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Unformatted text preview: SOLUTIONS FOR HOMEWORK 2 3.3. (iv) Applying part (iii) of Theorem 3.1 twice, we get: ( a )( b ) = ( a ( b ) ) = ( ( b ) a ) = ( ba ) = ab. (v) Multiply both sides of the equality ac = bc by c 1 . Then a = a ( cc 1 ) = ( ac ) c 1 = ( bc ) c 1 = b ( cc 1 ) = b. 3.4. (v) 1 = 1 1 = 1 2 0, by part (iv) of Theorem 3.2. However, 1 negationslash = 0 if the field F contains more than one element (this was proved in class). Thus, 1 > 0. (vii) Suppose 0 < a < b . By Theorem 3.2(vi), a 1 and b 1 are positive. As a 1 negationslash = b 1 (otherwise, we would have a = b ), it suffices to show that a 1 b 1 . By Axiom O5, a 1 b 1 a 1 0 = 0. Applying O5 once again, we see that a ( a 1 b 1 ) b ( a 1 b 1 ). By Axiom M1 (associativity of multiplication), a ( a 1 b 1 ) = b 1 . By Axioms M1 and M2 (commutativity and associativity of multiplication), b ( a 1 b 1 ) = a 1 . We conclude that a 1 b 1 . 3.5. (a) Suppose  b  a . Clearly, a 0. If b 0, then 0 b =  b  a , hence a b a . If b 0, then automatically, b a . On the other hand,  b  = b a , hence, by Theorem 3.2(i), b = ( b ) a . Thus, a b a . Now suppose a b a . If b 0, then  b  = b a . If b 0, then  b  = b ( a ) = a . (b) Let x = b and y = a b . Then, by the triangle inequality,  a  =  x + y   x  +  y  =  b  +  a b  ....
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 Winter '08
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