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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?

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  • $\begingroup$ Are you sure that you have stated the problem correctly? As stated, it follows immediately from the fact that there are infinitely many primes. $\endgroup$ – user247327 Apr 25 at 0:30
  • $\begingroup$ @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. $\endgroup$ – Tazim Taz Apr 25 at 0:32

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