Editing P versus NP problem (section)

====Example====
Let
:<math>\mathrm{COMPOSITE} = \left \{x\in\mathbb{N} \mid x=pq \text{ for integers } p, q > 1 \right \}</math>
:<math>R = \left \{(x,y)\in\mathbb{N} \times\mathbb{N} \mid 1<y \leq \sqrt x \text{ and } y \text{ divides } x \right \}.</math>
Whether a value of ''x'' is [[Composite number|composite]] is equivalent to of whether ''x'' is a member of COMPOSITE. It can be shown that COMPOSITE ∈ NP by verifying that it satisfies the above definition (if we identify natural numbers with their binary representations).

COMPOSITE also happens to be in P, a fact demonstrated by the invention of the [[AKS primality test]].<ref name="Agrawal">{{cite journal |first1=Manindra |last1=Agrawal |first2=Neeraj |last2=Kayal |first3=Nitin |last3=Saxena |url=http://www.cse.iitk.ac.in/users/manindra/algebra/primality_v6.pdf |archive-url=https://web.archive.org/web/20060926201057/http://www.cse.iitk.ac.in/users/manindra/algebra/primality_v6.pdf |archive-date=2006-09-26 |url-status=live |title=PRIMES is in P |journal=[[Annals of Mathematics]] |volume=160 |year=2004 |issue=2 |pages=781–793 |doi=10.4007/annals.2004.160.781 |jstor=3597229 |doi-access=free }}</ref>