measure on [0,1] or [0,1) since that’s easier to normalize. I’ve been told that this result also holds for any bell-curve distribution centered at 0 but I haven’t seen a proof of that and it isn’t at all obvious to me how to construct one.
Well, the quick way is to note that the bell-curve measure is absolutely continuous with respect to Lebesgue measure, as is any other measure given by an integrable distribution function on the real line. (If you want, you can do this by hand as well, comparing the probability of a small bounded open set in the bell curve distribution with its Lebesgue measure, taking limits, and then removing the condition of boundedness.)
Excellent, yes that does work. Thanks very much!