3 ms·

Isn't there a non-zero chance that given an infinite number of digits, the probability of finding repeats of pi, each a bit longer, increases until a perfect, e

Isn't there a non-zero chance that given an infinite number of digits, the probability of finding repeats of pi, each a bit longer, increases until a perfect, endless repeat of pi will eventually be found thus nullifying pi's own infinity?

Antipode 2y ago
The chance of that loop repeating forever is 0.
- robxorb 2y ago
  Infinity has entered the chat.
  - kevinventullo 2y ago
    In this case, the infinite sum 0+0+0+0+… is still zero.
djkorchi 2y ago
No, because it would create a contradiction. If a "perfect, endless repeat of pi" were eventually found (say, starting at the nth digit), then you can construct a rational number (a fraction with an integer numerator and denominator) that precisely matches it. However, pi is provably irrational, meaning no such pair of integers exists. That produces a contradiction, so the initial assumption that a "perfect, endless repeat of pi" exists cannot be true.
- robxorb 2y ago
  Yes and that contradiction is already present in my premise which is the point. Pi, if an infinite stream of digits and with the prime characteristic it is normal/random, will, at some point include itself, by chance. Unless, not random... This applies to every normal, "irrational" number, the name with which I massively agree, because the only way they can be not purely random suggests they are compressible further and so they have to be purely random, and thus... can't be. It is a completely irrational concept, thinking rationally.
  - linearrust 2y ago
    > Pi, if an infinite stream of digits and with the prime characteristic it is normal/random, will, at some point include itself, by chance. What you are essentially saying is that pi = 3.14....pi........... If that was the case, wouldn't it mean that the digits of pi are not countably infinite but instead is a continuum. So you wouldn't be able to put the digits of pi in one to one correspondence with natural numbers. But obviously we can so shouldn't our default be to assume our premise was wrong? > It is a completely irrational concept, thinking rationally. It is definitely interesting to think about.
    - traes 2y ago
      The belief that a normal number must eventually contain itself arises from extremely flawed thinking about probability. Like djkorchi mentioned above, if we knew pi = 3.14....pi..., that would mean pi = 3.14... + 10^n pi for some n, meaning (1 - 10^n) pi = 3.14... and pi = (3.14...) / (1 - 10^n), aka a rational number.
      - linearrust 2y ago
        > The belief that a normal number must eventually contain itself arises from extremely flawed thinking about probability. Yes. There is an issue with the premise as it leads to a contradiction. > Like djkorchi mentioned above, if we knew pi = 3.14....pi..., that would mean pi = 3.14... + 10^n pi for some n, meaning (1 - 10^n) pi = 3.14... and pi = (3.14...) / (1 - 10^n), aka a rational number. Yes. If pi = 3.14...pi ( pi repeats at the end ), then it is rational as the ending pi itself would contain an ending pi and it would repeat forever ( hence a rational number ). I thought the guy was talking about pi contain pi somewhere within itself. pi = 3.14...pi... ( where the second ... represents an infinite series of numbers ). Then we would never reach the second set of ... and the digits of pi would not be enumerable. So if pi cannot be contained within ( anywhere in the middle of pi ) and pi cannot be contained at the end, then pi must not contain pi.
    - thaumasiotes 2y ago
      > If that was the case, wouldn't it mean that the digits of pi are not countably infinite but instead is a continuum. No; combining two countably infinite sets doesn't increase the cardinality of the result (because two is finite). Combining one finite set with one countably infinite set won't give you an uncountable result either. The digits would still be countably infinite. Looking at this from another direction, it is literally true that, when x = 1/7, x = 0.142....x.... , but it is obviously not true that the decimal expansion of 1/7 contains uncountably many digits.

Antipode 2y ago

The chance of that loop repeating forever is 0.

robxorb 2y ago
Infinity has entered the chat.
- kevinventullo 2y ago
  In this case, the infinite sum 0+0+0+0+… is still zero.

djkorchi 2y ago

No, because it would create a contradiction. If a "perfect, endless repeat of pi" were eventually found (say, starting at the nth digit), then you can construct a rational number (a fraction with an integer numerator and denominator) that precisely matches it. However, pi is provably irrational, meaning no such pair of integers exists. That produces a contradiction, so the initial assumption that a "perfect, endless repeat of pi" exists cannot be true.

robxorb 2y ago
Yes and that contradiction is already present in my premise which is the point. Pi, if an infinite stream of digits and with the prime characteristic it is normal/random, will, at some point include itself, by chance. Unless, not random... This applies to every normal, "irrational" number, the name with which I massively agree, because the only way they can be not purely random suggests they are compressible further and so they have to be purely random, and thus... can't be. It is a completely irrational concept, thinking rationally.
- linearrust 2y ago
  > Pi, if an infinite stream of digits and with the prime characteristic it is normal/random, will, at some point include itself, by chance. What you are essentially saying is that pi = 3.14....pi........... If that was the case, wouldn't it mean that the digits of pi are not countably infinite but instead is a continuum. So you wouldn't be able to put the digits of pi in one to one correspondence with natural numbers. But obviously we can so shouldn't our default be to assume our premise was wrong? > It is a completely irrational concept, thinking rationally. It is definitely interesting to think about.
  - traes 2y ago
    The belief that a normal number must eventually contain itself arises from extremely flawed thinking about probability. Like djkorchi mentioned above, if we knew pi = 3.14....pi..., that would mean pi = 3.14... + 10^n pi for some n, meaning (1 - 10^n) pi = 3.14... and pi = (3.14...) / (1 - 10^n), aka a rational number.
    - linearrust 2y ago
      > The belief that a normal number must eventually contain itself arises from extremely flawed thinking about probability. Yes. There is an issue with the premise as it leads to a contradiction. > Like djkorchi mentioned above, if we knew pi = 3.14....pi..., that would mean pi = 3.14... + 10^n pi for some n, meaning (1 - 10^n) pi = 3.14... and pi = (3.14...) / (1 - 10^n), aka a rational number. Yes. If pi = 3.14...pi ( pi repeats at the end ), then it is rational as the ending pi itself would contain an ending pi and it would repeat forever ( hence a rational number ). I thought the guy was talking about pi contain pi somewhere within itself. pi = 3.14...pi... ( where the second ... represents an infinite series of numbers ). Then we would never reach the second set of ... and the digits of pi would not be enumerable. So if pi cannot be contained within ( anywhere in the middle of pi ) and pi cannot be contained at the end, then pi must not contain pi.
  - thaumasiotes 2y ago
    > If that was the case, wouldn't it mean that the digits of pi are not countably infinite but instead is a continuum. No; combining two countably infinite sets doesn't increase the cardinality of the result (because two is finite). Combining one finite set with one countably infinite set won't give you an uncountable result either. The digits would still be countably infinite. Looking at this from another direction, it is literally true that, when x = 1/7, x = 0.142....x.... , but it is obviously not true that the decimal expansion of 1/7 contains uncountably many digits.