8 ms·

Could you sketch a proof starting from some definition of the operation product of integers as "repeated addition" without using the distributive property, whic

by mnl 6y ago

Could you sketch a proof starting from some definition of the operation product of integers as "repeated addition" without using the distributive property, which would imply that we already have another binary operation besides the sum? I was thinking about the Peano axioms when I wrote that (hence the working backwards thing). Obviously if you start with a ring, you don't have to impose it, you get that as a property. I think I mentioned that in a later post.

thaumasiotes 6y ago
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.)
- guram11 6y ago
  > 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?
  - thaumasiotes 6y ago
    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.
    - guram11 6y ago
      > commutativity yes my mistake

by mnl 6y ago

thaumasiotes 6y ago

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.)

guram11 6y ago
> 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?
- thaumasiotes 6y ago
  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.
  - guram11 6y ago
    > commutativity yes my mistake