Solution of 2 Problems - Analysis I - Assignment | MATH 554, Assignments of Mathematics

Material Type: Assignment; Professor: Sharpley; Class: ANALYSIS I; Subject: Mathematics; University: University of South Carolina - Columbia; Term: Spring 2004;

Typology: Assignments

Pre 2010

Uploaded on 10/01/2009

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Math 554- 703 I - Analysis I
Solutions for Problems 1-2 of Homework Assignment # 1
1. Using the field axioms, prove that for each aโˆˆF, a 6= 0, the multiplicative
inverse of ais unique.
Solution: Let abe an arbitrary member of Fand suppose that ahas multiplicative inverses band c, that is
a b = 1 (1)
a c = 1 (2)
In this case, we can multiply both sides of equation (1) by cto obtain
(a b)c= 1 c
Using the field axioms, this reduces to
b(a c) = c
and so b=cusing equation (2) and the identity axiom that b1 = b.2
2. Using any of the results we proved in class (before the statement of this result)
carefully prove that (-1)(-1)=1.
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
โˆ’(โˆ’a) = a. (3)
We also know for each aโˆˆFthat
(โˆ’1) b= (โˆ’b)
the additive inverse of b. Therefore applying this with b=โˆ’1 and using equation (3) we obtain
(โˆ’1)(โˆ’1) = โˆ’(โˆ’1) = 1 2

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Math 554- 703 I - Analysis I

Solutions for Problems 1-2 of Homework Assignment # 1

1. Using the field axioms, prove that for each a โˆˆ F, a 6 = 0, the multiplicative

inverse of a is unique.

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

2. Using any of the results we proved in class (before the statement of this result)

carefully prove that (-1)(-1)=1.

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 โˆ’(โˆ’a) = a. (3) We also know for each a โˆˆ F that (โˆ’1) b = (โˆ’b) the additive inverse of b. Therefore applying this with b = โˆ’1 and using equation (3) we obtain (โˆ’1)(โˆ’1) = โˆ’(โˆ’1) = 1 2