Spyke

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3 replies

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.)

8

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.

3

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