Download 4 Questions on Analysis I With Solution - Midterm Exam | MATH 3150 and more Exams Mathematical Methods for Numerical Analysis and Optimization in PDF only on Docsity! Math 3150 Midterm 1 Instructions: You may refer to your notes and your textbook. No calculators are permitted. You have from 9:30am until 10:45am to complete the test. 1.[6 points] For x and n both in N define xn by setting x0 = 1 and xn+1 = (xn)x. Using only the results up to the end of Chapter 2 of your book, prove that xmxn = xm+n where x, n,m ∈ N. Solution: By induction. Fix m, induct on n, and let the statement P(n) be xmxn = xm+n. Base case: With n = 0 we have xmx0 = xm1 = xm = xm+0, so P(0) is true. Inductive step: Suppose P(n) is true. Then xmxn+1 = xmxnx = xm+nx = xm+n+1. 2.[6 points] Using only the results up to and including Definition 4.2.8 in your book, prove that if x, y, z ∈ Q and both x > y and z > 0 then xz > yz. Solution: By definition, x > y iff x − y > 0 iff x − y = rs where r, s ∈ N \ {0}. Also z > 0 iff z = p/q, p, q ∈ N \ {0}. Now xz − yz = (x − y)z = rs p q = rp sq where rp, sq ∈ N. Moreover, rp , 0 and sq , 0 by a result from the book (no zero divisors in N). Hence xz − yz is positive, so xz > yz. 3.[4 points] For the function f : Z → Z given by f (x) = x2, identify which of the statements below (if any) is/are false. For each false statement (if any) give examples of sets A and B that show the statement is false. I f (A ∪ B) = f (A) ∪ f (B). II f (A ∩ B) = f (A) ∩ f (B). III f (A \ B) = f (A) \ f (B). Solution: Statements II and III are false for this function. For both cases take A = {1}, B = {−1}. Then A ∩ B = ∅ but f (A) = f (B) = {1} so f (A) ∩ f (B) = {1} , ∅ = f (A ∩ B), so we have a counterexample for II. Also A \ B = A = {1} so f (A \ B) = {1}, but f (A) \ f (B) = ∅, so we have a counterexample to III. 4.[6 points] Suppose that w, x, y, z ∈ Q, and that x and z are 2-close, that x ≤ y ≤ z and that w is -close to y. Show that either w is -close to x or w is -close to z, or both. Solution: There are several ways to do this. One is to say |w−y| = d(w, y) ≤ implies − ≤ w−y ≤ which gives y − ≤ w ≤ y + (this is by a result in the text). Then using x ≤ y on the left and y ≤ z on the right, along with transitivity and invariance under addition implies x − ≤ w ≤ z + . Now let p = x+y2 , so d(x, p) = d(p, z) = |x−y| 2 ≤ . From this we have p− x ≤ |p− x| ≤ so p ≤ x+ and − ≤ p − z so z − ≤ p. Finally then p ≤ x + and z − ≤ p. From trichotomy, either w ≤ p or w ≥ p or both. Combining with the previous results we get that either x − ≤ w ≤ p ≤ x + , or z − ≤ p ≤ w ≤ z + , or both. In the first case, − ≤ w − x ≤ implies d(x,w) = |w − x| ≤ and in the second − ≤ w − z ≤ implies d(z,w) = |z − w| ≤ .
