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.
Since Pi is infinite and non-repeating, would that mean any finite sequence of non-repeating digits from 0-9 should appear somewhere in Pi in base 10?
…and this…
Does any possible string of infinite non-repeating digits contain every possible finite sequence of non repeating digits?
are equivalent statements.
The phrase “since X, would that mean Y” is the same as asking “is X a sufficient condition for Y”. Providing ANY example of X WITHOUT Y is a counter-example which proves X is NOT a sufficient condition.
The 1.010010001… example is literally one that is taught in classes to disprove OPs exact hypothesis. This isn’t a discussion where we’re both offering different perspectives and working towards a truth we don’t both see, thus is a discussion where you’re factually wrong and I’m trying to help you learn why lol.
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.
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
It was just an example of an infinite, non-repeating number that still does not contain every other finite number
Another example could be 0.10100100010000100000… with the number of 0’s increasing by one every time. It never repeats, but it still doesn’t contain every other finite number.
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.
Isnt this a stupid example though, because obviously if you remove all penguins from the zoo, you’re not going to see any penguins
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.
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In terms of formal logic, this…
…and this…
are equivalent statements.
The phrase “since X, would that mean Y” is the same as asking “is X a sufficient condition for Y”. Providing ANY example of X WITHOUT Y is a counter-example which proves X is NOT a sufficient condition.
The 1.010010001… example is literally one that is taught in classes to disprove OPs exact hypothesis. This isn’t a discussion where we’re both offering different perspectives and working towards a truth we don’t both see, thus is a discussion where you’re factually wrong and I’m trying to help you learn why lol.
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.
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
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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
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”
It was just an example of an infinite, non-repeating number that still does not contain every other finite number
Another example could be 0.10100100010000100000… with the number of 0’s increasing by one every time. It never repeats, but it still doesn’t contain every other finite number.
op’s question was focued very clearly on pi, but sure.
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