Can a Sequence of Consecutive Positive Integers Not Contain Any Primes?

[kai89]

Could someone help me with this induction proof. I know its true.

given any integer m is greater than or equal to 2, is it possible to find a sequence of m-1 consecutive positive integers none of which is prime? explain

any help is greatly appreciated thanks

 

[ZioX]

Is 5! prime?

 

[HallsofIvy]

I'm sorry, Ziox, I really don't see what that has to do with the problem. Please enlighten me.

 

[drpizza]

How do you know this is true?

 

[kai89]

My teacher only gave us true proofs so we wouldn't be able to prove it wrong by counter example.

 

[ZioX]

I'm sorry, Ziox, I really don't see what that has to do with the problem. Please enlighten me.

5!+i has divisors 2,3,4,5 for i=2,...,5. Paired up respectively.

 

[HallsofIvy]

Oh, of course! That answers the whole question. I was focusing on the "5" and didn't think about doing the same thing for n in general. Very nice.

kai89, do you understand what ZioX is saying?

 

[ZioX]

Oh, of course! That answers the whole question. I was focusing on the "5" and didn't think about doing the same thing for n in general. Very nice.

kai89, do you understand what ZioX is saying?

You're right though, I was being pretty vague and probably would've only made sense if someone has seen it before. Should've said something about divisibility when adding.

 

[HallsofIvy]

kai89, since it has been a couple of days now, I will give detail on what ZioX hinted at: For any positive integer n, n! is obviously divisible by every integer up to and including n. Therefore, n!+ 2 is divisible by 2, n!+ 3 is divisible by 3, up to n!+ n is divisible by by n. You have n-1 consecutive integers that are not prime.

As I said before, very nice!

 

[ZioX]

Primes are so unpredictable.

;-p