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# Need help for the proofs about number theory.

Note: $$p_{n}$$ is the n-th prime number ($$p_{1} = 2, p_{2} = 3, p_{3} = 5$$, etc).

1.) Prove that $$p_{n+1} < \sum\limits_{i=1}^{n} p_{i}$$ for any integer $$n > 2$$.

2.) Prove that $$\sum\limits_{i=1}^{n} p_{i} < \prod\limits_{i=1}^{n} p_{i}$$ for any integer $$n > 1$$.

3.) Prove or counterexample that $$\sqrt{n!} \notin \mathbb{N}$$ for any integer $$n > 1$$.

Note by Samuraiwarm Tsunayoshi
3 years ago

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Bertrand's Postulate · 3 years ago

You can have a one-liner inductive proof of Problem 1) using Bertrand's postulate, which is a well known proven result in number theory.

The base case for $$n=3$$ is easy. Assume the assertion to hold good for an integer $$n\geq 3$$. Using this postulate, we have $$p_{n+1} < p_{n+2} < 2 p_{n+1}= p_{n+1}+ p_{n+1} < \sum_{i=1}^{n+1}p_i$$, where the last inequality follows from the induction assumption. · 3 years ago

I agree with your q3 solution, it is very common in questions like these to use Bertrand's Postulate. In a moment I will post a very nice problem indeed. q2 is very susceptible to induction (just simple induction, maybe with a contradiction element if you want to simplify things). q1 is a very nice problem, as it combines techniques used in both q2 and q3 (use induction, with a contradiction, then use Bertrand's Postulate), so I thank you for these series of problems. If you want to know the problem I previously mentioned, just visit my page. · 3 years ago

I just proved number 3 in 3 seconds lol. Just Bertrand's Postulate that there exists a prime number from $$\frac{n}{2}$$ to $$n$$ if $$n$$ is even or from $$\frac{n-1}{2}$$ to $$n-1$$ if $$n$$ is odd.

For even number $$n$$, if $$p > \frac{n}{2}$$, then $$2p > n$$. Which means the next number that has a factor $$p$$ doesn't exist in $$n,n-1,n-2,...,3,2,1$$ any more. So $$n!$$ only have 1 factor of $$p$$ and makes it not perfect square.

For odd number $$n$$, if $$p > \frac{n-1}{2}$$, then $$2p > n-1$$. Same as above, there exists only 1 factor $$p$$.

How about $$n$$? Can $$n$$ have a factor of $$p$$? No. If $$2p = n$$, then $$n$$ is an odd number, which contradicts the given. So $$2p > n$$ or $$2p < n$$.

For $$2p > n$$, we're done. $$2p$$ doesn't exist in $$n,n-1,n-2,...,3,2,1$$ any more.

For $$2p < n$$, since $$2p > n-1$$, therefore $$2p$$ can't be integers, which contradicts the truth.

Therefore, $$\sqrt{n!}$$ can't be integers for any $$n > 1$$. · 3 years ago