As shown in the prior section,
the numbers $\,2, 3, 4,\, \ldots\,$ can be uniquely expressed as a product of
primes.
Just keep ‘breaking them down’ into smaller and smaller factors until only
prime numbers remain.
For example:

OR 

OR 

No matter how the number is ‘broken down’, you'll always get to the same place,
except for possibly different orderings of the factors.
In the example above, you always get $\,360 = 2\cdot 2\cdot 2\cdot 3\cdot 3\cdot 5\,$.
Three factors of $\,2\,$, two factors of $\,3\,$, and one factor of $\,5\,$.
This idea is formalized as:
THEOREM
the Prime Factorization Theorem
Every counting number greater than $\,1\,$ has a unique expression as a product of primes.

If we decided to call the number $\,1\,$ prime,
then we lose the uniqueness of representation
guaranteed in the Prime Factorization Theorem.
For example, here's what we'd have to contend with:
representation of number as a product of ‘primes’ 
factors of $\,2\,$  factors of $\,3\,$  factors of $\,1\,$ 
$6 = 2\cdot 3$  one factor of $\,2\,$  one factor of $\,3\,$  no factors of $\,1\,$ 
$6 = 2\cdot 3\cdot 1$  one factor of $\,2\,$  one factor of $\,3\,$  one factor of $\,1\,$ 
$6 = 2\cdot 3\cdot 1\cdot 1$  one factor of $\,2\,$  one factor of $\,3\,$  two factors of $\,1\,$ 
... and so on ... 
Given a counting number greater than $\,1\,$, if you can find any factor other than itself or $\,1\,$, then it's not prime.
For example, the number $\,217,695\,$ is definitely not prime. Since the last digit is $\,5\,$, it's divisible by $\,5\,$.
Or, the number $\,927,856\,$ is definitely not prime. It's even, so it's divisible by $\,2\,$.
However, factors aren't always so obvious.
For example, is the number $\,1{,}327\,$ prime?
If you take the following (thorough, but inefficient) approach, it will take you a LONG time to get the answer:
Let's start improving on this inefficiency.
For example, if a number is divisible by $\,10\,$ (which isn't prime), then it is also divisible by $\,2\,$ and $\,5\,$ (the prime factors of $\,10\,$).
Therefore:
Now we have far fewer questions:
For example, $\,4\cdot 11 = 44\,$, and $\,\sqrt{44}\approx 6.6\,$.
The first factor, $\,4\,$, is less than $\,\sqrt{44}\,$.
The second factor, $\,11\,$, is greater than $\,\sqrt{44}\,$.
A moment's reflection should convince you that this is true.
If two numbers are both less than $\,\sqrt{N}\,$, then their product is less than $\,N\,$.
If two numbers are both more than $\,\sqrt{N}\,$, then their product is more than $\,N\,$.
Therefore:
Now, back to our original question: is $\,1{,}327\,$ prime?
Note that $\,\sqrt{1327} \approx 36.4\,$.
There are only eleven primes less than $\,36.4\,$: $2$, $3$, $5$, $7$, $11$, $13$, $17$, $19$, $23$, $29$, and $31$.
It's pretty quick to check that none of these primes go into $\,1{,}327\,$ evenly.
Therefore, $\,1{,}327\,$ is prime!
There's a more convenient way to answer the question (but your teacher might not let you use it on a quiz or test)!
Zip up to WolframAlpha and type in: Is 1327 prime?
Voila!
CONCEPT QUESTIONS EXERCISE:
On this exercise, you will not key in your answer.However, you can check to see if your answer is correct. 
PROBLEM TYPES:
