<Dd> R = ((x, y) ∈ N × N ∣ 1 <y ≤ x and y divides x). (\ displaystyle R = \ left \ ((x, y) \ in \ mathbb (N) \ times \ mathbb (N) \ mid 1 <y \ leq (\ sqrt (x)) (\ text (and)) y (\ text (divides)) x \ right \).) </Dd> <P> Clearly, the question of whether a given x is a composite is equivalent to the question 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). </P> <P> COMPOSITE also happens to be in P . </P> <P> There are many equivalent ways of describing NP - completeness . </P>

If the solution to a problem is easy to check for correctness is the problem easy to solve