Hacker Newsnew | past | comments | ask | show | jobs | submitlogin

If you want to work up to it conceptually, then I'd say consider the meaning of something like

  5·3 + 2·7
We add 3 five times, and then we add 7 twice. It is then easy to extend this to

  5·3 + 2·7 + 0·4
and say, OK, add 3 five times, and then add 7 twice, and then add 4 zero times. And you get 29.

At that point you notice that when you say 5·3 means "add 3 five times", you forgot to say what you were adding it to. You're adding it to the identity, 0.

And finally we say that starting from 0 and adding something zero times leaves you where you started, at 0, so we can observe that 0·n = 0.

If you formalize that, you'll end up either deriving or postulating the distributive property, depending on what you start with. But it isn't arbitrary; it's not a coincidence that (as I mentioned above) exponentiation by 0 gives the multiplicative identity, and (as I haven't mentioned yet) exponentiation distributes over multiplication the same way multiplication distributes over addition.

(Asymmetry does start creeping in; aggregate exponentiation isn't as nice since exponentiation doesn't have the nice properties that addition and multiplication do.)



> exponentiation distributes over multiplication

things that irks me

    a**b * a**c == a**(b + c) != a**(b * c)
for exponentiation to distribute over multiplication, it must become addition, thus reinforcing the multiplication IS repeated addition?


Distributivity of exponentiation over multiplication looks like

  (ab)^x = (a^x)(b^x)
You're talking about "being exponentiated", not exponentiation. This one isn't a commutative operation.


> commutativity

yes my mistake




Guidelines | FAQ | Lists | API | Security | Legal | Apply to YC | Contact

Search: