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.
> 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.
