# How to prove that $\Pi_{\substack{1\leq i\leq m}\\{m+1\leq j\leq n}}\dfrac{1}{i-j}=\Pi_{\substack{1\leq i\leq m}\\{m+1\leq j\leq n}}\dfrac{1}{j-i-n}$?

While attempting to prove that the Hodge *-operator an an oriented real inner product vector space is involutive up to sign, I came across the following problem which I needed to complete the proof. The problem is if given any two natural numbers $$m,n$$ with $$m\leq n$$, then the following products are equal: $$\prod_{\substack{1\leq i\leq m}\\{m+1\leq j\leq n}}\dfrac{1}{i-j}=\prod_{\substack{1\leq i\leq m}\\{m+1\leq j\leq n}}\dfrac{1}{j-i-n}.$$ I have no clue how to proceed, any help is appreciated.

## 1 Answer

Let $$u=m-i+1\quad\text{and}\quad v=n-j+m+1.$$ Then, $$\prod_{1\le u\le m\atop m+1\le v\le n}\frac1{u-v}=\prod_{1\le i\le m\atop m+1\le j\le n}\frac1{j-i-n}.$$

• Wow, you’re so fast. I wonder how do you instantly come up with these substitutions? Commented Jul 23 at 16:35
• I looked for a substitution leaving invariant the two sets $\{1,\dots,m\}$ and $\{m+1,\dots,n\}$, and the first one I thought of happened to work. Commented Jul 23 at 17:28