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.
Please read it all again. They didn’t rely on the conversion. It’s just a convenient way to create a counterexample.
Anyway, here’s a simple equivalent. Let’s consider a number like pi except that wherever pi has a 9, this new number has a 1. This new number is infinite and doesn’t repeat. So it also answers the original question.
“please consider a number that isnt pi” so not relevant, gotcha. it does not answer the original question, this new number is not normal, sure, but that has no bearing on if pi is normal.
The digits of pi are not proven to be uniform or randomly distributed according to any pattern.
Pi could have a point where it stops having 9’s at all.
If that’s the case, it would not contain all sequences that contain the digit 9, and could not contain all sequences.
While we can’t look at all the digits of Pi, we could consider that the uniform behavior of the digits in pi ends at some point, and wherever there would usually be a 9, the digit is instead a 1. This new number candidate for pi is infinite, doesn’t repeat and contains all the known properties of pi.
Therefore, it is possible that not any finite sequence of non-repeating numbers would appear somewhere in Pi.
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.
Please read it all again. They didn’t rely on the conversion. It’s just a convenient way to create a counterexample.
Anyway, here’s a simple equivalent. Let’s consider a number like pi except that wherever pi has a 9, this new number has a 1. This new number is infinite and doesn’t repeat. So it also answers the original question.
“please consider a number that isnt pi” so not relevant, gotcha. it does not answer the original question, this new number is not normal, sure, but that has no bearing on if pi is normal.
OK, fine. Imagine that in pi after the quadrillionth digit, all 1s are replaced with 9. It still holds
“ok fine consider a number that still isn’t pi, it still holds.” ??
Prove that said number isn’t pi.
isnt, qed
Hmm, ok. Let me retry.
The digits of pi are not proven to be uniform or randomly distributed according to any pattern.
Pi could have a point where it stops having 9’s at all.
If that’s the case, it would not contain all sequences that contain the digit 9, and could not contain all sequences.
While we can’t look at all the digits of Pi, we could consider that the uniform behavior of the digits in pi ends at some point, and wherever there would usually be a 9, the digit is instead a 1. This new number candidate for pi is infinite, doesn’t repeat and contains all the known properties of pi.
Therefore, it is possible that not any finite sequence of non-repeating numbers would appear somewhere in Pi.
i am aware nobody has proven pi is normal.
They didn’t convert anything to anything, and the 1.010010001… number isn’t binary
then it’s not relevant to the question as it is not pi.
The question is
Since pi is infinite and non-repeating, would it mean…
Then the answer is mathematically, no. If X is infinite and non-repeating it doesn’t.
If a number is normal, infinite, and non-repeating, then yes.
To answer the real question “Does any finite sequence of non-repeating numbers appear somewhere in Pi?”
The answer depends on if Pi is normal or not, but not necessarily