# Thread: Prime Number on an interval

1. ## Prime Number on an interval

Can someone help me prove the following:

If n is an integer and n is greater than 2, then there exists a prime p such that n is less than p which is less than n!.

Thanks so much for any help.

2. Originally Posted by JimmyT
Can someone help me prove the following:

If n is an integer and n is greater than 2, then there exists a prime p such that n is less than p which is less than n!.

Thanks so much for any help.
Suppose that for some number n there is no prime between n and n!.

Now consider (n!-1), this is not divisible by any number 2<=k<=n, and so
is not divisible by any prime <=n. Hence either (n!-1) is prime or there is
some prime q, n<n<n!, which divides (n!-1). Either case is a contradiction
and so there must be a prime netween n and n!.

RonL

3. Originally Posted by JimmyT
Can someone help me prove the following:

If n is an integer and n is greater than 2, then there exists a prime p such that n is less than p which is less than n!.

Thanks so much for any help.
Use the Betrand Postulate (proven by Chebyshev).

Given any integer n>2 there exists a prime,
n<p<2n.

Then, since n!>2n we have that it too contains a prime.

4. Originally Posted by ThePerfectHacker
Use the Betrand Postulate (proven by Chebyshev).

Given any integer n>2 there exists a prime,
n<p<2n.

Then, since n!>2n we have that it too contains a prime.
Prove the weaker result from the stronger? Doesn't that suggest to
you that maybe they have not covered the stronger result?

RonL

5. Originally Posted by CaptainBlack
Doesn't that suggest to
you that maybe they have not covered the stronger result?

RonL
Are you saying that there is a flaw in my flawlwss proof written by my flawless hand?

It seems you are saying I should not use that since I did not proof it. But the result is well known.

6. Originally Posted by ThePerfectHacker
Are you saying that there is a flaw in my flawlwss proof written by my flawless hand?

It seems you are saying I should not use that since I did not proof it. But the result is well known.
No I am suggesting that you could prove the result in a manner that
the student might understand, and could reproduce in their own words.

RonL

7. Originally Posted by ThePerfectHacker
Use the Betrand Postulate (proven by Chebyshev).

Given any integer n>2 there exists a prime,
n<p<2n.

Then, since n!>2n we have that it too contains a prime.
and yet n can equal 2 and it would work

8. Originally Posted by Quick
and yet n can equal 2 and it would work
Yes it does.
I should have said n>1 but that is trivial.

9. Originally Posted by ThePerfectHacker
Yes it does.
I should have said n>1 but that is trivial.
Why would that be trivial?

10. Originally Posted by Quick
Why would that be trivial?
When a mathemation uses the phrase trivial it might refer to something which is easily seen to be true. Like that x(x^2+1)=0 it is trivial that x=0, but not that x=i. Also, many theorem do not work for some integers but work for sufficiently large integers. Like this theorem over here that n=1 does not work, a mathemation would say that is trivial. When you learn more math you will see that many theorem do not work for the simplest case, usually invlovling 1 or 0. Thus, when someone says that is does not work a mathemation might respond that what he said is trivial. Those are the two meanings of the word.