Prove that if ## n>2 ##, then there exists a prime ## p ## satisfying....

[Math100]

Homework Statement

Prove that if $n>2$, then there exists a prime $p$ satisfying $n<p<n!$.

Relevant Equations

None.

Proof:

Assume to the contrary that $n!-1$ is not prime.
That is, $n!-1$ is composite.
Let $p$ be a prime factor of $n!-1$ such that $n<p$ for some $n\in\mathbb{N}$
where $n>2$.
Since $p\mid (n!-1)$, it follows that $p\nmid n!$.
This means $p\nleq n$.
Note that $p\leq n!-1$,
because $p\mid (n!-1)$.
Thus $n<p<n!$.
Therefore, if $n>2$, then there exists a prime $p$ satisfying $n<p<n!$.

 

[fishturtle1]

Homework Statement:: Prove that if $n>2$, then there exists a prime $p$ satisfying $n<p<n!$.
Relevant Equations:: None.

Proof:

Assume to the contrary that $n!-1$ is not prime.
That is, $n!-1$ is composite.
Let $p$ be a prime factor of $n!-1$ such that $n<p$ for some $n\in\mathbb{N}$
where $n>2$.

It's not clear to me why such a prime $p$ exists. (referring to "
Let $p$ be a prime factor of $n!-1$ such that $n<p$ for some $n\in\mathbb{N}$
where $n>2$." which won't quote for some reason).

Since $p\mid (n!-1)$, it follows that $p\nmid n!$.
This means $p\nleq n$.

Here is where you've shown such a prime $p$ exists.

Note that $p\leq n!-1$,
because $p\mid (n!-1)$.
Thus $n<p<n!$.

Looks good to me. But what about the case where $n! - 1$ is prime?

Therefore, if $n>2$, then there exists a prime $p$ satisfying $n<p<n!$.

Note: In your first sentence you say 'assume to the contrary' but never reach a contradiction. Consider constructing the proof as:

Proof: Let $n > 2$ be an integer. We consider two cases:

Case 1: $n! - 1$ is composite. [insert your work]

Case 2: $n! - 1$ is prime. [more work to do] edit: I'm a dope, but it's still work mentioning (imo) why this case is 'obvious'.

 

[PeroK]

Homework Statement:: Prove that if $n>2$, then there exists a prime $p$ satisfying $n<p<n!$.
Relevant Equations:: None.

Proof:

Assume to the contrary that $n!-1$ is not prime.
That is, $n!-1$ is composite.
Let $p$ be a prime factor of $n!-1$ such that $n<p$ for some $n\in\mathbb{N}$
where $n>2$.
Since $p\mid (n!-1)$, it follows that $p\nmid n!$.
This means $p\nleq n$.
Note that $p\leq n!-1$,
because $p\mid (n!-1)$.
Thus $n<p<n!$.
Therefore, if $n>2$, then there exists a prime $p$ satisfying $n<p<n!$.

I'm sorry to say that none of this makes any sense to me.

 

[Mark44]

Prove that if $n>2$, then there exists a prime $p$ satisfying $n<p<n!$.
Proof:
Assume to the contrary that $n!-1$ is not prime.

Starting at the beginning of what @PeroK said made no sense, your assumption is not the negation of the original premise. The original premise says that the prime p must be somewhere between n and n!, not just n! - 1.

 

[nuuskur]

This can be proved via induction. Bertrand's postulate gives much better bounds, so could be used to indirectly prove the statement in the OP.

Either way, your supposition is not to the contrary. The claim is
<br /> \forall n&gt;2,\quad \exists p\in\mathbb P,\quad n&lt;p \quad\mbox{and}\quad p&lt;n!<br />
When negated, it becomes
<br /> \exists n&gt;2,\quad \forall p\in\mathbb P,\quad p&lt; n! \Rightarrow p\leqslant n<br />
and it's not obvious at all, why the negation should be false.

Spoiler

I teach my students to get comfortable with playing around with propositions, so they would understand what it means (among other things) to consider "the contrary" to a given statement. I extend the same recommendation to you, @Math100 . Learn or revise propositional calculus.

P. Suppes "Introduction to logic" is an excellent source for that. Available online for free, among the first results in a google search.

