proving result from monoid theory An element $x$ of a monoid $M$ is called an atom if $x \neq 1$ and $x=ab$, for some $a,b \in M$, implies that $x=a$ or $x=b$. Furthermore, $M$ is said to be atomic if every element $a \in M$ can be written as a product of atoms. If the number of factors is bounded, then $M$ is called bounded atomic. 
Given two elements $a,b \in M$, one defines a relation $\leq$ on $M$ as follows: $a \leq b$ if and only if there exists an $x \in M$ such that $b=xa$. It is clear that if $M$ is right cancellative and has trivial unit group $M^{\times}$, then $\leq$ is a partial order on $M$.
$M$ is called noetherian if every ascending chain $a_1 \leq a_2 \leq ...$  becomes stationary, that is, there exists some $n \in \mathbb{N}$ such that $a_{n+1} \leq a_n$.
Can someone help me with proving the following result?
$\textbf{Proposition:}$ Every bounded atomic monoid $M$ is noetherian. Conversely, a right and left cancellative noetherian monoid $M$ with trivial unit group is atomic.
 A: Neither of your propositions are true so far as I understand the definitions involved.
For the first one note that in the definition of a bounded atomic monoid the bound is a local one, not a global one.  That is, it is not the case that factorizations of all elements are less than some bound, instead for each element there is a bound on the lengths of factorizations of that element (see the bottom of page 3 here).
Now $\mathbb N$ under addition is bounded atomic because $1$ is the only atom and every element has a unique factorization, but $1 \leq 2 \leq \cdots$ doesn't become stationary.  I couldn't find a definition of a Noetherian monoid, but I suspect your definition is backwards.  Instead of no infinite ascending chains it should be no infinite descending chains; at least that's what I'd expect coming from ring theory.
If we take "no infinite descending chains" as the definition of Noetherian then the statement is quite easy to prove.  Assume, by way of contradiction, that $M$ is bounded atomic and non-Noetherian.  Then there is an infinite descending chain $\cdots \leq a_2 \leq a_1$ that does not stabilize.  Write $a_i = a_{i + 1}x_{i + 1}$.  As $M$ is bounded atomic there is some $N$ such that no factorization of $a_1$ has more than $N$ factors.  But
$$a_1 = a_2x_2 = a_3x_3x_2 = \cdots = a_{N + 1}x_{N + 1}x_N\cdots x_3x_2$$
expresses $a_1$ as a product of $N + 1$ elements.  If we exand each element in that product into a product of atoms then we've exressed $a_1$ as a product of at least $N + 1$ atoms, a contradiction.
For a counterexample to your converse take $\mathbb R^{\geq 0}$ to be the non-negative reals.  This is cancellative with trivial unit group, but it's not atomic because positive integers are not units and $x$ positive can always be written as $\frac12x + \frac12x$.
For the converse did you mean to assume that $M$ is left and right-cancellative and Noetherian?  In that case I do believe you can prove that $M$ is atomic, though it's getting late for me so I'll let you fill in the proof for that one.
