# Discriminating Function

Does any function $f(x)$ exists? such that

$$f(x) = \begin{cases} a , & \text{if x is rational} \\ b , & \text{if x is irrational} \end{cases}$$ where $a \neq b$

• It is not very clear... you mean $a$ for irrational values of $x$ and $b$ for rational values of $x$? If this is so, yes, it does exist. You just define it that way... Oct 10, 2014 at 16:08

Your definition is practically the same as Dirichlet Function.

It is a function, and can be defined analytically as:

$D(x)=\lim\limits_{m \to \infty }\lim\limits_{n \to \infty}cos^{2n}(m!\pi x)$

Yes..except that you probably mean $f(x) = b$ for rational values of $x$. And your description defines the function.

There's no polynomial expression that has this property, but there are many ways to define a function, and polynomials aren't the only one.

You might try to find a good set theory book and read about how functions are really defined.

One formal definition is that a function is a triple, $f = (D, C, R)$, where $D$ is a set called the domain, $C$ is a set called the codomain, and $R$ is a subset of $D \times C$ having two properties:

1. Every element $d$ of $D$ is the first element of some pair $(d, c) \in R$

2. Every element $d$ of $D$ is the first element of exactly ONE pair in $R$; more formally, if $(d, c_1) \in R$ and $(d, c_2) \in R$, then $c_1 = c_2$.

In your case, the function can be described by $$D = \mathbb R \\ C = \mathbb R \\ R = (\mathbb Q \times \{a\} ) \cup ((\mathbb R \setminus \mathbb Q) \times \{b\})$$ where $\mathbb Q$ denotes the rationals, and $(\mathbb R \setminus \mathbb Q)$ denotes the irrationals.