The decimal representation of any irrational number (e.g. pi, e) contains the decimal representation of every natural number somewhere. (Proof by diagonalization.) What you say here isn't quite true. The number with decimal rep 0.10100100000010000000000000000000000001.... where the number of zero's is going 1!, 2!, 3!, 4!, ... is transcendatal, and hence irrational, but clearly doesn't contain
the decimal representation of every natural number. i'm sure the above fact is believed about e, pi & other such ``important transcendentals'' - i can't recall if there is a proof or how it goes. diagonalization is used to prove that there are uncountably many irrationals. if you want to argue the ludicrosity of trying to ban certain numbers, just consider the function f(n) = n + 1. Iterating this function yields all natural numbers, so the increment operation should clearly be banned. I'm not sure how much programming you can do without increment. - robbie -- ---------------------------------------------------------------------- robbie gates | it's not a religion, it's just a technique. apprentice algebraist | it's just a way of making you speak. pgp key available | - "destination", the church.