# what are the domain(s) of self inverse functions?

A self inverse function is a function $$f$$, such that $$y=f(x)$$, with the special property that $$ff(x)=x$$, or written another way, $$f(x) = f^-1(x)$$

Example:

Imagine there's a function $$f$$, such that $$y = 1/x$$

$$f^-1(x) = 1/x$$, which means that $$f(x) = f^-1(x)$$, therefore this specific function is said to be a self-inverse function.

Another example:

Let the function $$f$$ be such that $$y = 3-x$$

$$ff(x) = 3-x$$, { so, $$ff(x) = x$$ } therefore this specific function is a self-inverse function.

What is/are the domain(s) of these types of functions?

• The domain must equal the range. Other than that it's up to the definition of the involution. – dxiv Oct 9 '16 at 0:36
• @dxiv, No idea how I missed that, wow. Not everything that's simple is also obvious, I guess. Consider posting this as an answer and I'll accept it. – Buffer Over Read Oct 9 '16 at 0:44

Such self-inverse functions are called involutions. Since the function $f : X \to Y$ and its inverse $f^{-1} : Y \to X$ coincide for an involution, the domain and codomain must be the same $X = Y$. Moreover, since $f$ must be a bijection (in order to have an inverse), the range must equal the codomain. So in the end, the domain of an involution $f$ must equal its range $f(X) = X$.
• $f : \mathbb{R} \to \mathbb{R} \quad f(x) = x$
• $f : [-1,1] \cap \mathbb{Q} \to [-1,1] \cap \mathbb{Q} \quad f(x) = -x$
• $f : \mathbb{R} \setminus \{0\} \to \mathbb{R} \setminus \{0\} \quad f(x) = \frac{1}{x}$