[linearrust]

> 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]

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]

> 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]

> 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]

> 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.