Is this expression ${R^2 = R}$ correct? If relation R if reflexive and transitive so ${R^2 = R}$

So, let's take an example, we have A = {1, 2} and relation R = {(1, 1), (2, 2), (1, 2)} this relation reflexive and transitive. 
I don't understand what does it mean ${R^2}$, how to find ${R^2}$ if R is R = {(1, 1), (2, 2), (1, 2)} ??
But anyway as far as I see, it is not logical that ${R^2 = R}$
 A: $R^2$ in this context is in reference to $R\circ R$.
If you have two relations $S\subset A\times B$ and $T\subset B\times C$, the composition of the two relations $S\circ T\subset A\times C$ (very similar to the composition of two functions) is the set of pairs:
$$S\circ T = \{(a,c)\in A\times C~:~\exists b\in B~(a,b)\in S\wedge (b,c)\in T\}$$
If you were to have drawn yourself a directed graph where elements of $A$ appear in the first column, elements of $B$ appear in the second column, $C$ in the third column, and you draw an arrow from an element in one set to an element in the next iff the corresponding pair appears in the corresponding relation, you have $S\circ T$ is the resulting relation that occurs if you consider the paths of "length two."

For your example, you have $R = \{(a,a),(a,b),(b,b)\}$.  To make this clearer, I'll use colors.  If we have $\color{red}{R} = \{(\color{red}{a},a),(\color{red}{a},b),(\color{red}{b},b)\}$ and rewriting again with a different color, $\color{blue}{R}=\{(a,\color{blue}{a}),(a,\color{blue}{b}),(b,\color{blue}{b})\}$, we would have:
$$\color{red}{R}\circ\color{blue}{R} = \{(\color{red}{a},\color{blue}{a}),(\color{red}{a},\color{blue}{b}),(\color{red}{b},\color{blue}{b})\}$$
For instance, we have $(\color{red}{a},\color{blue}{a})$ since we have both $(\color{red}{a},a)$ in the relation on the left and $(a,\color{blue}{a})$ in the relation on the right and the element $a$ which connects the two.
On the other hand we have $(\color{red}{a},\color{blue}{b})$ because of either $(\color{red}{a},b)$ in the relation on the left and $(b,\color{blue}{b})$ in the relation on the right where $b$ was used as the connection between them, or alternatively we have $(\color{red}{a},a)$ from the relation on the left and $(a,\color{blue}{b})$ in the relation on the right where $a$ was used as the connection between them.

As for proving that if $R$ is reflexive and transitive that $R^2=R$, we should probably approach with double-inclusion.  Let us first show that $R^2\subseteq R$ and then show that $R\subseteq R^2$.
So, suppose that $(a,c)\in R^2$.  That means that we have some $b$ such that $(a,b)$ and $(b,c)$ are both elements of $R$.  Since $R$ is transitive this further implies that $(a,c)\in R$.  Therefore $R^2\subseteq R$ is true.
Now, suppose instead that $(a,b)\in R$ and we wish to show that $(a,b)\in R^2$.  How can we do that?
Hint:

 we haven't used all of our hypotheses yet.

Solution:

 Since $R$ is reflexive, $(a,a)\in R$.  Given that $(a,a)\in R$ and $(a,b)\in R$ it follows that $(a,b)\in R^2$, proving that $R\subseteq R^2$

