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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?

sigsegv
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    Perhaps the question should be, what is special about Boolean algebra that allows you to distribute addition over multiplication? –  Mar 31 '18 at 15:10
  • What do you mean "in boolean algebra"? Do you mean in a boolean ring? – MPW Mar 31 '18 at 15:10
  • @MPW By boolean algebra, I mean, the one which involve only zeros and ones, like what we do in digital electronics. – sigsegv Mar 31 '18 at 15:17
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    The better question is why would that be true outside of a Boolean ring? –  Mar 31 '18 at 15:19
  • @Rahul I think that since, multiplication distributes over addition, it is only fair and natural for addition to distribute over multiplication, as in boolean algebra. So, why should reals break this beautiful symmetry? – sigsegv Mar 31 '18 at 15:21
  • It's not true only in Boolean algebra; it's true in any distributive lattice. That is, in lattices each of the two distributive laws $a\land(b\lor c)=(a\land b)\lor(a\land c)$ and $a\lor(b\land c)=(a\lor b)\land(a\lor c)$ implies the other one. – Andreas Blass Mar 31 '18 at 16:59

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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.

MJD
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