Given that ## p\nmid n ## for all primes ## p\leq \sqrt[3]{n} ##

[Math100]

Homework Statement

Given that $p\nmid n$ for all primes $p\leq \sqrt[3]{n}$, show that $n>1$ is either a prime or the product of two primes.
[Hint: Assume to the contrary that $n$ contains at least three prime factors.]

Relevant Equations

None.

Proof:

Suppose for the sake of contradiction that $n$ contains at least three prime factors.
Let $n=p_{1} p_{2}\dotsb p_{X}$ for $x\geq 3$.
Note that $n=p_{1} p_{2} p_{3}$.
Then we have $p_{1}\nleq \sqrt[3]{n}$, $p_{2}\nleq \sqrt[3]{n}$ and $p_{3}\nleq \sqrt[3]{n}$.
Thus $p_{1} p_{2} p_{3}\nleq (\sqrt[3]{n})(\sqrt[3]{n})(\sqrt[3]{n})$.
This means $p_{1} p_{2} p_{3}>(\sqrt[3]{n})(\sqrt[3]{n})(\sqrt[3]{n})$,
which implies that $n>n$.
This is a contradiction because $n\ngtr n$.
Therefore, given that $p\nmid n$ for all primes $p\leq \sqrt[3]{n}$,
$n>1$ is either a prime or the product of two primes.

 

[PeroK]

Suppose for the sake of contradiction that $n$ contains at least three prime factors.
Let $n=p_{1} p_{2}\dotsb p_{X}$ for $x\geq 3$.
Note that $n=p_{1} p_{2} p_{3}$.

Which is it? Does $n$ have precisely three prime factors, or at least three?

Then we have $p_{1}\nleq \sqrt[3]{n}$, $p_{2}\nleq \sqrt[3]{n}$ and $p_{3}\nleq \sqrt[3]{n}$.
Thus $p_{1} p_{2} p_{3}\nleq (\sqrt[3]{n})(\sqrt[3]{n})(\sqrt[3]{n})$.

What's the difference between $\nleq$ and $>$?

This is a contradiction because $n\ngtr n$.

You really need to say that?

On a general note, I think you need to move on. You're spending too much time on easy problems. You need to try some harder questions.

In this case, it's clear that a number cannot have three or more prime factors all greater than its cube root. That's something that barely needs a formal proof.

 

[nuuskur]

Note that $n=p_{1} p_{2} p_{3}$.

Ok, what if it was $n=p_1p_2p_3p_4p_5$? I don't get instructions for this case from the rest of your proof.

Then we have $p_{1}\nleq \sqrt[3]{n}$

Why? We know that $\neg (p\mid n)$. So, since we're working in $\mathbb N$, we could say something like $a\mid b$ implies $a\leqslant b$, but that still doesn't answer, why..

And again you are exhibiting this unfortunate quality that you are elaborating needlessly on obvious details, but then suddenly become vague and make it look like what you say is obviously true. If that was the case, why would you not apply the same logic during the obviously obvious steps?

I have asked you before to explain certain obvious steps and you have, for lack of a better expression, given me non-sense/irrelevant answers. This indicates that you don't really understand the arguments and all the moving parts. If I'm being particularly uncharitable, I would say you are trying to compensate for lack of detail in some part of your proof attempt by writing out things that are obvious (and even over-explain them) in some other part assuming that this "balances things out". It doesn't. You can explain in great detail 99 obvious points. If you don't justify the core arguments in a proof, you haven't completed the proof.

I would advise the symbols $\leqslant$ and $>$ as opposites for the usual ordering on $\mathbb R$. The notation $p\nleqslant q$ is correct, but it's unnecessary here.

Edit:

To be more specific. Which do you think is more obvious and so may be omitted?

We have that $n>n$, which is a contradiction, because $n\nleq n$ is false

OR..

Let $p$ be a prime factor of $n$. By assumption, since $p\mid n$, we must have $p>\sqrt[3]{n}$..

 

[Math100]

Suppose for the sake of contradiction that $n$ contains at least three prime factors
such that $n=p_{1} p_{2}\dotsb p_{X}$ for $x\geq 3$.
Let $n=p_{1} p_{2} p_{3} m$ where $p_{1}, p_{2}$ and $p_{3}$ are
distinct primes and $m$ is an integer for $m>1$.
Note that $p\nmid n$ for all primes $p\leq \sqrt[3]{n}$.
Then we have $p_{1}>\sqrt[3]{n}, p_{2}>\sqrt[3]{n}$ and $p_{3}>\sqrt[3]{n}$.
Thus $p_{1} p_{2} p_{3}>(\sqrt[3]{n})^{3}$,
and so $p_{1} p_{2} p_{3}>n$.
But then $n>p_{1} p_{2} p_{3}$ because $n=p_{1} p_{2} p_{3} m$ for $m>1$,
which implies that $n>p_{1} p_{2} p_{3}>n$,
now we have a contradiction.
Therefore, given that $p\nmid n$ for all primes $p\leq \sqrt[3]{n}$, $n>1$ is
either a prime or the product of two primes.

 

[PeroK]

Suppose for the sake of contradiction that $n$ contains at least three prime factors
such that $n=p_{1} p_{2}\dotsb p_{X}$ for $x\geq 3$.
Let $n=p_{1} p_{2} p_{3} m$ where $p_{1}, p_{2}$ and $p_{3}$ are
distinct primes and $m$ is an integer for $m>1$.

That should be $m \ge 1$.

Thus $p_{1} p_{2} p_{3}>(\sqrt[3]{n})^{3}$,

This line is unnecessary.

and so $p_{1} p_{2} p_{3}>n$.

At this point you should simply state:

Hence $mp_1p_2p_3 > n$, which is a contradiction. QED

But then $n>p_{1} p_{2} p_{3}$ because $n=p_{1} p_{2} p_{3} m$ for $m>1$,
which implies that $n>p_{1} p_{2} p_{3}>n$,
now we have a contradiction.
Therefore, given that $p\nmid n$ for all primes $p\leq \sqrt[3]{n}$, $n>1$ is
either a prime or the product of two primes.

This is "laboured".