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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.
Exceptions are infinite. Is that rare?
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.
This is what allows pifs to work!
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.
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.
you are absolutely right.
it just proves that even if Pi contains all finite sequences it's not "since it oa infinite and non-repeating"
That was quite an elegant proof
it's actually unknown. It looks like it, but it is not proven
Also is it even possible to prove it at all? My completely math inept brain thinks that it might be similar to the countable vs uncountable infinities thing, where even if you mapped every element of a countable infinity to one in the uncountable infinity, you could still generate more elements from the uncountable infinity. Would the same kind of logic apply to sequences in pi?
Man, you're giving me flashbacks to real analysis. Shit is weird. Like the set of all integers is the same size as the set of all positive integers. The set of all fractions, including whole numbers, aka integers, is the same size as the set of all integers. The set of all real numbers (all numbers including factions and irrational numbers like pi) is the same size as the set of all real numbers between 0 and 1. The proofs make perfect sense, but the conclusions are maddening.
A number for which that is true is called a normal number. Itβs proven that almost all real numbers are normal, but itβs very difficult to prove that any particular number is normal. It hasnβt yet been proved that Ο is normal, though itβs generally assumed to be.
I love the idea (and it's definitely true) that there are irrational numbers which, when written in a suitable base, contain the sequence of characters, "This number is provably normal" and are simultaneously not normal.
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 ...
It has not been proven either way but if pi is proven to be normal then yes. https://en.m.wikipedia.org/wiki/Normal_number
My guess would be that - depending on the number of digits you are looking for in the sequence - you could calculate the probability of finding any given group of those digits.
For example, there is a 100% probability of finding any group of two, three or four digits, but that probability decreases as you approach one hundred thousand digits.
Of course, the difficulty in proving this hypothesis rests on the computing power needed to prove it empirically and the number of digits of Pi available. That is, a million digits of Pi is a small number if you are looking for a ten thousand digit sequence
But surely given infinity, there is no problem finding a number of ANY length. It's there, somewhere, eventually, given that nothing repeats, the number is NORMAL, as people have said, and infinite.
The probability is 100% for any number, no matter how large, isn't it?
Smart people?
no. it merely being infinitely non-repeating is insufficient to say that it contains any particular finite string.
for instance, write out pi in base 2, and reinterpret as base 10.
11.0010010000111111011010101000100010000101...
it is infinitely non-repeating, but nowhere will you find a 2.
i've often heard it said that pi, in particular, does contain any finite sequence of digits, but i haven't seen a proof of that myself, and if it did exist, it would have to depend on more than its irrationality.
Isnt this a stupid example though, because obviously if you remove all penguins from the zoo, you're not going to see any penguins
The explanation is misdirecting because yes they're removing the penguins from the zoo. But they also interpreted the question as to if the zoo had infinite non-repeating exhibits whether it would NECESSARILY contain penguins. So all they had to show was that the penguins weren't necessary.
By tying the example to pi they seemed to be trying to show something about pi. I don't think that was the intention.
i just figured using pi was an easy way to acquire a known irrational number, not trying to make any special point about it.
Yeah i got confused too and saw someone else have the same distraction.
It makes sense why you chose that.
This kind of thing messed me up so much in school π
Its not stupid. To disprove a claim that states "All X have Y" then you only need ONE example. So, as pick a really obvious example.
Yes.
And if you're thinking of a compression algorithm, nope, pigeonhole principle.