integer n as a product of primes, 2, 3,...P, P is less than or equal to $\sqrt{n}$

Question

I have no idea how should I start to answer this question.

Show that to write integer n as a product of primes, 2, 3,...P, P is less than or equal to $\sqrt{n}$.

Can you give me a clue to show it?

Answer 1

Let $q$ be the penultimate of the primes. By Bertrand's postulate, $q<p<2q$. Except for the cases $n=2$ (where $2>\sqrt 2$) and $n=2\cdot 3$ (where $3>\sqrt 6$), the three distinct factors $2$, $q$, $p$ occur in $n=2\cdot\ldots\cdot q\cdot p$, making $n\ge 2\cdot q\cdot p>p^2$, or: $p<\sqrt n$.

Answer 2

Edit: the OP appears to be asking a different question than what I interpreted. I'll leave this answer here to remove any doubts that this (different) claim might be true.

---

The claim is that any integer $n$ can be expressed as a product of primes each less than or equal to $\sqrt n$. Well, you certainly cannot prove this claim, since it is false. Here are a few examples whose only factorisations do not satisfy the claim. $$\begin{split} 17 & = 17 \quad \text{but }17>\sqrt{17}\\ 23 & = 23 \quad \text{but }23>\sqrt{23}\\ 14 & = 2\times 7\quad \text{but }7=\sqrt{49}>\sqrt{14}\\ 55 & = 5\times 11 \quad \text{but }11=\sqrt{121}>\sqrt{55} \end{split}$$

Answer 3

This isn't true for $P= 2$ and $2 > \sqrt{2}$ and for $P = 3$ and $3 > \sqrt {2*3}$ but for $P \ge 5$ it is true. For example: $5 < \sqrt {2*3*5}$.

Suppose $p_1, .... ,p_k$ are the first $k$ primes.

Then this is statement is $p_k \le \sqrt {p_1p_2....p_k}$.

This true iff and only if $p_k^2 \le p_1p_2....p_k$ which is true if and only if

$p_k \le p_1p_2....p_{k-1}$.

Which is easy to prove:

I claim that for $k\ge 3$ then $p_{k} < p_1p_2.... p_{k-1}$ (assuming $p_k \ge p_3 = 5$).

Pf: Consider $M=2\cdot 3.... p_{k-1} -1 \ge 2*3 - 1 = 5 > 1$. None of the $p_i; i\le k-1$ divide into $M$. So either some other prime, $p_m$, does. So there exists a $p_{m}\ge p_k > p_{k-1}$ that divides into and is therefore less than or equal to $M= 2\cdot 3.... p_{k-1} -1$.

(Note: we needed the condition $k \ge 3$ to assure that $M = p_1p_2...p_{k-1} -1 > 1$.)

(Also note: the $p_m$ that divides into $M$ doesn't need to be $p_k$. But if it isn't $p_k$ then it must be a prime larger than $p_k$. Example: $M = 2*3*5-1 = 29$ and so $2,3,5$ do not divide $29$ so there is a prime greater than $5$ that divides $29$. That prime is $29$. So the next prime after $5$ (which is $7$) must be $\le 29$. And $7 \le M = 2*3*5 -1$. )

Also note: $p_k$ is strictly less than $\sqrt {p_1...p_k}$ (which is an irrational number anyway....)