Irreducible in $2{\bf N}$ divides product of terms doesn't imply it divides a term Lemma $3$. if $p$ is prime and $p$ divides $bc$, then $p$ divides $b$ or $p$ divides $c$.
proof: we use corollary $8$ an application of Bezout's identity, namely: if $a$ divides $bc$ and $(a,b)=1$, then a divides $c$.
Let $2 \mathbb{N}$ denote the even integers > 0. Say that a number $a$ in  $2\mathbb{N}$ is irreducible if there are no numbers $b, c$ in $2 \mathbb{N}$ so that $a = bc$
question show that the anologue of lemma 3 fails in $2 \mathbb{N}$
this was 4 parts question. other 3 parts i understood and was able to finish them. but last part seem bit hard for me.
 A: First make sure that you understand what’s being asked. The analogue of Lemma $3$ is:

if $a\mid bc$, and $a$ is irreducible, then $a\mid b$ or $a\mid b$.

You’re asked to show that this is false in $2\Bbb N$. To show this, you must find $a,b,c\in 2\Bbb N$ such that 


*

*$a$ is irreducible;  

*$a\mid bc$;  

*$a\nmid b$; and  

*$a\nmid c$.


So what elements of $2\Bbb N$ are irreducible? It might be easier to see what elements of $2\Bbb N$ are not irreducible. Suppose that $a\in 2\Bbb N$ is not irreducible; then $a=bc$ for some $b,c\in 2\Bbb N$. Since $b$ and $c$ are in $2\Bbb N$, there are $m,n\in\Bbb N$ such that $b=2m$ and $c=2n$, so $a=(2m)(2n)=4mn$. What members of $2\Bbb N$ are not multiples of $4$?
Now, under what circumstances can we have $a\nmid b$? If $a=2k$, and $b=2m$, then $a\mid b$ if and only if $k\mid m$, so we need to arrange matters so that $k\nmid m$. Similarly, if $c=2n$, we need to arrange matters so that $k\nmid n$. But on the other hand we do want to have $a\mid ab$, i.e., $2k\mid 4mn$; that’ll certainly be the case if $k\mid mn$. Can you find a way to choose $k,m$, and $n$ so that $k\mid mn$, $k\nmid m$, and $k\nmid n$?
A: Suppose that our universe consists of even numbers only (that's the $2\mathbb{N}$ of the post).
Call a number in our universe prime if it cannot be expressed as the product of two numbers in our universe.
So for example $30$ is in this sense prime, since it cannot be expressed as the product of two numbers in our universe. But $180$ is not prime, for $180=(10)(18)$, and $10$ and $18$ are in our universe.
Now we show that the analogue of Lemma 3 does not hold in our universe. Let $p=30$, $b=18$ and $c=10$. Then $30$ divides $bc$, since $180=(30)(6)$ and $6$ is in our universe, but $30$ does not divide $18$ or $10$.
Remark: This is meant to show that some of the basic properties of the integers are perhaps somewhat less obvious than they appear to be.  Later, you may see more sophisticated examples of important number-theoretic "universes" where whether Lemma 3 holds or fails is of great number-theretic significance.
One of the first examples you will meet is the Gaussian Integers, which are complex numbers of the form $a+bi$, where $a$ and $b$ are ordinary integers. The analogue of Lemma 3 holds for the Gaussian integers. This is a very useful fact. From it we can show that an analogue of the Unique Factorization Theorem holds in the universe of Gaussian integers.
However, if we look at the universe of numbers of the form $a+b\sqrt{-7}$, where $a$ and $b$ are integers, it turns out that the analogue of Lemma 3 fails, so in that sense the behaviour is like the one we saw for $2\mathbb{N}$. 
