Is it true that
$$\left\{\frac{m^2}{n!}:m,n\in\mathbb{N}\right\}=\mathbb{Q}^{+} \tag{1}$$
My intuition comes from the fact that $m$ and $n$-increases $m^2=p_1^{2e_1} p_2^{2e_2}…p_s^{2e_s}$ where $p_1,…,p_s$ are the first $s$ primes and $n!=q_1^{f_1} q_2^{f_2}\cdot\cdot\cdot q_r^{f_r}$ where $q_1,…,q_r$ are the first $r$ primes
The following answer already shows:
$$\left\{\frac{m^a}{n^b}:m,n\in\mathbb{N}\right\}=\left\{\frac{c^{\gcd(a,b)}}{d^{\gcd(a,b)}}:c,d\in\mathbb{N}\right\}$$
Since
$$\left\{\frac{m^2}{n!}:m,n\in\mathbb{N}\right\}=\left\{\frac{p_1^{2e_1} p_2^{2e_2}…p_s^{2e_s}}{q_1^{f_1} q_2^{f_2}\cdot\cdot\cdot q_r^{f_r}}:e_1,…,e_r,f_1,…,f_r\in\mathbb{N}\right\}$$
and each element in $2e_{1},…,2e_r$ is relatively prime to each element in $f_{1},…,f_{r}$
We can assume that
$$\left\{\frac{m^2}{n!}:m,n\in\mathbb{N}\right\}=\mathbb{Q}^{+} \tag{1}$$
holds true.
If this is correct can we generalize this to:
$$\left\{\frac{m^{p}}{n!}:m,n\in\mathbb{N}\right\}=\mathbb{Q}^{+}$$
Where $p\in\mathbb{Z}$
Is this also true?
Best Answer
By Bertrand, there is always a prime $p$ which divides $n!$ exactly once. Hence for every rational of the form $\frac {m^2}{n!}$ there must be a prime which divides either the numerator or the denominator an odd number of times. Thus you can't get every rational this way.
As a concrete example, $\frac 14$ can not be written as $\frac {m^k}{n!}$ for any $m,k,n\in \mathbb N$ with $k≥2$. Note that it is easy to see that $n$ would have to be $>2$ But then we can use Bertrand to produce a prime $p$ between $\frac n2$ and $n$ and such $p$ would have to divide either the numerator or the denominator of $\frac 14$, which is clearly not possible.