I don't understand Why the number of reflexive relations is $2^{n^2-n}$ instead of $2^n$
I thought reflexive relations are elements in form of (a,a). So I understand that there are $2^{n^2}$ Relations in total. But the reflexive ones should only be on the main diagonal right? Why are they excluding those and including only the ones that are not reflexive? The textbook didn't explain this
If $\{x_1, x_2, \ldots, x_n\}$ is your $n$-element set, then things get really simple when you represent a relation $R$ with a matrix $(a_{ij})$, in the sense that $(x_i, x_j) \in R$ if $a_{ij} = 1$, and $(x_i, x_j) \notin R$ if $a_{ij} = 0$.
A relation $R$ is reflexive if and only if $a_{ii} = 1$ for all $i = 1, \ldots, n$. Other $n^2 - n$ elements can be either $0$ or $1$.
$$\begin{array}{ccc} & & \begin{array}{ccc} \, x_1 & \!\! x_2 & \!\! x_3 & \!\!\cdots & x_n \end{array}\\ & \begin{array}{c} x_1\\ x_2\\ x_3 \\ \vdots \\ x_n\end{array} & \left(\begin{array}{ccccc} 1 & * & * & \cdots & *\\ * & 1 & * & \cdots & *\\ * & * & 1 & \cdots & *\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ * & * & * & \cdots & 1\\ \end{array}\right) \end{array}$$
Therefore, we make $n^2 - n$ binary choices, which amounts to $2^{n^2 - n}$ different matrices, i.e. reflexive relations.