How about ANY FINITE SEQUENCE AT ALL?
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.
