# 3-4 - SOLUTIONS FOR HOMEWORK 2 3.3(iv Applying part(iii of...

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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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3-4 - SOLUTIONS FOR HOMEWORK 2 3.3(iv Applying part(iii of...

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