I'd be inclined to define $a - b = a + (-b)$ and division by $\frac{a}{b} = a \cdot b^{-1}$.
That's still not the full story, but a reasonable step. What you still have to do is to define what $-b$. Is supposed to mean.
As the context is presumably some field $(K,+,\,\cdot\,)$, this means that $(K,+)$ is an abelian (=commutative) group. This means that every element $b\in K$ has an inverse, denoted as "$-b$". Now the existence of unique, inverse elements and associativity means that a cancelling rule holds. This means that $$a+b=c\quad \Longleftrightarrow\quad a=c+(-b)\tag 1$$ for all $a$, $b$, $c$ in the group. That is, one can "undo" additions. Here, "$-b$" is just the symbol for the inverse of $b$ in the group, that is the unique element associated to $b$ that satisfies $$b+(-b) = (-b)+b = 0\tag 2$$ where "$0$" denotes the neutral element of $(K,+)$.
Whether one preferes $(1)$ or $(2)$ is a matter of taste; the group and field axioms usually use $(2)$, whereas the formulation $(1)$ is more useful for manipulating equations.
The case for multiplication is similar with two exceptions:
- It's only that $(K\!\setminus\!\{0\},\,\cdot\,)$ is a group, i.e. $0$ has no multiplicative inverse and one cannot undo multiplications with 0.
- Multiplication need not be commutative. An example of such a skew-field is Hamilton's quaternions.
If the latter is the case, one must be more careful, though. For example, one should avoid notations like $\frac ab$ because in the skew case, one usually has $a\cdot(b^{-1})\neq (b^{-1})\cdot a$. "$b^{-1}$" is one common notation for the inverse of $b$ if the group notation is as a multiplicative group. A different notation can be "$1/b$" where "1" is the neutral element in that group. This notation works and is unique because $1\cdot (b^{-1}) = (b^{-1})\cdot 1 = b^{-1}$ even if multiplication is not commutative.
