Why is reflexivity a feature of multi-step β-reduction, and not of β-reduction? I understand the need to distinguish between two algorithms, even when one is enclosed in another. Transitivity as a feature of multi-step reduction makes perfect sense, since we literally need multiple steps of $\beta$-reduction execute it. What about reflexivity?
If I had to answer this question by myself, I would use a mathematician's point of view, saying that it is convenient to not have reflexivity in the definition of $\beta$-reduction, in order to prove certain propositions without having to deal with the case of reflexivity.
What would be the programmer's point of view?
EDIT: I am referring to untyped $\lambda$-calculus, sorry for not mentioning it.
 A: From the "computer scientist's point of view", a single step $\beta$-reduction roughly corresponds to an elementary step of computation. Thus, reaching a normal form (i.e. a term in the $\lambda$-calculus where $\beta$-steps cannot apply) means to evaluate a program to its output, i.e. to compute the final result.
Suppose that $\beta$-reduction were defined so as to be reflexive. What would be a normal form with such a definition? It would be really clumsy to define the notion of normal form, because for every term $M$, we have $M \to_\beta M$, which means that every term $M$ is not normal. And saying that $M$ is normal if and only if $M \to_\beta N$ implies $M = N$ does not solve the problem at all, because according to this attempt of definition the term $(\lambda x.xx) \lambda x.xx$ would be normal but actually  it $\beta$-reduces to itself by performing a "real" $\beta$-step, not thanks to reflexivity.
So, if the definition of $\beta$-reduction included reflexivity, it would be really clumsy to define the notion of normal form, which is the crucial notion corresponding to the idea of result of a computation.
Moreover, if you are interested in counting the number of steps to reach the normal form (i.e. to estimate the evaluation length), it would be impossible to do that if $\beta$-reduction were reflexive. For instance, if $M \to_\beta N$ by performing a "real" $\beta$-step and $\to_\beta$ is reflexive by definition, then
\begin{align*}
M &\to_\beta N \\
M &\to_\beta N \to_\beta N \\
M &\to_\beta N \to_\beta N \to_\beta N \\
M &\to_\beta N \to_\beta N \to_\beta N \to_\beta \dots\\
\end{align*}
where $N \to_\beta N$ because $\to_\beta$ is reflexive.
Said differently, if $M \to_\beta N$ by performing a "real" $\beta$-step, then $M$ $\beta$-reduces to $N$ with an arbitrary number of $\beta$-steps. Any quantitative analysis would be simply impossbible.
