5 ms·

No, it's called distributive property: https://en.m.wikipedia.org/wiki/Distributive_property https://en.m.wikipedia.org/wiki/Distributive_property

by konjin 6y ago

henryfjordan 6y ago
After some further reading, I don't think anyone in this thread is correct. Modulo in math is not an operator like a programmer might think of it. See: https://math.stackexchange.com/questions/2832649/modular-arithmetic-and-the-distributive-property https://math.stackexchange.com/questions/2832649/modular-ari...
- vitus 6y ago
  Right. This is a consequence of compatibility with exponentiation. https://en.wikipedia.org/wiki/Modular_arithmetic#Properties https://en.wikipedia.org/wiki/Modular_arithmetic#Properties It should be easy to see that n ≡ (n % 15) (mod 15) which, applying compatibility of exp, then gives us n^4 ≡ (n % 15)^4 (mod 15) which can be rewritten in Python's notation as n**4 % 15 == (n % 15) ** 4 % 15
- edflsafoiewq 6y ago
  The injection mod15: Z -> Z/15 is an operation and the basic idea here is mod15(pow4(x)) = pow4(mod15(x)), so I think it's correct to call it commutativity.
  - vitus 6y ago
    Well, not quite. You really want to say that mod15(pow4(x)) == mod15(pow4(mod15(x))). Example: (3 ** 4) % 15 = 81 % 15 = 6. But, (3 % 15) ** 4 = 3 ** 4 = 81. (That said, the two functions do commute when you restrict your domain and range to Z/15.) edit: also, I wouldn't consider mod15 to be an injection, as it's, um, not injective (it maps multiple inputs to the same output).
    - edflsafoiewq 6y ago
      pow4(mod15(x)) is already in Z/15, you can't apply mod15 to it. The point is pow4ing in Z and then reducing mod 15 is the same as reducing mod 15, then pow4ing in Z/15. It's not actually the same pow4 on the LHS and RHS (the LHS is in Z, the RHS in Z/15), but I think "commutative" still fits. > as it's, um, not injective Er, surjection :)
      - vitus 6y ago
        Well, how about this, if we use pow4' to be modular exponentiation, i.e. mod15 o pow4? (of course then changing mod15 to take Z -> Z so the types line up) Then the desired statement showing commutativity is pow4'(mod15(x)) = pow4'(x) = mod15(pow4'(x)) where that last equality is trivial because mod15 o mod15 = mod15 so mod15 o pow4' = mod15 o mod15 o pow4 = mod15 o pow4 = pow4' per associativity of function composition
- ChrisLomont 6y ago
  I'm a PhD in math. Modulo is a function in that it maps integer to integers. Power is a function mall mg integers to integers. Both are operators in the math sense, and one can talk about operators commuting. So the terms are correct. As far as programmers, modulus is an operator, complete with operator overloading in many languages and satisfying operator precedence. So operator is the correct word there also. Here's the idea from math https://mathoverflow.net/questions/20968/rules-for-operator-commutativity https://mathoverflow.net/questions/20968/rules-for-operator-...

No, it's called distributive property: https://en.m.wikipedia.org/wiki/Distributive_property https://en.m.wikipedia.org/wiki/Distributive_property

by konjin 6y ago

No, it's called distributive property: https://en.m.wikipedia.org/wiki/Distributive_property https://en.m.wikipedia.org/wiki/Distributive_property

henryfjordan 6y ago

After some further reading, I don't think anyone in this thread is correct. Modulo in math is not an operator like a programmer might think of it. See: https://math.stackexchange.com/questions/2832649/modular-arithmetic-and-the-distributive-property https://math.stackexchange.com/questions/2832649/modular-ari...

vitus 6y ago
Right. This is a consequence of compatibility with exponentiation. https://en.wikipedia.org/wiki/Modular_arithmetic#Properties https://en.wikipedia.org/wiki/Modular_arithmetic#Properties It should be easy to see that n ≡ (n % 15) (mod 15) which, applying compatibility of exp, then gives us n^4 ≡ (n % 15)^4 (mod 15) which can be rewritten in Python's notation as n**4 % 15 == (n % 15) ** 4 % 15
edflsafoiewq 6y ago
The injection mod15: Z -> Z/15 is an operation and the basic idea here is mod15(pow4(x)) = pow4(mod15(x)), so I think it's correct to call it commutativity.
- vitus 6y ago
  Well, not quite. You really want to say that mod15(pow4(x)) == mod15(pow4(mod15(x))). Example: (3 ** 4) % 15 = 81 % 15 = 6. But, (3 % 15) ** 4 = 3 ** 4 = 81. (That said, the two functions do commute when you restrict your domain and range to Z/15.) edit: also, I wouldn't consider mod15 to be an injection, as it's, um, not injective (it maps multiple inputs to the same output).
  - edflsafoiewq 6y ago
    pow4(mod15(x)) is already in Z/15, you can't apply mod15 to it. The point is pow4ing in Z and then reducing mod 15 is the same as reducing mod 15, then pow4ing in Z/15. It's not actually the same pow4 on the LHS and RHS (the LHS is in Z, the RHS in Z/15), but I think "commutative" still fits. > as it's, um, not injective Er, surjection :)
    - vitus 6y ago
      Well, how about this, if we use pow4' to be modular exponentiation, i.e. mod15 o pow4? (of course then changing mod15 to take Z -> Z so the types line up) Then the desired statement showing commutativity is pow4'(mod15(x)) = pow4'(x) = mod15(pow4'(x)) where that last equality is trivial because mod15 o mod15 = mod15 so mod15 o pow4' = mod15 o mod15 o pow4 = mod15 o pow4 = pow4' per associativity of function composition
ChrisLomont 6y ago
I'm a PhD in math. Modulo is a function in that it maps integer to integers. Power is a function mall mg integers to integers. Both are operators in the math sense, and one can talk about operators commuting. So the terms are correct. As far as programmers, modulus is an operator, complete with operator overloading in many languages and satisfying operator precedence. So operator is the correct word there also. Here's the idea from math https://mathoverflow.net/questions/20968/rules-for-operator-commutativity https://mathoverflow.net/questions/20968/rules-for-operator-...