How about ANY FINITE SEQUENCE AT ALL?
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.
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.
it’s not a good example because you’ve only changed the symbolic representation and not the numerical value. the op’s question is identical when you convert to binary. thir is not a counterexample and does not prove anything.
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?”
that number is no longer pi… this is like answering the question “does the number “3548” contain 35?” by answering “no, 6925 doesnthave 35. qed”
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
