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I've seen that there is a problem in the definition of the functions if the domain is the reals because there are real numbers that does not have a value. So does it make sense to say that the only way for a function to be really defined in Intuitionism is if it's domain is the natural numbers?

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    Given a domain $A$, surely $f:A\rightarrow A$ given by $f(x)=x$ for all $x\in A$ is well defined intuitionistically? – 2017-01-06
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    Could we get you to check the spelling and do something about "Intuitism", "defenition", "deffined" and "if's" ? – 2017-01-06
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    @Hans: Could we get you to lighten up? – 2017-01-06
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    Sorry, it's late, changed everything as fast as possible. – 2017-01-06
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    Not an expert, but I think nice functions from the intuistionistic reals to the intuitionistic reals are well defined. Arbitrary functions, I don't think so – 2017-01-06

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There is a vast literature on intuitionistic and constructive approaches to the foundations of analysis. These approaches start by viewing a constructive real number $x$ as a process that produces (hopefully increasingly better) approximations to $x$. Bishop's work on constructive analysis and work by Weihrauch and others on computable analysis are good places to start in the literature on this subject.

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    So the answer is *NO*? – 2017-01-07
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    Constructive mathematics can certainly deal with functions whose domains are much more general than the natural numbers. – 2017-01-07
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    That's a *YES*, then? – 2017-01-07
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    @Thumbnail: what do you think? – 2017-01-07
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The various approaches to constructive mathematics all introduce the real numbers in some form. According to Brouwer every real number is represented by a so-called choice sequence. A choice sequence is a function from the set of natural numbers to a set of finite mathematical objects (which may be natural numbers) determined by free will. One can then speak of functions over the real numbers, but their definition must rely on choice sequences.

Moreover, the definition of continuity for functions over the reals is completely different from that used in classical mathematics. One consequence is that certain functions are not acceptable (as in being ill-defined). For instance,

$$f(x) = \begin{cases} 0 & \text{if $x$ is rational} \\ 1 & \text{if $x$ is irrational} \end{cases}$$

is not a function, since the property of being a rational number is not decidable. Another consequence is that every acceptable function over the reals is continuous.

See https://plato.stanford.edu/entries/intuitionism/ for a good introduction.