Question

(1) Does there exist a sequence of 10100 consecutive composite numbers? (2) Let Pn be the nth prime number. Show that pn < 22".

Answer 1

2. Hint
The idea is that you use induction, and that you expand on Euclid's argument that there are infinitely many primes to show that pn is bounded above by 2^2n.

10loo composite (1+2)! + (N+2) 13 multi bication @ yes, 3 a sequence of composite numbesy, let is possible to have N. consecutive numbers. The serie (N+2)! +2. (N+2)+3. (N+2)!+4--- it is to Prove it easy Because N! OF All numbesy Fyom 1 to you can see, (N+2)! +a (N+2)! + a will give you a common factan (N+2)! + 3(N+2)! +3 will give you common factori في as showing a þrti . (N+2)! + (N+2).(N+2)! + (N+2) will give u LN+2) as common factor N consecutive combositive number finite no. of prime. suppose there only are say Pro P2 bn let N = Prope mean have. OF the perime which Bir pn divided N But As we know that, n>1 Every natustal number hos At least one Prime divisor be ditter N nay a prime Factor p. which from þrope po = Protis hopz Prtl Now. To show that, By Inductive Hyphothesie Pikaal pzca 27 Poille galt '+ 92? + -tgan g? +22+ tana 2 n+1 Hence Pntia com" By induction an t na i must pnsaan pocoon hey ) ८२ ht/ en <a