 

[Math100]

Let $n>2$ be a natural number.
Then we consider two cases.
Case #1: Suppose $n!-1$ is prime.
Then we have $p=n!-1$ for some $p\in\mathbb{P}$.
Thus $n<p<n!$ implies $n<n!-1<n!$ for $n>2$.
Case #2: Suppose $n!-1$ is not prime.
That is, $n!-1$ is composite.
Let $p$ be a prime factor of $n!-1$ such that $n<p$ for some
$n\in\mathbb{N}$ where $n>2$.
Then we have $p\mid (n!-1)$.
Note that $p\leq n$ implies $p\mid n!$.
Since $p\mid n!$ and $p\mid (n!-1)$,
it follows that $p\mid 1$.
Thus $p=\pm 1$.
This is a contradiction because $p$ must be a prime number
where $p\neq \pm 1$.
Therefore, if $n>2$, then there exists a prime $p$ satisfying $n<p<n!$.

 

[fishturtle1]

Let $n>2$ be a natural number.
Then we consider two cases.
Case #1: Suppose $n!-1$ is prime.
Then we have $p=n!-1$ for some $p\in\mathbb{P}$.
Thus $n<p<n!$ implies $n<n!-1<n!$ for $n>2$.

ok so far. I'd rewrite the last two sentences as "Choose $p = n! - 1$. Then $n < p < n!$ since $n > 2$.

Case #2: Suppose $n!-1$ is not prime.
That is, $n!-1$ is composite.
Let $p$ be a prime factor of $n!-1$ such that $n<p$ for some
$n\in\mathbb{N}$ where $n>2$.

Since the argument below shows such a prime $n < p$ exists, you shouldn't assume its existence here. The above sentence should be "Let $p$ be a prime factor of $n! - 1$". We don't need to redefine $n$ because we did so at the beginning.

Then we have $p\mid (n!-1)$.

Right; so $p \le n! - 1 < n!$.

Note that $p\leq n$ implies $p\mid n!$.
Since $p\mid n!$ and $p\mid (n!-1)$,
it follows that $p\mid 1$.
Thus $p=\pm 1$.
This is a contradiction because $p$ must be a prime number
where $p\neq \pm 1$.

Therefore $p > n$.

Therefore, if $n>2$, then there exists a prime $p$ satisfying $n<p<n!$.

 

[nuuskur]

Let $n>2$ be a natural number.
Then we consider two cases.
Case #1: Suppose $n!-1$ is prime.
Then we have $p=n!-1$ for some $p\in\mathbb{P}$.
Thus $n<p<n!$ implies $n<n!-1<n!$ for $n>2$.
Case #2: Suppose $n!-1$ is not prime.
That is, $n!-1$ is composite.
Let $p$ be a prime factor of $n!-1$ such that $n<p$ for some
$n\in\mathbb{N}$ where $n>2$.
Then we have $p\mid (n!-1)$.
Note that $p\leq n$ implies $p\mid n!$.
Since $p\mid n!$ and $p\mid (n!-1)$,
it follows that $p\mid 1$.
Thus $p=\pm 1$.
This is a contradiction because $p$ must be a prime number
where $p\neq \pm 1$.
Therefore, if $n>2$, then there exists a prime $p$ satisfying $n<p<n!$.

It's not necessary to "beat around the bush" so to speak.

Case 1. If $n!-1$ is prime, we're done. This is clear, the reader is eager to read about Case 2, already.

Case 2. Assume $n!-1$ is composite and let $p$ be a prime factor. If $p\leqslant n$, then what you say follows. Thus, $n<p<n!-1$ and we're done.

What you write is mathematically correct, though. Some stylistic observations.

Let $p$ be a prime factor of $n!-1$ such that $n<p$ for some
$n\in\mathbb{N}$ where $n>2$.
Then we have $p\mid (n!-1)$.
Note that $p\leq n$ implies $p\mid n!$.
Since $p\mid n!$ and $p\mid (n!-1)$,
it follows that $p\mid 1$.

What follows from "Note that.." is what would happen if some prime factor of $n!-1$ was smaller than $n$. So every prime factor must be between $n$ and $n!-1$. Goes back to what I told you about quantifying your statements. It helps the reader and it helps you.

The same thing could be expressed more clearly as follows.

Suppose $n!-1$ is composite and let $p$ be a prime factor. Then necessarily $n<p$, otherwise $p\mid n!$, whence $p\mid 1$, which is impossible.

 

[PeroK]

Here's how I would do it:

Let $n > 2$. We will show that there exists a prime $p$ such that $n < p < n!$.

Note that every prime $\le n$ divides $n!$ and a prime cannot divide two consecutive integers. Therefore, any prime factor of $n! - 1$ must be greater than $n$. And, if $n! - 1 > 1$ then it must have a prime factor, $p$, which must then satisfy $n < p < n!$.

Note that $n! - 1> 1$ is satisfied by all $n > 2$.