Digits of irrationals I've been studying floating point arithmetic and I've read somewhere that numbers with infinitely many decimal digits without recursion are irrational.
But since we can't know all the digits of such a number then how did we come to the conclusion that its digits have no recursion? Does it have anything to do with formulae used to compute the $n$-th digit of a number?
(This is a question simply out of curiosity.)
 A: The simpler (and ancient) way to know if a number $a$ is irrational is to explicitly show that it cannot be expressed as a quotient $\frac{n}{m}$ of two integers $n,m$.
But there are numbers, as the number $\pi+e$,  for which we don't know if they are rational or irrational.
A: Floating-point (as well as fixed-point) arithmetic is just unable to represent irrationals and most rationals.
Actually, the floating-point numbers are essentially integers in a finite range, with a movable point, and can't have more than 16 (significant) decimal digits.
A: 
I've read somewhere that numbers with infinitely many decimal digits
  without recursion are irrational.

This property can indeed be used to determine whether some numbers are irrational. For instance, the prime constant, defined by 
$$
\rho =\sum _{{p}}{\frac  {1}{2^{p}}}=\sum _{{n=1}}^{\infty }{\frac  {\chi _{{{\mathbb  {P}}}}(n)}{2^{n}}},
$$
where the sum goes over all primes $p$ and ${\displaystyle \chi _{\mathbb {P} }}$ is the characteristic function of the primes, has a $1$ as its $k^{\rm th}$ digit if $k$ is prime, and $0$ otherwise. It can be shown using an argument similar to yours that $\rho$ is irrational. Another example is the constant whose every $2^{k^{\rm th}}$ digit is $1$ and every other is $0$:
$$
0.0101000100000001\ldots,
$$
which is also irrational.

But since we can't know all the digits of such a number then how did
  we come to the conclusion that its digits have no recursion?

As you rightly observe, in other cases irrationality is proved by other means. For instance, the irrationality of $\sqrt 2$ is proved by assuming first that it is rational, i.e., equal to some $\frac ab$, where $a\neq b$ and $a,b>1$ are positive integers, and coming to a contradiction (for a nice presentation see here). A slightly more difficult proof using different techniques is used to show that the number denoted by $\zeta(3)$, where $\zeta$ is the Riemann zeta function, is irrational.
A: The proof is that if the number had a series of recurring digits, then it would be rational.  Suppose the number is a.bbbrrrr  where bbb is the leader digits and rrrr is a recuring patten.
Then this number is $a + \frac b{10^B} + \frac r{10^{B+R}-10^B}$, where b is the digit string bbb, and B is the number of places it occupies, and r, R are the recurring digits, and the period length.
Since this number is rational in every case, there is no recurring pattern that is not rational.
