Andy my friend claims that $0=X\cdot 0$ where $X$ is an arbitrary value, $0\div 0=X$, and thus $0/0$ can be assigned to an arbitrary value. But he mistakenly considers $0÷0=1$ or the multiplication by $0$ is always reversible. As far as I know, $0/0$ is undefined and cannot be assigned to any arbitrary value $X$ by any valid mathematical operations.
But why doesn't math introduce an axiom $0\div 0=0$ as a means to claim the multiplication by $0$ is irreversible? The mathematical definition of fractions does exclude cases of $0$ dividers, so it seems that this axiom doesn't violate the established math rules.
What paradoxes, flaws, or inconsistencies will arise from this pseudo-axiom $0\div 0=0$ in whole established math system? Or does it change the true value of any equation?
ps: I am sorry if I have made the wrong statements.
Edits:
I think the pseudo-axiom should be clarified as, $$0∈φ$$, where every element of $φ$ is every solution to $0/0$, and $0$ is not necessarily the unique solution.
With the following perspectives:
- The division is meant to be the inverse of multiplication. Multiplying X by 0 is a multiple-to-1 function, so $0$ divider map $0$ to every real number, or $0/0$ have the range set containing elements of every real number.
- The law of associative is satisfied if $0/0$ is assigned to $0$ or $k=1$, so that $(k×0)/0=k×(0/0)$, where k is a constant.
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- It holds true if $0/0$ is assigned to $0$, so that $$k×(0/0)=m×(0/0)$$, where $k$ and $m$ are constants.
- In conclusion, $0/0$ have solutions of every real number in the perspective of the inverse of the multiplication by $0$, or mathematically called $undefined$. But if a non-1 constant is factored out from $0/0$ while holding the true value of equating with the initial value, $0/0$ is assigned to $0$ by this behaviour.
ps: I am sorry if I have the wrong statements or logical falsities.
Edits:
However, in the most general sense, $0/0$ is undefined mathematically, as the pseudo-axiom states that dividing $0$ by $0$ is a futile endeavour. From a strict mathematical perspective $0/0$ is undefined.
My favorite answer to this is: $0$ doesn't belong to the multiplicative group so that the question does not even arise.
A bit longer is the answer, that multiplication and addition are only connected via the distributive law. There is no other connection. This law allows us to conclude that $$0\cdot x=(1+(-1))\cdot x= 1\cdot x+ (-1)\cdot x=0.$$ Now, what is division? Division isn't really a mathematical concept. It is only the abbreviation of "multiply by the inverse element". Hence the question is, what would $0^{-1}$ be? Any inverse $x^{-1}$ of an element $x$ is defined as the solution to $x\cdot x^{-1}=1.$ Finally assume that there would be a reasonable solution to $0\cdot 0^{-1}=1.$ Then $$ 1=0\cdot 0^{-1}= 0 $$
by the above argument with the distributive law. But if we add the equation $1=0$ to our number system, then it will be pretty useless, or in mathematical terms: the real numbers wouldn't be a field anymore.