I am trying to understand a proof and got stuck at a certain point, which is supposedly easy. I don't see why this works, could you help me out?
Let $A = [\frac{p_{i,j}}{q_{i,j}}] \in \mathbb{Q}^{nxn}$ be a rational-valued square matrix with $p_{i,j},q_{i,j} \in \mathbb{Z}$ co-prime and $q_{i,j} > 0$ for every pair $(i,j) \in [n]^2$.
Further let $det(A) = \frac{p}{q}$, where $p,q \in \mathbb{Z}$ co-prime and $q > 0$. The following inequality seems to follow directly from Leibniz Determinant Formula:
$$q \leq \prod_{1 \leq i,j \leq n}q_{i,j}$$
What am I missing here?
The determinant is linear in each row and column. That means if you multiply a row/column of a matrix by a number, you multiply the determinant by the same number.
If you multiply each row $i$ of the matrix with $\prod_{j=1}^n q_{i,j}$ you will get a matrix with only integer entries.
Hence by the Leibniz formula, the determinant of that matrix is an integer, $P$. To get the original determinant $D$, you need to divide that by the product of all the factors you used:
$$D = \frac{P}{\prod_{i=1}^n\prod_{j=1}^n q_{i,j}}$$
This proves what you want proven, as your $q$ is the denominator in the most shortened form of $D$.