4 ms·

Any proof you cannot have inverse. If so, may I wonder if your drop rule 2 and insist on having for most vector u there exist a v so that u*v = 1. As an icing

• gpm 3y ago
  This is a theorem that under approximately those rules* in R^n you cannot have inverses except in the cases of the reals, complexes, and quaternions: https://en.wikipedia.org/wiki/Frobenius_theorem_%28real_division_algebras%29 https://en.wikipedia.org/wiki/Frobenius_theorem_%28real_divi... *It doesn't rely on rule 2, it does however rely on (u*v)*w = u*(v*w). Adding rule 2 excludes the quaternions.
• gpm 3y ago
  By the way, the definition of a vector space means we have a 0 vector, and the distribute property means that multiplication has to respect it. v * w = v * (w + 0) # Vector space property that w + 0 = w = v * w + v * 0 # Distributive property Another vector space property guarantees that there exists -(v * w) such that -(v * w) + (v * w) = 0. Add it to both sides -(v * w) + v * w = -(v * w) + v * w + v * 0 0 = 0 + v * 0 # And then reduce based on the definition of -(v * w) 0 = v * 0 # Vector space property that 0 + x = x So you have your icing :)