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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
by 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 agoWell, 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