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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

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.
    - linearrust 2y ago
      > No; combining two countably infinite sets doesn't increase the cardinality of the result (because two is finite). Agreed. But pi = 3.14...pi... isn't combing 2 infinite sets. It 'combining' infinite amounts of infinite sets and not in a linear fashion either. You have to keep in mind the 2nd pi in the equation can be expanded to 3.14...pi... pi = 3.14...pi... when expanded is pi = 3.14...(3.14...pi...)... and you can keep expanding the inner pi forever. > The digits would still be countably infinite. How can you ever reach the first number after the inner pi in (pi = 3.14...pi...). Or put another way how do you get to the 4th '.'? You can't. This is a classical example of countably infinite and a continuum.
- jcranmer 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. A normal number would mean that every finite sequence of digits is contained within the number. It does not follow that the number contains every infinite sequence of digits. In general, something that holds for all finite x does not necessarily hold for infinite x as well.
  - orbillius 2y ago
    > A normal number would mean that every finite sequence of digits is contained within the number. Is that true? I don't see how that could be true. The sequence 0-9 repeated infinitely is, by definition, a normal number (in that the distribution of digits is uniform) ...and yet nowhere in that sequence does "321" appear ...or "654" ...or "99" There are an infinite number of combinations of digits that do not appear in that normal number I've just described. So, I don't think your statement is true.
    - moefh 2y ago
      It depends on your definition of "normal number". You seem to be using what wikipedia[1] calls "simply normal", which is that every digit appears with equal probability. What people usually call "normal number" is much stronger: a number is normal if, when you write it in any base b, every n-digit sequence appears with the same probability 1/b^n. [1] https://en.wikipedia.org/wiki/Normal_number https://en.wikipedia.org/wiki/Normal_number
    - jumhyn 2y ago
      IIRC the property ‘each single digit has the same density’ is the definition for a ‘simply normal number’ (in a given base), while ‘each finite string of a particular length has the same density as all other strings of that length’ is the definition for a ‘normal number’ (in a given base). And then ‘normal in all bases’ is sometimes called ‘absolutely normal’, or just ‘normal’ without reference to a base.
    - thaumasiotes 2y ago
      > I don't see how that could be true. The sequence 0-9 repeated infinitely is, by definition, a normal number (in that the distribution of digits is uniform) Well, your first problem is that you don't know the definition of a normal number. Your second problem is that this statement is clearly false. Here's Wolfram Alpha: > A normal number is an irrational number for which any finite pattern of numbers occurs with the expected limiting frequency in the expansion in a given base (or all bases). For example, for a normal decimal number, each digit 0-9 would be expected to occur 1/10 of the time, each pair of digits 00-99 would be expected to occur 1/100 of the time, etc. A number that is normal in base-b is often called b-normal. Your "counterexample" is not a normal number in any sense, most obviously because it isn't irrational, but only slightly less obviously because, as you note yourself, the sequences "321", "654", and "99" do not ever appear.

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.
  - linearrust 2y ago
    > No; combining two countably infinite sets doesn't increase the cardinality of the result (because two is finite). Agreed. But pi = 3.14...pi... isn't combing 2 infinite sets. It 'combining' infinite amounts of infinite sets and not in a linear fashion either. You have to keep in mind the 2nd pi in the equation can be expanded to 3.14...pi... pi = 3.14...pi... when expanded is pi = 3.14...(3.14...pi...)... and you can keep expanding the inner pi forever. > The digits would still be countably infinite. How can you ever reach the first number after the inner pi in (pi = 3.14...pi...). Or put another way how do you get to the 4th '.'? You can't. This is a classical example of countably infinite and a continuum.
jcranmer 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. A normal number would mean that every finite sequence of digits is contained within the number. It does not follow that the number contains every infinite sequence of digits. In general, something that holds for all finite x does not necessarily hold for infinite x as well.
- orbillius 2y ago
  > A normal number would mean that every finite sequence of digits is contained within the number. Is that true? I don't see how that could be true. The sequence 0-9 repeated infinitely is, by definition, a normal number (in that the distribution of digits is uniform) ...and yet nowhere in that sequence does "321" appear ...or "654" ...or "99" There are an infinite number of combinations of digits that do not appear in that normal number I've just described. So, I don't think your statement is true.
  - moefh 2y ago
    It depends on your definition of "normal number". You seem to be using what wikipedia[1] calls "simply normal", which is that every digit appears with equal probability. What people usually call "normal number" is much stronger: a number is normal if, when you write it in any base b, every n-digit sequence appears with the same probability 1/b^n. [1] https://en.wikipedia.org/wiki/Normal_number https://en.wikipedia.org/wiki/Normal_number
  - jumhyn 2y ago
    IIRC the property ‘each single digit has the same density’ is the definition for a ‘simply normal number’ (in a given base), while ‘each finite string of a particular length has the same density as all other strings of that length’ is the definition for a ‘normal number’ (in a given base). And then ‘normal in all bases’ is sometimes called ‘absolutely normal’, or just ‘normal’ without reference to a base.
  - thaumasiotes 2y ago
    > I don't see how that could be true. The sequence 0-9 repeated infinitely is, by definition, a normal number (in that the distribution of digits is uniform) Well, your first problem is that you don't know the definition of a normal number. Your second problem is that this statement is clearly false. Here's Wolfram Alpha: > A normal number is an irrational number for which any finite pattern of numbers occurs with the expected limiting frequency in the expansion in a given base (or all bases). For example, for a normal decimal number, each digit 0-9 would be expected to occur 1/10 of the time, each pair of digits 00-99 would be expected to occur 1/100 of the time, etc. A number that is normal in base-b is often called b-normal. Your "counterexample" is not a normal number in any sense, most obviously because it isn't irrational, but only slightly less obviously because, as you note yourself, the sequences "321", "654", and "99" do not ever appear.