Let (A,<) be a strictly ordered set and b not in A. Define a relation $\displaystyle \prec$ in B=A$\displaystyle \cup${b} as follows:
x$\displaystyle \prec$y if and only if (x,y in A and x<y) or (x in A and y=b).
Show that $\displaystyle \prec$ is a strict ordering of B and $\displaystyle \prec \cap A^2$ =<. (Intuitively, $\displaystyle \prec$ keeps A ordered in the same way as < and makes b greaters than every element of A.)

Can anyone help with this? Thanks!