No, this is true, proof by induction: (x+y)^2 = x^2 + y^2 For (x,y) = (0,0) : (x+y)^2 = (0+0)^2 = (0)^2 = 0 x^2 + y^2 = 0^2 + 0^2 = 0 + 0 = 0 Q. E. D. Thus it should follow that it applies to every other case. (The other cases are left as an exercise for the reader.)
From the first one you get (a = -1/sqrt(2), b=1/sqrt(2)) that 0=1. Thus the field of real numbers is trivial. It's thus not surprising that every function into it is linear lol.
No, this is true, proof by induction:
(x+y)^2 = x^2 + y^2
For (x,y) = (0,0) :
(x+y)^2 = (0+0)^2 = (0)^2 = 0
x^2 + y^2 = 0^2 + 0^2 = 0 + 0 = 0
Q. E. D.
Thus it should follow that it applies to every other case. (The other cases are left as an exercise for the reader.)
save