this is my first question on this site.
I recently, study on NP. I have some confusion about this Topic, and want to propose my inference and some one verify me.
I) each NP problem can be solved in Exponential Time.
II) if P=NP then NP=NP-Complete.
III) The following problem is in NP: given a natural number, determine whether it is the product of two prime factors.
IV) if problem X can reduce to a known NP-Hard problem, then X must be NP-HARD.
anyone can verify my inference and learn me?
I) each NP problem can be solved in Exponential Time.
Correct. This is covered by an answer to our reference question (search for the heading "Brute-Force/Exhaustive-Search Algorithms for NP and NP$\,\subseteq\,$ExpTime").
II) if P=NP then NP=NP-Complete.
Aaaaaaaalmost true. But false. The problems $\emptyset$ and $\Sigma^*$ are the only two that would not be NP-complete if P$\,=\,$NP. These two are the problem where the answer is always "no" and the problem where the answer is always "yes". These two problems can't be NP-complete because a reduction from $X$ to $Y$ is required by definition to map "yes" instances of $X$ to "yes" instances of $Y$, and map "no" instances to "no" instances. But $\emptyset$ has no "yes" instances and $\Sigma^*$ has no "no" instances so you can't define a reduction from any other problem to one of those.
III) The following problem is in NP: given a natural number, determine whether it is the product of two prime factors.
True. This can be shown using the certificate definition of NP: a problem is in NP if there is a polynomial $p$ such that every "yes instance" $x$ of the problem has a certificate $y$ with $|y|\leq p(|x|)$ and there's a deterministic polynomial-time algorithm that, given an instance and a certificate, checks that the instance really is a "yes" instance.
In this case, the certificate is the two prime numbers. Given $x$ and prime numbers $y_1$ and $y_2$, we can check in polynomial time that $y_1$ and $y_2$ really are prime (using the AKS algorithm) and that $x$ really is $y_1y_2$.
IV) if problem X can reduce to a known NP-Hard problem, then X must be NP-HARD.
False. Informally, "$X$ reduces to $Y$" means "If I can solve $Y$, then I can solve $X$." However, if $Y$ is hard and solving $Y$ lets you solve $X$, that doesn't necessarily mean that solving $X$ was hard: going via $Y$ might have been a really dumb move. (Practical example: suppose you're hungry and have no food. If you had a Ferrari, you could sell the Ferrari and use the money to buy food. This is a reduction from the problem of getting food to the problem of getting a Ferrari. But that doesn't mean that getting food is as hard as getting a Ferrari.)
If you define exponential time as $O(2^{n^k})$ for some $k$, then (i) is correct.
(ii) is almost correct – you have to exclude trivial problems (those in which all instances are yes instances or all instances are no instances).
What you describe (iii) is not quite a decision problem, so as stated it is not strictly true. The language consisting of all tuples $(n,k,i,b)$ such that the $k$th smallest prime factor of $n$ has $i$th bit equal to $b$ is in NP.
(iv) is incorrect – the empty language, for example, reduces to all NP-hard problems, but is not NP-hard itself. If some NP-hard problem Q reduces to a problem X then X is NP-hard: indeed any problem in NP reduces to Q (by definition of NP-hard), and through your reduction to X, and so by definition X is NP-hard.
