## Ordering relations question

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

Can anyone help with this? Thanks!