Why is the distributive law, $A+BC=(A+B)(A+C)$ not true in real algebra?

Question

It's only true in boolean algebra. But why is that? At what point does that break? What is special about real algebra that breaks that rule?

Answer 1

If you had both distribution laws, you would be able to prove:

$$\begin{align} a + bc & = & (a + b)(a + c) \\\\ & = & a^2 + ab +ac + bc \\\\ a & = & a^2 + ab + ac \end{align} $$

for all $a,b,c$. It should be clear that this places a severe constraint on how the operations and the elements can behave, and so there must be very few systems that can do it.

It should not be surprising that the real numbers happen nor to be one of these very few systems. The real numbers form a field, which is itself a very severe constraint. The combination of the two constraints rules out almost every possible system. In a field, if $a = a^2 + ab + ac$ then (unless all three are zero) we can divide by $a$ to show that $1 = a + b + c$ for all $a,b,c$, and perhaps you can see how this system is so constrained that it applies to very few actual problems.

There are many other elegant symmetries not possessed by the reals. For example, in some systems (even some fields) one has $a=-a$ for all $a$; the reals are not one of them.