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The following is an example from a textbook on logic and proof.

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I question the viability of choosing $n=(k+1)!+2$. First, if $k=1$, it would be meaningless to say $n+1$ is divisible by $3$. Second, with this choice, we should have $$n+k=(k+1)!+k+2$$ in the third line of the proof. Third, is it really okay to address the numbers $n$ through $n+k$ as $k$ consecutive numbers instead of $k+1$ consecutive numbers? Similarly, for the statement to be proved, is it wrong to rephrase the statement as "No matter how large $k$ is, some consecutive primes differ by at least $k+2$."? Lastly, if the choice for $n$ doesn't work, is there any way to prove this existence theorem?

Thank you.

Bill Dubuque
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Boar
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    If $k=1$ then you're just trying to find $1$ composite number, so you stop at the first step: $n=(1+1)!+2=4$ is divisible by $2$. – jjagmath Nov 20 '24 at 10:43

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Change the proof to "$n+k-1 = (k+1)! + k + 1$" and the proof will become valid.

rafilou2003
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  • Thank you, but does your choice make any difference? It's just a rearrangement of the book's choice. – Boar Nov 20 '24 at 11:24
  • @Boar Indeed we are not proving exactly what the textbook wanted to prove. If we were trying to prove that for any $k$, there exists $n$ such that $n,...,n+k$ are not prime, then we should have taken $n = (k+2)!+2$. $n$ is still divisible by $2$, $n+1$ divisible by $3$, up until $n+k = (k+2)! + k + 2$ is divisible by $k+2$. Thus none of those $k+1$ consecutive numbers is prime. – rafilou2003 Nov 20 '24 at 14:22
  • Please comment on (don't answer) questions boiling down to typos (esp. when they are dupes of FAQs). Answering them greatly increases community effort needed to process them. – Bill Dubuque Nov 20 '24 at 22:21
  • @rafilou2003 I got your point. Thank you. – Boar Nov 21 '24 at 13:03