MAA 3200 Midterm: MAA 3200 FIU Exam 3k
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The true-false was better this time, and the part 5) proofs were good. The proofs for 3) and 4) were weak - make sure to get these [and other problems from hw6] right before the. Q is an ordered eld, but is not complete. s2(n) = n + 2 misses 0 and 1. : [(a, b)][(c, d)] = [(ac + bd, ad + bc)] = [(ca + db, cb + da)] = And n such that n, m > n |sn sm| < /2. So, if m n then |sm l| |sm snk| + |snk l| < /2 + /2 = . 5 a) assume l > m to get a contradiction. Use the def"n of limit (with /2) to get sn l > /2. Then add l = m + to both sides, to get a contradiction: draw a b as we did for q, and a path in it, see goldberg.
