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.
They also say “and reinterpret in base 10”. I.e. interpret the base 2 number as a base 10 number (which could theoretically contain 2,3,4,etc). So 10 in that number represents decimal 10 and not binary 10
I don’t think the example given above is an apples-to-apples comparison though. This new example of “an infinite non-repeating string” is actually “an infinite non-repeating string of only 0s and 1s”. Of course it’s not going to contain a “2”, just like pi doesn’t contain a “Y”. Wouldn’t a more appropriate reframing of the original question to go with this new example be “would any finite string consisting of only 0s and 1s be present in it?”
They just proved that “X is irrational and non-repeating digits -> can find any sequence in X”, as the original question implied, is false. Maybe pi does in fact contain any sequence, but that wouldn’t be because of its irrationality or the fact that it’s non-repeating, it would be some other property
this is correct but i think op is asking the wrong question.
at least from a mathematical perspective, the claim that pi contains any finite string is only a half-baked version of the conjecture with that implication. the property tied to this is the normality of pi which is actually about whether the digits present in pi are uniformly distributed or not.
from this angle, the given example only shows that a base 2 string contains no digits greater than 1 but the question of whether the 1s and 0s present are uniformly distributed remains unanswered. if they are uniformly distributed (which is unknown) the implication does follow that every possible finite string containing only 1s and 0s is contained within, even if interpreted as a base 10 string while still base 2. base 3 pi would similarly contain every possible finite string containing only the digits 0-2, even when interpreted in base 10 etc. if it is true in any one base it is true in all bases for their corresponding digits
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.
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.
It does contain a 2 though? Binary ‘10’ is 2, which this sequence contains?
They also say “and reinterpret in base 10”. I.e. interpret the base 2 number as a base 10 number (which could theoretically contain 2,3,4,etc). So 10 in that number represents decimal 10 and not binary 10
I don’t think the example given above is an apples-to-apples comparison though. This new example of “an infinite non-repeating string” is actually “an infinite non-repeating string of only 0s and 1s”. Of course it’s not going to contain a “2”, just like pi doesn’t contain a “Y”. Wouldn’t a more appropriate reframing of the original question to go with this new example be “would any finite string consisting of only 0s and 1s be present in it?”
They just proved that “X is irrational and non-repeating digits -> can find any sequence in X”, as the original question implied, is false. Maybe pi does in fact contain any sequence, but that wouldn’t be because of its irrationality or the fact that it’s non-repeating, it would be some other property
Then put 23456789 at the start. Doesn’t contain 22 then but all digits in base 10.
Like the other commenter said its meant to be interpreted in base10.
You could also just take 0.01001100011100001111… as an example
this is correct but i think op is asking the wrong question.
at least from a mathematical perspective, the claim that pi contains any finite string is only a half-baked version of the conjecture with that implication. the property tied to this is the normality of pi which is actually about whether the digits present in pi are uniformly distributed or not.
from this angle, the given example only shows that a base 2 string contains no digits greater than 1 but the question of whether the 1s and 0s present are uniformly distributed remains unanswered. if they are uniformly distributed (which is unknown) the implication does follow that every possible finite string containing only 1s and 0s is contained within, even if interpreted as a base 10 string while still base 2. base 3 pi would similarly contain every possible finite string containing only the digits 0-2, even when interpreted in base 10 etc. if it is true in any one base it is true in all bases for their corresponding digits