Prove that every element of E is equal to a product of primas. I initially had not problem understanding what it wanted from me. However, earlier in the book it said 

Let E be the set of all positive even integers. We call a number e in E "prima" if e cannot be expressed as a product of two other members of E.

Is this not contradicting what it is asking me to prove? I used 6 as an example as 6=2*3, while 2 is a prima, 3 is not as it is not in the set of E. I emailed my instructor and he messaged me this:

Be careful about how things are defined.
E is the set of all positive even integers, and there is a subset of E that consist of the prima integers.
For instance, both 4 and 6 are in E, but 6 is prima while 4 is not.
You want to show that every element of E (that is, every positive even integer) is a product of primas.
4=2x2, and 2 is prima.  6 is prima so it is a product of one prima (namely, itself).
It is analogous to noting that even though 10 is not prime, it is product of two primes (2x5) and 7 is a product of one prime (namely itself).

This did not provide any clarification to me. Any help is greatly appreciated. Thank you.
 A: $E$ being a set of all positive even numbers means each element has at least one factor of $2$. Thus, the product of $2$ elements of $E$ would have at least $2$ factors of $2$ and, as such, any element with only $1$ factor of $2$ cannot be a product of $2$ elements. By definition, this means these numbers are all "prima", including $2$, $6 = 2 \times 3$, $10 = 2 \times 5$, etc. These elements are all products of primas, namely just the $1$ prima of themselves.
Each $e \in E$, where $e$ has $2$ or more factors of $2$, can be written as $e = 2^n m$, where $n \ge 2$ and $m$ is an odd integer. In these cases, $e = 2 \times 2^{n-1} m$, where $2$ and $2^{n-1} m$ are each a member of $E$. Thus, this shows that, by the definition, these elements are not considered to be prima, e.g., $4 = 2 \times 2, 8 = 2 \times 4, 12 = 2 \times 6$, etc. Also, note that $e = \left(\Pi_{i=1}^{n-1} 2\right)\left(2m\right)$, where $2$ and $2m$ are each primas, so $e$ is a product of primas, e.g., $24 = 2 \times 2 \times 6$.
As this covers all of the cases, it shows that all elements of $E$ can be written as product of primas. However, note that unlike with primes, the set of prima factors is not unique for any integers where $2$ or more factors of $2$ and $m$ having $2$ or more prime factors. For example, $60 = 2 \times 30 = 10 \times 6$.
