Let there be > included in AxB as a binary relation. What does (x)>^2(y) mean? What is the meaning of an order relation raised to a power? My first tought was that >^2 = >x> which is a cartesian product of > included in AxB. I have a solution in a Linear Algebra book : (x)>^2(y) = x+2 >= y , wich is a bit confusing; I see that the digit 2 summed from nowhere has to do something with the exponent but I don't quite comprehend the logical flow.

If $R$ and $S$ are binary relations, then the composition relation $R\circ S$ is usually defined by $a\mathrel{R\circ S} c$ if and only if there is $b$ such that $a\mathrel{R}b$ and $b\mathrel{S}c$. In the special case where the relations are the (graphs) of functions $f$ and $g$, this produces the (graph) of the usual composition function $f\circ g$, since $z=(f\circ g)(x)\iff \exists y\ z=f(y)$ and $y=g(x)$. (But if one understands the graph with the variables in the order $(x,y)$, as is usual, then the composition relation technically is $g\circ f$.) In the case of an order $\lt$, what the relation $\lt^2$ would mean is ${\lt}\circ{\lt}$, which would be defined by $a\mathrel{\lt^2} c$ if and only if there is $b$ such that $a\lt b\lt c$. In a discrete order, such as the usual order $\lt$ on the integers $\mathbb{Z}$, this means that $r\mathrel{\lt^2} t$ if and only if there is $s$ with $r\lt s\lt t$, which is the same relation as $r+2\leq t$, which may explain the confusing $+2$ issue you mention. Similarly, in the integers we have $r\mathrel{\lt^n} t$ if and only if $r+n\leq t$. Meanwhile, in a dense order $\lt$, such as the order on the rationals, we have $\lt^2=\lt$, since $a\lt c\iff\exists b\ a\lt b\lt c$. Also, in any reflexive relation $\leq$, we have $\leq^2=\leq$, since $a\leq b\iff a\leq a\leq b$. 

