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hw1-sol

hw1-sol - Modern Analysis Homework 1 Possible solutions 1(1...

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Modern Analysis, Homework 1 Possible solutions 1. (1 pt) Let k be a field and x k be non-zero. Show that 1 / (1 /x ) = x . Solution: First of all, notice that multiplicative inverses are unique in fields: if x k \ { 0 } 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 α R such that for all x A we have x α . Therefore for all - x ∈ - A we have - x ≤ - α . 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 . Let’s 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 let’s check item 2. Let β be a lower bound for A . Then by similar arguments from above, - β is an upper bound for - A . Since sup( - A ) is the least upper bound for - A , we have - β sup( - A ) and hence β ≤ - 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 nq -th 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 0 = 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

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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 . With this definition of integral powers, the following is true Fact 1. Let ( F, < ) be an ordered field and a > 1 . Then for any n N , a n > 1 . Proof of Fact 1. This is true for the following reason: if n = 1 it is true by assumption. But then since F is ordered, it follows that a · a > a , which in turn is greater than 1. Similarly we get that a · a · a > a · a > a > 1, again since ( F, < ) is an ordered field. And so on...
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