Handling division by zero axiomatically Suppose we define the multiplicative inverse function on real numbers as follows:
$\forall{x \in \mathbb{R}}(x \neq 0 \implies x \times \frac{1}{x} = 1)  $.
Consider this truth table.
\begin{array} {|c|c|c|c||c|}
\hline
x \neq 0 & x \times \frac{1}{x} = 1 & x \neq 0 \implies x \times \frac{1}{x} = 1 \\ \hline
T & T & T  \\ 
T & F & F  \\ 
F & T & T  \\ 
F & F & T  \\ \hline
\end{array}
If the implication always holds we know that there will be a case when $x = 0$ and $x \times \frac{1}{x} = 1$ is true!  This seems to fly in the face of common sense.  Is this the correct way to handle division axiomatically?
 A: No, you can't conclude that there will be such a case. Simply knowing that the implication holds means that we will always find ourselves in one of the lines of the truth table that ends with T. It does not mean that each of those three lines has to occur in practice.
As long as the line that ends with F does not occur, it doesn't matter whether it's all the others that occur, or only some of them.
The axiom taken in itself certainly allows for a model where $0\times \frac10=1$. But there are (or should be) other axioms that reject such a model.
A: No, it isn't.
First of all, if $x=0$ and $x\times\frac1x=1$, the implication is true, but that does not mean that a value for $x$ that satisfies both of these conditions exists. So this conclusion doesn't necessarily invalidate your operation.
However, you may want to ensure that such a value for $x$ can't exist. To do this, you could specify that $x\ne0 \Longleftrightarrow x\times\frac1x$ instead of the one-way implication.
A: If you look at this link on constructing real numbers. It looks like the correct axiom is:
For every $x \neq 0 \in \mathbb{R}$, There exists an element $x^{-1} \in \mathbb{R}$, such that $x * x^{-1} = 1$.  
This shows the existence of a multiplicative inverse, only if $x \neq 0$, thus division by zero will never happen.
