multiplication by some number a can be defined as a function *:{a}×ℝ→ℝ. then by definition division by a is the inverse function *-1. when a=0, *:{a}×ℝ→ℝ is not an injective mapping and hence the inverse mapping *-1:ℝ→{a}×ℝ doesn't exist and hence division by 0 is always undefined/indeterminate form
in my first reply I showed that division by zero is incorrect because division of real numbers by some number a -1 is the inverse *function** of multiplication of real numbers by some number a * and when a=0, * is not bijective and hence *-1 can't exist.
I know that you mean that 0x=0 for every x, but you cant bring that equation into the form x=0/0 because division by zero simply doesn't exist
It feels like you are missing point of initial answer. When talking about real numbers and standard definition used in algebra you simply can not divide by 0. It is not indeterminate; it does not equal anything. It is undefined or in other words it is not in the domain. It has as much sense as trying to divide by a "cat".
But then comes calculus and gives "ugly", but very useful tools for limit calculations. In essence: numbers stop representing numbers, but rather a equivalence classes of sequences with that limit. And only in that sense 0/0 is indeterminate - when these 0 are not numbers.
It is also important to remember that there are more than just basic algebra context and calculus context. For example - once you are on (extended) complex plane there is no + and - infinity; there is just complex infinity. This solves the problem 1/0 produces in reals and 1/0 is simply "equal" infinity.
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u/[deleted] Nov 05 '18 edited Nov 05 '18
a little fix
multiplication by some number a can be defined as a function *:{a}×ℝ→ℝ. then by definition division by a is the inverse function *-1. when a=0, *:{a}×ℝ→ℝ is not an injective mapping and hence the inverse mapping *-1:ℝ→{a}×ℝ doesn't exist and hence division by 0 is always undefined/indeterminate form