Can non-units be multiplied together to form units? An irreducible element of a ring is a non-unit such that it cannot be written as a product of two non-units. 
A UFD is a domain such that each non-unit $x\in R \backslash \{0\}$ can be written as a product of irreducible elements.
So if we take an irreducible element $r$ in a UFD, then it's not a unit and so admits a unique factorization into irreducible elements $r_1\dots r_n$ (which are also non-units). As it's irreducible it cannot be written as a product of two non-units so if $r = ab$ then $a$ or $b$ is a unit.
Does this mean that irreducible elements can be multiplied together to form units? And therefore non-units can be multiplied together to form units? 
 A: Hint $ $ Units = divisors of $1,$ so divisors of units are units $\rm\:a\:|\:b\:|\:1\:\Rightarrow\:a\:|\:1\:$ by transitivity of $ $ '|'.
Thus the unit group $\rm\,U\,$ is a saturated monoid, i.e $\rm\:ab\in U\iff a,b\in U.$
Hence nonunit $\rm\:\!a\:\Rightarrow\: ab\:\!$ nonunit, another example of a complementary view of a subgroup.
A: I don't understand your "Does that mean..."
In a UFD, the factorization into irreducibles of an irreducible is just itself. Just like, in $\mathbb{Z}$, the way to express $2$ as a product of primes is by writing $2=2$. 
In a commutative ring, any divisor of a unit is itself a unit:
If $x$ divides $u$ and $u$ is a unit, then there exists $y$ such that $xy=u$, and there exists $v$ such that $uv=1$. Then $x(yv) = (xy)v = uv = 1$, so $x$ is a unit.
In particular, in a commutative ring, a product that contains at least one nonunit factor is a nonunit; and a product of units is a unit (easy to verify).
In a noncommutative ring, however, it is possible for a non-unit to divide a unit (on one side). Recall that in a noncommutative ring, a unit is an element $u$ that has a two-sided inverse, that is an element $v$ such that $uv=vu=1$. An element may have a one-sided inverse and not be a unit.
Here's a standard example:
Let $V$ be the vector space of all real sequences, with pointwise addition. Let $R$ be the ring of all linear transformation $V\to V$, with multiplication being composition of functions. Then we can let $\lambda\colon V\to V$ be the left-shift operator that maps the sequence $(a_0,a_1,a_2,\ldots)$ to $(a_1,a_2,a_3,\ldots)$, and $\rho\colon V\to V$ be the right-shift operator that maps the sequence $(a_0,a_1,a_2,\ldots,)$ to $(0,a_0,a_1,a_2,\ldots)$.
Then $\lambda\rho=1$, so $\lambda$ is a left divisor of a unit, and $\rho$ is a right divisor of a unit. However, neither $\lambda$ nor $\rho$ are units: a unit in $R$ must be a bijective linear transformation on $V$, but $\lambda$ is not one-to-one, and $\rho$ is not onto, so neither element is a unit.
A: In a factorization, note that $n$ can be 1 (or zero!).
In general if $ab$ is a unit, then there exists $c$ so that $(ab)c=1$. Now, what can you say about $a$?
A: The product of two non-zero elements is a unit iff the elements themselves are units.
If $a$ and $b$ are irreducible this means that $a$ and $b$ are non-units, and hence $ab$ is non-unit; so the product of two or more irreducibles cannot be a unit.
