Can we find a well ordering on an infinite set with no largest element? According to the well ordering theorem "Any set can be well ordered". Whenever we have a well ordering on a set, it is not difficult to construct a new well ordering with a largest element.
My question is:

For a given infinite set, can we find a well ordering on it such that there is no largest element? 

 A: If $A$ is infinite then $A$ can be put in bijection with $A \cup \mathbb{N}$. Now from the well ordering of $A$ you can ge a well ordering of $A \cup \mathbb{N}$ without a largest element ( in the obvious way ) and then back, a new well-ordering of $A$ without a largest element. 
A: In the usual development of ZFC, a cardinal is defined to the the
least ordinal with a given cardinality. Infinite cardinals are "limit
ordinals" --- as ordered sets they have no largest element.
A: Yes. In fact, there is a canonical way to transform a well-ordering into a well-ordering of the same set with no greatest element.
Suppose I have an infinite well-order $X$. Let $F_X$ be the set of elements of $X$ with finitely many things bigger than them - for instance, the greatest element of $X$ (if such an element exists) is in $F_X$, while (if $X$ is infinite) the least element of $X$ (which always exists, since $X$ is a well-ordering) is not in $F_X$.
It's not hard to show that $F_X$ is finite, since otherwise we could build an infinite descending chain in $X$ (exercise). Now let $I_X=X\setminus F_X$; this is a well-ordered set with no greatest element. 
If $X$ is infinite, $I_X$ and $X$ have the same cardinality. If you want an explicit isomorphism, we can "move $F_X$ to the back" - define a new ordering $\prec$ on $X$, given by $a\prec b$ iff


*

*$a, b\in I_X$ and $a<b$ in the original sense of $X$; or

*$a, b\in F_X$ and $a<b$ in the original sense of $X$; or

*$a\in F_X, b\in I_X$.
