Your proof attempts to demonstrate that for any $N > n$, there are at least $n$ consecutive integers that are not prime, starting from $N! + 2$. Let's go through your argument step by step to check its validity and clarity:

Restating the Proof:

1. You are considering a natural number $N$ such that $N > n$, where $n$ is some natural number.
2. For $N > 1$, you claim that $N!$ (the factorial of $N$) is even, which is correct because $N!$ includes a factor of 2 (since $N! = 1 \cdot 2 \cdot 3 \cdot \cdots \cdot N$).
3. Since $N!$ is even, $N! + 2$ is also even, making $N! + 2$ non-prime (unless it's 2, but here $N > 2$, so $N! + 2$ is not prime).
4. You then claim that $N! + 3$ is divisible by 3. This is correct because $N!$ contains the factor 3 (for $N \geq 3$), so $N!$ is divisible by 3, and therefore $N! + 3$ is divisible by 3 and not prime.
5. This reasoning is extended for $N! + k$, where $2 \leq k \leq N$. Since $N!$ contains the factor $k$ (for any $k \leq N$), $N! + k$ will be divisible by $k$, ensuring that $N! + k$ is not prime.
6. Finally, you conclude that since $N! + 2, N! + 3, \dots, N! + (n+1)$ are all composite (non-prime), you have found at least $n$ consecutive integers that are not prime.

Analysis:

• Correctness of Argument:

  • Your reasoning is correct. For each $k \in \{2, 3, \dots, N\}$, the term $N! + k$ is divisible by $k$, ensuring that $N! + k$ is composite for each $k$. Hence, you have indeed demonstrated that the numbers $N! + 2, N! + 3, \dots, N! + (n+1)$ are not prime.
  • The choice of $N!$ is effective because it ensures that each $N! + k$ contains the factor $k$, thereby making $N! + k$ divisible by $k$.

• Logical Flow:

  • The logic flows correctly from the assumption that $N!$ contains all factors up to $N$, leading to the conclusion that the integers $N! + 2$ to $N! + (n+1)$ are not prime.

• Conclusion:

  • Your proof is correct, and you have successfully demonstrated that for any $N > n$, there are at least $n$ consecutive integers that are not prime. This technique is a common approach to finding sequences of consecutive composite numbers.

Suggestions for Improvement:

1. You could clarify that the proof works for any $N > n$ by explicitly stating that since $N!$ includes all factors from 2 to $N$, every $N! + k$ for $k \leq N$ will necessarily be composite.
2. Emphasize that this method generates $n$ consecutive composite numbers starting from $N! + 2$.
3. Although implied, it might help to state clearly that since $N > n$, the set $\{2, 3, \dots, N\}$ covers $n$ distinct values, ensuring $n$ consecutive non-prime numbers.

Would you like more details or have any other questions?

Relative Questions:

1. How does this method ensure the sequence $N! + 2, N! + 3, ..., N! + (n+1)$ contains exactly $n$ consecutive non-prime numbers?
2. Can this proof be generalized for other starting points besides factorials?
3. Why is $N! + 2$ guaranteed to be even for all $N > 2$?
4. How does the concept of prime gaps relate to this proof?
5. What are the implications of this proof in the context of large prime numbers?

Tip:

Using factorials is a powerful tool in constructing sequences of consecutive composite numbers because $N!$ guarantees divisibility by all integers less than or equal to $N$.