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Unformatted text preview: Modern Analysis, Homework 1 Possible solutions 1. (1 pt) Let k be a field and x k be nonzero. Show that 1 / (1 /x ) = x . Solution: First of all, notice that multiplicative inverses are unique in fields: if x k \ { } and y,z k satisfy xy = xz = 1, then y =1 y =( xy ) y =( xz ) y =( zx ) y by commutativity of k = z ( xy ) by associativity of k = z (1) = z Therefore, there is a unique element, which we denote by (1 /x ), in k such that x (1 /x ) = 1. Notice that this equation shows that (1 /x ) 6 = 0 (if it were, then x (1 /x ) = x 0 = 0, but 1 6 = 0). It also shows that x satisfies the defining property of (1 / (1 /x )), and by uniqueness of inverses we get x = 1 / (1 /x ). 2. (3 pts) Rudin, Ch 1, problem 5. Solution: Let A be a nonempty set of real numbers which is bounded below. Let A be the set of all numbers x where x A . By assumption, there exists R such that for all x A we have x . Therefore for all x  A we have x  . Therefore A is bounded above. A is also nonempty since A is. By the completeness of R , sup( A ) exists. We claim that sup( A ) is actually inf( A ). To justify this, we need to check two things: 1. sup( A ) is a lower bound for A , and 2. sup( A ) is the greatest lower bound for A . Lets check item 1 first. Let x A . Then x  A and since sup( A ) is an upper bound for A we have that x sup( A ) and hence x  sup( A ). By definition, this means that sup( A ) is a lower bound for A . Now lets check item 2. Let be a lower bound for A . Then by similar arguments from above, is an upper bound for A . Since sup( A ) is the least upper bound for A , we have sup( A ) and hence  sup( A ) and item 2 is proved. 3. (5 pts) Rudin, Ch 1, problem 6. Solution: Fix b > 1. (a) Let m,n,p,q Z , n > 0, q > 0, and r = m/n = p/q . Then, by definition, ( b m ) 1 /n is the unique positive real such that (( b m ) 1 /n ) n = b m . Similarly, ( b p ) 1 /q is the unique positive real such that (( b p ) 1 /q ) q = b p . To prove the equality ( b m ) 1 /m = ( b p ) 1 /q , we will show that both sides are equal to the nqth root of b mq = b pn (these are equal since m/n = p/q ). To prove this, we will use some facts about integral powers. Begin digression: let F be a field and n N . Then we define x n to mean x x x ( n times). Similarly, we define x = 1 F . If n Z is negative, we now define x n to be 1 /x n . One can check that the following always hold: 1 1. For all m,n Z and x F , x mn = ( x m ) n 2. For all m,n Z and x F , x m + n = x m x m 3. For all n Z and x,y F , ( xy ) n = x n y n ....
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This note was uploaded on 10/18/2011 for the course MATH S4061X taught by Professor Peters during the Spring '11 term at Columbia.
 Spring '11
 Peters

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