>>12697286technically dividing by zero is undefined, since that's the precise mathematical definition that comes with any set of "numbers" (look up "fields" under algebra), but to see why you can look at a couple of cases, each with it's own examples:
one case, technically being the algebraic view, comes when you consider what the inverse of zero would be with respect to some number. Let's denote the inverse of zero with respect to 1 to be [math\] x [\math].
So we have [math\] 0 \times x = 1 [\math]. so far it looks like we could just rebrand it as [math\] x = \frac{1}{0} [\math]. but now consider [math\] (3 \times 0) \times x [\math] and [math\] (5 \times 0) \times x [\math].
Both of the brackets in these are zero, so the answer should be 1 right? but then if you move the brackets around, you have [math\] 3 \times ( 0 \times x) = 3 \times 1 = 3 [\math] and [math\] 5 \times ( 0 \times x ) = 5 [\math]. So what gives? we can move brackets around when multiplying all other numbers, why can't we here? this inconsistency is inherent to what one may think of as "numbers" in some sense, so multiplying by [math\] x [\math] can't be well defined, since you don't have a straight up output for multiplying 0 by it. This is particularly a problem for our [math\] x [\math] here, since we literally defined it in terms of what it does when it's multiplied by zero. since you don't have a unique output for multiplying [math\] x [\math] by zero, you just don't have any sense to it. the definition of the thing breaks down in the system itself.
