Non Standard Prime number

Problem : Prove that for any $$m\in^* \mathbb{N}$$ there exists $$n \in ^* \mathbb{N}$$ such that $$n\geq m$$ and $$n$$ is prime .

My Attempt :

If n is prime, we can write as : $$( \forall m \in ^*N)(m|n \implies m =1 \vee m =n).$$

The above formula holds in the suberstructure $$V(N)$$ of $$\mathbb{N}.$$ So by transfer, its counterpart $$( \forall n \in ^* \mathbb{N} - (0,1) )( \forall m \in ^*N)(m|n \implies m =1 \vee m =n) \Longleftrightarrow$$ x in the set of prime number ( non standard ) and $$n\ge m )$$ holds in the non-standard superstructure $$V(^*\mathbb{N})$$.

Is my work good?

• Are you sure that you have stated the problem correctly? As stated, it follows immediately from the fact that there are infinitely many primes. – user247327 Apr 25 at 0:30
• @user247327 Problem stated like this: A nonstandard number n∈* N is called prime if it has no (nonstandard integer) divisors beyond 1 and n . Prove that for any m∈* N there exists n∈* N such that n≥m and n is prime. – Tazim Taz Apr 25 at 0:32