# Bertrand's postulate Strong proof

Bertrand's postulate

claim :

for all $$5< n \in \mathbb{N}$$ there is at least two prime numbers different in the opening interval $$(n,2n)$$.

Use reinforcement of Bertrand's postulate in order to prove that :

for all $$10 Exists that has at least two Prime factors in the factorization of $$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 $$n = 2t$$

if we use the claim we can get $$(2t,4t)$$

Case (2): $$n$$ odd

if $$n$$ is odd we can rewrite $$n = 2t + 1$$

if we use the claim we can get $$(2t+1,2(2t+1))$$

You want to find primes that divide $$n!$$ exactly once, i.e., primes such that $$p\le n$$ but $$2p>n$$. On the other hand, Bertrand will give us primes with $$t for suitable $$t>5$$. So in our situation, we want $$2t\ge n$$ (to make $$2p>n$$) and $$2t\le n+1$$ (because $$p means the same as $$p\le n$$). This suggests taking $$t=\lceil \frac n2\rceil$$. This will make $$t>5$$ as soon as $$n>10$$.