Prove, using only the field axioms of real numbers, that $0/0$ is undefined.

Question

Prove, using only the field axioms of real numbers, that $0/0$ is undefined.
I have thought about it for a while and come up with an idea how to solve this. First, I would like to prove (using field axioms - again) that $x \cdot \frac{1}{y} = \frac{x}{y}$, (but I am still working on this - any hints would be most appreciated),therefore $\frac{0}{0} = 0 * \frac{1}{0}$ and then, say that - by virtue of one of the axioms there exists only one number $b$ satisfying the equation $ab = 1$ for all $a \ne 0$. Since in our case we want to take the reciprocal of 0 - the axiom clearly states that it does not exist.
Is this a good way to prove that the expression in question is undefined?

Answer 1

Let $R$ be any ring with $1$. Suppose that $0^{-1}$ makes sense. Then that means $0 \cdot 0^{-1}=1 = 0^{-1}\cdot 0$. Let $a \in R$ be arbitrary. Then $$ a = 1a = 0^{-1}0a = 0^{-1}0=0. $$ Hence $R=\{0\}$.

Thus, if you want your ring to have more than one element, you better not let $0$ have an inverse. So, not only should $\frac{0}{0}$ not make sense, no "fraction" of the form $\frac{a}{0}$ should make sense.

In case you do not believe that $a0=0$, we have $$ a0 = a(0+0) = a0+a0. $$ Now cancel an $a0$ from each side. If you do not believe in subtraction/cancellation someone else will have to help you.

Answer 2

Suppose $\frac{0}{0}$ is defined.

Then, by the Existence of a Multiplicative Inverse, $\frac{0}{0}=1$

By the Existence of an Additive Inverse, we have $a+(-a)=0$. Using this with $a=1$, we can write:

$1-1=0 \Leftrightarrow \frac{0}{0} - \frac{0}{0} =0 \Leftrightarrow \frac{0-0}{0}=0 \Leftrightarrow \frac{0}{0}=0$

which leads to a contradiction by uniqueness of your multiplicative inverse.

Answer 3

The expression $\frac xy$ is not immediately in the language of fields, which only has the symbols $+,\times,-,0,1$. Thus we must define what $\frac xy$ means, and the fastest thing to do is as follows: if the equation $ya = x$ has a unique solution $a$, then $a = \frac xy$. (Typically you would define $y^{-1}$ first, and then set $\frac xy = xy^{-1}$, and we would have to rule out the existence of $0^{-1}$, but this route also works.)

Thus, let us show that $0 \times a = 0$ does not have a unique solution. In fact, for any $a$ we have $$ 0 \times a = (0 + 0) \times a = 0 \times a + 0 \times a. $$ Now subtract $0 \times a$ from each side to obtain $0 = 0 \times a$. This means that the equation defining $\frac 00$ does not have a unique solution -- it has at least the solutions $0,1$ -- and thus $\frac 00$ is not well-defined.