Wikipedia says that the [intermediate value theorem](https://en.wikipedia.org/wiki/Intermediate_value_theorem) “depends on (and is actually equivalent to) the completeness of the real numbers.” It then offers a simple counterexample to the analogous proposition on ℚ and a proof of the theorem in terms of the completeness property.


Does an analogous result hold for the computable reals (perhaps assuming that the function in question is computable)? If not, is there a nice counterexample?



Intermediate value theorem on computable reals
Let me assume that you are speaking about computable reals and functions in the sense of
[computable analysis](https://en.wikipedia.org/wiki/Computable_analysis), which is one of the most successful approaches to the topic. (One must be careful, since there are several incompatible notions of computability on the reals.)


In computable analysis, the [computable real numbers](https://en.wikipedia.org/wiki/Computable_real_number) are those that can be computed to within any desired precision by a finite, terminating algorithm (e.g. using Turing machines). One should imagine receiving rational approximations to the given real. In this subject, functions on the reals are said to be computable, if there is an algorithm that can compute, for any desired degree of accuracy, the value of the function, for any algorithm that produces approximations to the input with sufficient accuracy. That is, if we want to know $f(x)$ to within $\epsilon$, then the algorithm is allowed to ask for $x$ to within any $\delta$ it cares to.


The **Computable Intermediate Value Theorem** would be the assertion that if $f$ is a computable continuous function and $f(a)\lt c\lt f(b)$ for computable reals $a$, $b$, $c$, then there is a computable real $d$ with $f(d)=c$.


The book [Computable analysis: an introduction](https://books.google.com/books?id=OPolVWVFDJYC&dq=computable+analysis&printsec=frontcover&source=bl&ots=_QlUjkXhSy&sig=5m7E0aXQ-CVl8HcF77tzW2AW5Pk&hl=en&ei=Eh8pS93oN4Oj8AbDoIW1DQ&sa=X&oi=book_result&ct=result&resnum=10&ved=0CEIQ6AEwCQ#v=onepage&q=&f=false) by Klaus Weihrauch discusses exactly this question in Example 6.3.6.


The basic situation is as follows. The answer is **Yes**. If f happens to be increasing, then the usual bisection proof of existence turns out to be effective. For other $f$, however, one can use a trisection proof. Theorem 6.3.8 says that if $f$ is computable and $f(x)\cdot f(z)\lt 0$, then f has a computable zero. This implies the Computable Intermediate Value theorem above.


In contrast, the same theorem also says that there is a non-negative computable continuous function f on $[0,1]$, such that the sets of zeros of $f$ has Lebesgue measure greater than $\frac{1}{2}$, but $f$ has no computable zero.


In summary, if the function crosses the line, you can compute a crossing point, but if it stays on one side, then you might not be able to compute a kissing point, even if it is kissing on a large measure set.


