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Analysis Homework #1 Solutions 1. Determine max A when A = {y ∈ R : y = −3x2 + 2x − 1 for some x ∈ R}. • Completing the square, we find that every element of A has the form ( ) ( )2 2x 1 1 1 1 2 2 2 −3x + 2x − 1 = −3 x − + − + = −3 x − − . 3 9 9 3 3 3 Using this fact, it is now easy to see that −2/3 is the largest element of A. n−1 2. Show that the set B = { 2n+1 : n ∈ N} has no maximum. • To see that B has no maximum, suppose x = y= n−1 2n+1 is the maximum and let (n + 1) − 1 n = . 2(n + 1) + 1 2n + 3 Then y is an element of B, and it is actually larger than x because n−1 n < 2n + 1 2n + 3 ⇐⇒ 2n2 + n − 3 < 2n2 + n ⇐⇒ −3 < 0. As no element of B can be larger than the maximum of B, this is a contradiction. n−1 : n ∈ N} is such that sup B = 1/2. 3. Show that the set B = { 2n+1 • To show that 1/2 is an upper bound of the given set, we note that n−1 1 ≤ 2n + 1 2 ⇐⇒ 2n − 2 ≤ 2n + 1 ⇐⇒ −2 ≤ 1. Since the rightmost inequality is true, the leftmost one must also be true. • To show that 1/2 is the least upper bound, we show that no number x < 1/2 is an upper bound. Let us then fix some x < 1/2 and try to find an element of B which is n−1 bigger than x. Since every element of B has the form 2n+1 , we need to ensure that n−1 >x 2n + 1 ⇐⇒ n − 1 > 2nx + x ⇐⇒ (1 − 2x)n > x + 1. Noting that 1 − 2x is positive whenever x < 1/2, we end up with the inequality n> x+1 . 1 − 2x According to one of our theorems, we can always find an integer n that satisfies this n−1 inequality. Then, our computation above shows that 2n+1 is an element of B which is bigger than x. This also means that x is not an upper bound of B, as needed. 4. Show that the set C = {x2 : x ∈ R} has no upper bound. • If y ∈ R is an upper bound of C, then y ≥ y 2 and also y ≥ 22 > 1. Multiplying the last inequality by the positive number y, we now get y 2 > y, a contradiction.