Answer
Showing that F(x, y) = x ⊕ y is not a threshold function.
Work Step by Step
Suppose it were with weights a and b.
--Then there would be a real number T such that xa + yb ≥ T
for (1,0) and (0,1),
-but with xa + yb < T for (0,0) and (1,1).
Hence,
-- a ≥ T, b ≥ T, 0 < T, and a + b < T.
Thus, a and b are positive,
-- which implies that a + b > a ≥ T , a contradiction.
