Another proof that dividing by $0$ does not exist -- is it right?

Solution 1:

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 $a = c\cdot b$)
• $0\neq1$

Solution 2:

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

Solution 3:

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.

Solution 4:

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 $d=0$).