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

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?

• As far as I'm concerned, you're right: if $1_{0,1}(x) = 0$ then the quotient is undefined, so the expression can't be generally true. Your professor might have a convention that $\frac{0}{0} = 1$ in this situation, or similar. – Patrick Stevens Apr 8 '18 at 8:05
• no he doesn't which is weird but if the convetion is 0/0=1 then thats why we dont have x between (0,1) on the RHS? since they can cancel out? – james black Apr 8 '18 at 8:18
• maybe the function on the LHS has been defined (only) on $\{(x,y):0<x,y<1\}$? – user52227 Apr 8 '18 at 9:01
• the full question is $f_{x,y}$ is defined on $0<y<x<1$ so we calculate $f_{y given x}=f_{x,y}/f_x$ and if we integrate, we would get f_x defined on (0,1); then we derive the result, which i dont think would be defined on anything? – james black Apr 8 '18 at 9:04
• its a bivariate probability density based on nomral distribution but that shouldnt matte – james black Apr 8 '18 at 9:05

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.