As I understand it (layman alert), the definition of computable numbers is binary: either a number is or is not computable.
Is it meaningful to imagine a function telling how computable (or accessible) a number is ? Maybe such a function could be related to number of steps that are needed to compute it.
If so, which number is the least accessible, though computable ?
On
There is also the useful notion of left-computability and right-computability. i A real number $x$ is left-computable if for any computable real $a$ you can answer YES within finite time to the question : is $x$ greater than $a$ ?
A real number $x$ is right-computable if $-x$ is left-computable.
A real number $x$ is computable if and only if it is left and right-computable.
A real number $x$ is left-computable if and only if an algorithm can give you a non-decreasing sequence of rational numbers which tends to $x$.
For example, given a polynomial with real computable coefficients. Then its largest root is not a computable function of the coefficients. However, it is right-computable.
ADDENDUM
Following the friendly controversy with Carl Mummert, I realized that I was unclear.
As an example, consider the polynomial $f(x)=(x+1)x^2-\epsilon$, with $\epsilon$ a small computable real number. The largest root of $f$ is near $\sqrt\epsilon$ if $\epsilon\geq 0$, and near $-1$ if not.
Given the information that $\epsilon=0$ or not, this root is computable. However, as a function of $\epsilon$, it is not. If it were, it would imply that we are able to decide if $\epsilon\geq 0$ or not. But we can't : if you make run the Turing machine which approximate $\epsilon$ and it gives you only zero, you will never be able to state after a finite number of steps that $\epsilon$ is positive, or negative, or null.
This problem is quite general, and in fact, the Grzegorczyk states that a computable function from real numbers to real numbers is continuous. And obviously the function of $\epsilon$ "largest root of f" is not.
However, this largest root is right-computable : the algorithm answer zero as long as it as not enough precision to disprove $\epsilon=0$, and then it is able chose the correct location. But if the algorithm answers zero, you can never be sure that there is indeed a root near zero.
Carl Mummert, please correct me if I'm wrong.
The notion you seem to want is Kolmogorov complexity, which informally speaking measures the size of the smallest program needed to compute an object.