Does the division of two indicator functions cancel out? (intuition behind $\frac00=1$)

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This is on my professor's notes: $$\frac {1_{0<y<x<1} (x,y)} {1_{0,1}(x)}=1_{0,x}(y).$$ I can provide more context if you want but I believe this is sufficient for my question.

I am ultimately wondering why would $x$ not need to be in $(0,1)$ in the RHS? Is there a general way to derive this?

The reason I tagged is because I want to know what happens if ${1_{0,1}(x)}=0$? Would it then be undefined?

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Division by $0$ is undefined. This means that for $x\neq 0,1$ your LHS is undefined, no matter your conventions or other definitions.

Also, you cannot just define $\frac00=1$, for this will get you into trouble. For instance it would imply that $$\frac00\cdot0=1\cdot0 =0,$$ and that $$\frac00\cdot0=\frac{0\cdot0}0=\frac00=1.$$ This implies $0=1$. This is either a contradiction or a very hard constraint on the algebraic set $x$ and $y$ live in.