I know I've said this before but I think I might of thought of some proof that can end all of this division by zero shit forever. For those who don’t know what logarithms are, $\log{(a)} b$ is the same as saying "With ${\displaystyle a^{x}}$, what does $x$ have to be to get $b$?". Lets look at $\log{(2)} 256$ as an example, with ${\displaystyle 2^{x}}$, what does $x$ have to be to get $256$? The answer is $8$, because ${\displaystyle 2^{8}} = 256$. Now lets look at $\log{(1)} 1$, with ${\displaystyle 1^{x}}$, what does $x$ have to be to get $1$? $x$ can of course be any number, so I will just keep the answer as $x$. Because $\log{(a)} b = \log{(b)}/\log{(a)}$ (with $\log{}$ being $\log{(10)}$, it still works the same with any other positive logarithm) we get $$x = \log{(1)} 1 = \log{(1)} / \log{(1)} = 0/0$$
We can use a similar method to prove that $1/0$ doesn't have an answer, with ${\displaystyle 1^{x}}$, what does $x$ have to be to get $10$? $$\log{(1)} 10=\log{(10)} / \log{(1)}=1/0$$
If you don’t understand the above proof you can get the same answers by asking "If you are travailing at $0$mph, how long does it take you to travel $0$ meters?" or simply "How many $0$'s go into $0$?" for $0/0$ and "If you are travailing at $0$mph , how long does it take you to travel $1$ meter?" or "How many $0$'s go into $1$?" for $1/0$. Well okay maybe '$x$' isn't a suitable answer for the first one, but you have to at least accept that the answer can be any number, and that it deserves at least a small mention in this article.