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Solutions – §4.2 8. The set of all ordered pairs of real numbers with the operations (x, y)⊕(x0, y 0) = (x+x0, y +y 0) and r (x, y) = (x, ry). The distributive property (r + s) (x, y) = r (x, y) ⊕ s (x, y) does not hold as can be seen by the counterexample below. (2 + 3) (2, 3) = 5 (2, 3) = (2, 15) (2 (2, 3)) ⊕ (3 (2, 3)) = (2, 6) ⊕ (2, 9) = (4, 15) 12. Let V be the set of all positive real numbers; define ⊕ by u ⊕ v = uv and define by c v = v c . Prove that V is a vector space. Let a, b and c be positive real numbers and let s and t be real numbers. If a and b are positive real numbers then a ⊕ b = ab is a positive real number. Thus V is closed with respect to ⊕. Similar as is a positive real number, so V is closed with respect to scalar multiplication. (a) a ⊕ b = ab = ba = b ⊕ a. (b) a ⊕ (b ⊕ c) = a ⊕ bc = a(bc) = (ab)c = (a ⊕ b) ⊕ c (c) a ⊕ 1 = a(1) = a. Thus the 0 element in V is 1. (d) a ⊕ 1 a = 1. Thus the negative of a is a1 . (a) s (a ⊕ b) = s ab = (ab)s = as bs = (s a) ⊕ (s b) (b) (s + t) a = as+t = as at = (s a) ⊕ (t a) (c) s (t a) = s at = (at )s = ast = (st) a (d) 1 a = a1 = a There V is a vector space. 24. Prove that if u ⊕ v = u ⊕ w, then v = w. u⊕v=u⊕w −u ⊕ u ⊕ v = −u ⊕ u ⊕ w 0⊕v = 0⊕w v=w