Another proof that dividing by $0$ does not exist -- is it right? Ok I am in grade 9 and I am maybe too young for this.
But I thought about this, why dividing by $0$ is impossible.
Dividing by $0$ is possible would mean $1/0$ is possible, which would mean $0$ has a multiplicative inverse.
So if we multiply a number by $0$ then by $1/0$ we get the same number.
But that's impossible because all numbers multiplied by $0$ give $0$ therefore we can’t have an inverse for $0$, as that gives us the initial number and thus division by $0$ is impossible
Is this right?
 A: That's the most basic reason that division by $0$ is usually considered to be a Bad Thing, yes. Because if we did allow dividing by $0$, we would have to give up at least of one of the following things (these are usually considered Very Nice):

*

*What $1$ means ($1\cdot a = a$ for any $a$)

*What $0$ means ($0 \cdot a = 0$ for any $a$) (actually a consequence of $0+a=a$ and $(a+b)\cdot c=a\cdot c+a\cdot b$, two other Nice Things)

*What division means ($\frac ab = c$ means $a = c\cdot b$)

*$0\neq1$
A: That is quite right. However, I would like you to have a higher point of view.
Mathematicians derive theorems from axioms and definitions. And here is the definition of a field.
A field is a set $F$ equipped with two binary operations $+,\times$, such that there exists $e_+, e_\times$, such that for all $a,b,c\in F$,
- $a+b=b+a$,
- $(a+b)+c=a+(b+c)$,
- $e_++a=a$,
- there exists $a'$ such that $a'+a=e_+$,
- $(a\times b)\times c=a\times (b\times c)$,
- $e_\times\times a=a$,
- there exists $a''$ such that $a''\times a=e_\times$ if $a\ne e_+$.
Now verify that the rationals and the reals are fields.
Try and prove that if there exists $x$ such that $x\times e_+=e_\times$, the set $F$ can only have one element.
A: Yes . . . and no.
You might be interested in, for example, Wheel Theory, where division by zero is defined.
See Lemma 2 of the 1997 article "Wheels," by A. Setzer for tables describing addition, multiplication, and their inverses in what is called $R_\bot^\infty$, the wheel given by adjoining special symbols and rules to an arbitrary integral domain $R$ in order to allow division by zero, even $\frac{0}{0}=:\bot$.
A: You are quite right.
There is a simpler way, though (which spares the concept of multiplicative inverse):
By definition, $q$ is the quotient of the division of $d$ by $0$ if the following equation is satisfied:
$$0\cdot q=d.$$
But we know that $0\cdot q=0$, so the equation has no solution (unless $d=0$).
