Argthtjtr

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?

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 ($1cdot a = a$ for any $a$)


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


• What division means ($frac ab = c$ means $a = ccdot b$)

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 $frac{0}{0}=:bot$.

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 $a,b,cin F$,

- $a+b=b+a$,

- $(a+b)+c=a+(b+c)$,

- $e_++a=a$,

- there exists $a'$ such that $a'+a=e_+$,

- $(atimes b)times c=atimes (btimes c)$,

- $e_timestimes a=a$,

- there exists $a''$ such that $a''times a=e_times$ if $ane e_+$.

Now verify that the rationals and the reals are fields.

Try and prove that if there exists $x$ such that $xtimes e_+=e_times$, the set $F$ can only have one element.

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:

$$0cdot q=d.$$

But we know that $0cdot q=0$, so the equation has no solution (unless $d=0$).