Let n be a positive integer and let p1p2pt be the primes not

Let n be a positive integer and let p₁,p₂,...,p_t be the primes not exceeding n. Show

that p₁p₂ . . . p_t < 4ⁿ by filling in the following outline of a proof.

(a) Show the result is true for n = 1, 2, 3.

(c) Let k < p ?n where p is prime. Show that p / k byshowing p / n!, but p does not divide k!

and p does not divide (n?k)!. For the latter, note tha tn?k = k?1 or k+1.

positive integers.

case as well.

Solution

Solution :

a)

Let n be a positive integer and let p1,p2,...,pt be the primes not exceeding n.

First we need to consider a lowest bound value of n and for that purpose is can be stated as pt above.

so we want to show that p1 * p2 * p3...* pt < 4^pt

consider the natural log of both sides

log (p1 * p2...) = log(4^pt)

consider the left hand side

log(p1 * p2 * p3...) = log(p1) + log(p2) + log(p3) ...

Now consider the prime number theorem which can be restated such that :

log(p1) + log(p2) + log(p3) +...+ log(pt) ~ pt

Now the right hand side log(4^pt) = pt * log 4 = 1.386 * pt

Therefore this proves the original statement since pt < 1.386 * pt

For n = 1, we have p1 < 4^n

That is, 2 < 4^(1)

So, 2 < 4 => true

For n = 2, we have p1 * p2 < 4^n

2 * 3 < 4^2

So , 6 < 16 => true

Also for n = 3, we have p1 * p2 * p3 < 4^n

2 * 3 * 5 < 4^3

That is, 30 < 64 => true