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.
That was quite an elegant proof
Doesn't the sequence "01" repeat? Or am I misunderstanding the term.
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).
yeah, but non-repeating in terms of decimal numbers usually mean: you cannot write it as 0.(abc), which would mean 0.abcabcabcabc...
What about in the context of Pi?
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 repeating (like pi) but doesn't contain all finite numbers
You mean infinite and non- repeating?
"2" is a finite sequence that doesn't exist in the example number
https://github.com/philipl/pifs
πfs is a revolutionary new file system that, instead of wasting space storing your data on your hard drive, stores your data in π! You'll never run out of space again - π holds every file that could possibly exist! They said 100% compression was impossible? You're looking at it!
I enjoyed this linked text:
If you compute it, you will be guilty of:
- Copyright infringement (of all books, all short stories, all newspapers, all magazines, all web sites, all music, all movies, and all software, including the complete Windows source code)
- Trademark infringement
- Possession of child pornography
- Espionage (unauthorized possession of top secret information)
- Possession of DVD-cracking software
- Possession of threats to the President
- Possession of everyone's SSN, everyone's credit card numbers, everyone's PIN numbers, everyone's unlisted phone numbers, and everyone's passwords
- Defaming Islam. Not technically illegal, but you'll have to go into hiding along with Salman Rushdie.
- Defaming Scientology. Which IS illegal--just ask Keith Henson.
Not just any all finite number sequence appear in pi
0.101001000100001000001 . . .
I'm infinite and non-repeating. Can you find a 2 in me?
Are you trying to say the answer to their question is no? Because if so, you're wrong, and if not I'm not sure what you're trying to say.
The conclusion does not follow from the premises, as evidenced by my counterexample. It could be the case that every finite string of digits appears in the decimal expansion of pi, but if that's the case, a proof would have to involve more properties than an infinite non-repeating decimal expansion. I would like to see your proof that every finite string of digits appears in the decimal expansion of pi.
Well that's just being pointlessly pedantic, obviously they fucking know that a repeating number of all zeros and ones doesn't have a two in it. This is pure reddit pedantry you're doing
It's a math proof. Chill.
Are you Pi?
You can't prove that there isn't one somewhere
You can, it's literally the way the number is defined.
Defined where ?
It's implicitly defined here by its decimal form:
0.101001000100001000001 . . .
The definition of this number is that the number of 0s after each 1 is given by the total previous number of 1s in the sequence. That's why it can't contain 2 despite being infinite and non-repeating.
Why couldn’t you?
Because you'd need to search through an infinite number of digits (unless you have access to the original formula)
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.
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.
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.
Very rare in the sense that they have a probability of 0.
There are lot that fit that pattern. However, most/all naturally used irrational numbers seem to be normal. Maths has, however had enough things that seemed 'obvious' which turned out to be false later. Just because it's obvious doesn't mean it's mathematically true.
Yes
Can you prove this? Or link a proof?
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 ...
Yes.
And if you're thinking of a compression algorithm, nope, pigeonhole principle.
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