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Material Type: Assignment; Professor: Sharpley; Class: ANALYSIS I; Subject: Mathematics; University: University of South Carolina - Columbia; Term: Spring 2004;
Typology: Assignments
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Solution: Let a be an arbitrary member of F and suppose that a has multiplicative inverses b and c, that is a b = 1 (1) a c = 1 (2) In this case, we can multiply both sides of equation (1) by c to obtain (a b) c = 1 c Using the field axioms, this reduces to b (a c) = c and so b = c using equation (2) and the identity axiom that b 1 = b. 2
Solution: We know from our previous results which were established in class that the additive inverse of the additive inverse of an element is the element itself โ(โa) = a. (3) We also know for each a โ F that (โ1) b = (โb) the additive inverse of b. Therefore applying this with b = โ1 and using equation (3) we obtain (โ1)(โ1) = โ(โ1) = 1 2