Ive been trying to find a proof of the fact that for each na

 I\'ve been trying to find a proof of the fact that for each natural number N, there exists a prime number Q such that N is less than Q and Q is less than or equal to 1 + N! 

Solution

we will try to proof here by contradiction

Assume that there is no prime number in n and1+n !

let b = 2*3*..............p

p is a prime number and b is mult. of prime number

here b < n so b+1< n!+1

we have assumed that b+1 cannot be a prime number So there exist a prime number which divides b+1 (note this nunber has to be smaller than b+1 and also be one of the primes)

this means that q divides b, b+1 its difference 1 , but this cannot be the case since q is prime number , hene it is false .