Bertrand's postulate claim : for all 5 &lt; n &#x2208;<!-- ∈ --> <mrow cla

Micah Haynes

Micah Haynes

Answered question

2022-05-09

Bertrand's postulate

claim :

for all 5 < n N there is at least two prime numbers different in the opening interval ( n , 2 n ).Use reinforcement of Bertrand's postulate in order to prove that :
for all 10 < n N Exists that has at least two Prime factors in the factorization of n ! which appear with a power of 1.
Example : n = 11 the primes 7,11 are such prime's meet the conditions.
But , for n = 10 it dosen't meet the conditions because only 7 appears with power 1 in the factorization of 10!.

Hint : Consider two cases, n even or n odd

Attempt:
we need to use the claim in order to Implement the solution

Case (1): n even
if n is even we can rewrite n = 2 t
if we use the claim we can get ( 2 t , 4 t )

Case (2): n odd
if n is odd we can rewrite n = 2 t + 1
if we use the claim we can get ( 2 t + 1 , 2 ( 2 t + 1 ) )

Answer & Explanation

Cristal Obrien

Cristal Obrien

Beginner2022-05-10Added 16 answers

You want to find primes that divide n ! exactly once, i.e., primes such that p n but 2 p > n. On the other hand, Bertrand will give us primes with t < p < 2 t for suitable t > 5. So in our situation, we want 2 t n (to make 2 p > n) and 2 t n + 1 (because p < n + 1 means the same as p n). This suggests taking t = n 2 . This will make t > 5 as soon as n > 10.

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