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
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.