How about ANY FINITE SEQUENCE AT ALL?
It’s almost sure to be the case, but nobody has managed to prove it yet.
Simply being infinite and non-repeating doesn’t guarantee that all finite sequences will appear. For example, you could have an infinite non-repeating number that doesn’t have any 9s in it. But, as far as numbers go, exceptions like that are very rare, and in almost all (infinite, non-repeating) numbers you’ll have all finite sequences appearing.
Rare in this context is a question of density. There are infinitely many integers within the real numbers, for example, but there are far more non-integers than integers. So integers are more rare within the real.
Yes, compared to the infinitely more non exceptions. For each infinite number that doesn’t contain the digit 9 you have an infinite amount of numbers that can be mapped to that by removing all the 9s. For example 3.99345 and 3.34999995 both map to 3.345. In the other direction it doesn’t work that way.
A number for which that is true is called a normal number. It’s proven that almost all real numbers are normal, but it’s very difficult to prove that any particular number is normal. It hasn’t yet been proved that π is normal, though it’s generally assumed to be.
No, the fact that a number is infinite and non-repeating doesn’t mean that and since in order to disprove something you need only one example here it is: 0.1101001000100001000001… this is a number that goes 1 and then x times 0 with x incrementing. It is infinite and non-repeating, yet doesn’t contain a single 2.
This proves that an infinite, non-repeating number needn’t contain any given finite numeric sequence, but it doesn’t prove that an infinite, non-repeating number can’t. This is not to say that Pi does contain all finite numeric sequences, just that this statement isn’t sufficient to prove it can’t.
A nonrepeating number does not mean that a sequence within that number never happens again, it means that the there is no point in the number where you can predict the numbers to follow by playing back a subset of the numbers before that point on repeat. So for 01 to be the “repeating pattern”, the rest of the number at some point would have to be 010101010101010101… You can find the sequence “14” at digits 2 and 3, 104 and 105, 251 and 252, and 296 and 297 (I’m sure more places as well).
But didn’t you just give a counterexample with an infinite number? OP only said something about finite numbers.
They were showing that another Infinite repeating sequence 0.1010010001… is infinite and non-repeating (like pi) but doesn’t contain all finite numbers
Wouldn’t binary ‘10’ be 2, which it does contain? I feel like that’s cheating, since binary is just a mode of interpreting information …all numbers, regardless of base, can be represented in binary.
They’re not writing in binary. They’re defining a base 10 number that is 0.11, followed by a single 0, then 1, then two 0s, then 1, then three 0s, then 1, and so on. The definition ensures that it never repeats, but because it only contains 1 and 0, it would never contain any sequence with the numbers 2 through 9.
The jury is out on whether every finite sequence of digits is contained in pi.
However, there are a multitude of real numbers that contain every finite sequence of digits when written in base 10. Here’s one, which is defined by concatenating the digits of every non-negative integer in increasing order. It looks like this:
0 . 0 1 2 3 4 5 6 7 8 9 10 11 12 ...
Thats very cool. It brings to mind some sort of espionage where spies are exchanging massive messages contained in 2 numbers. The index and the Metadata length. All the other spy has to do is pass it though pifs to decode. Maybe adding some salt as well to prevent someone figuring it out.
I’m a layman here and not a mathematician but how does it store the complete value of pi and not rounded up to a certain amount? Or do one of the libraries generate that?
You generate it when needed, using one of the known sequences that converges to π. As a simple example, the pi() recipe here shows how to compute π to arbitrary precision. For an application like pifs you can do even better and use the BBP formula which lets you directly calculate a specific hexadecimal digit of π.
I can’t tell if this is a joke or real code… like for this sentence below.
The cat is back.
Will that repo seriously run until it finds where that is in pi? However long it might take, hours, days, years, decades, and then tell you, so you can look it up quickly?
I can’t tell if this is a joke or real code
Yes.
Will that repo seriously run until it finds where that is in pi?
Sure. It’ll take a very long while though. We can estimate roughly how long - encoded as ASCII and translated to hex your sentence looks like 54686520636174206973206261636b. That’s 30 hexadecimal digits. So very roughly, one of each 16^30 30-digit sequences will match this one. So on average, you’d need to look about 16^30 * 30 ≈ 4e37 digits into π to find a sequence matching this one. For comparison, something on the order of 1e15 digits of pi were ever calculated.
so you can look it up quickly?
Not very quickly, it’s still n log n time. More importantly, information theory is ruthless: there exist no compression algorithms that have on average a >1 compression coefficient for arbitrary data. So if you tried to use π as compression, the offsets you get would on average be larger than the data you are compressing. For example, your data here can be written written as 30 hexadecimal digits, but the offset into pi would be on the order of 4e37, which takes ~90 hexadecimal digits to write down.
