### Students' answers to review problems for the first exam in Math 311, spring 2003

 #1 Beckhorn #2 Benson #3 Chan #4 Chang #5 #6 Cohen #7 #8 Greenbaum #9 #10 Hedberg #11 #12 #13 #14 Oleynick #15 #16 #17

 1.Prove that 2n-3 <= 2^(n-2) for all n >= 5, n in N. Base Case: Take n=5. Then it is true that 2(5)-3 <= 2^(5-2), or 7<=2^3=8. Induction Step: Assume the inequality in the proposition (denote it P(n)) is true for n=k>=5, so 2(k)-3<=2^(k-2). Multiply each side of the last inequality by 2. Note that the direction of the inequality doesn't change because 2>0. I get 2(2k-3)<=2^(k-2)*2=2^((k+1)-2). In the induction step I want to show P(k+1) is true, or 2(k+1)-3<=2^((k+1)-2). Since I know 2(2k-3)<=2^((k+1)-2), it suffices to show 2(k+1)-3<=2(2k-3). Simplifying this last inequality I get 2k-1<=4k-6, or k>=5/2. I know that k>=5>5/2, so I have arrived at a true statement and P(k+1) is therefore true. Sent by Graham Beckhorn on Tue, 25 Feb 2003 21:57:57

 2. Prove using the definition of limit that lim((1+2n^2)/(3+4n^2))=1/2. According to definition 3.1.3. A sequence is said to converge to x if for every eps > 0, there exists k in N s.t. for all n>=k, the terms xn satisfy |x_n-x| (1/(8eps))^(1/2) > 0. Note that since eps > 0, the square root of 1/(8eps) exists, and by the Archimedean Property, we can find an integer greater than it. Now if k > (1/(8eps))^(1/2), then for all n >= k, n > (1/(8eps))^(1/2). Now, by 2.1.13 n^2 > (1/(8eps)). Multiplying through by 8, we have 8n^2 > 1/eps (since 8>0). Note that this implies that 8n^2 + 6 > 1/eps. Since eps and 8n^2+6 are positive, this implies that eps > 1/(8n^2+6). Since 1/(8n^2+6) is positive, eps > |(1/(8n^2+6))|, by definition of absolute value. Now, note that 1/(8n^2+6) = -[ (1+2n^2)/(3+4n^2) - 1/2 ]. So, eps > |(1+2n^2)/(3+4n^2)-1/2| for all n >=k. Thus, by the definition of the limit, (1+2n^2)/(3+4n^2) converges to 1/2. Sent by Michael Benson on Sun, 2 Mar 2003 21:32:19

 3. Suppose a,b are in R and suppose that for every epsilon>0 we have a <= b+ epsilon. a) Prove that a <= b. b) Show by example that it does not follow that a < b. Since we want to prove that a <= b, we will try to contradict using a>b. Using a <= b + epsilon, we let epsilon be (a-b)/2. Plug this in and we get: a <= b + (a-b)/2 so that a <= (b+a)/2. (#) Suppose a > b [assumption]. Then a/2 > b/2 (*) [multiplication by a non negative number, so inequality holds)]. Since a/2 + a/2 = a [another way of writing a] using * we get a/2 + b/2 < a (**). (a+b)/2 < a (rearranged **) [this contradict #] we have a contradiction. So a <= b. Q.E.D. b) Since 3<=3+epsilon for all positive epsilon, we can take a=3 and b=3, so that a does not have to be less than b for the hypotheses to hold. Sent by Man Wai Chan on Tue, 4 Mar 2003 22:17:52

 4.If A and B are any nonempty sets of numbers which have the property that a <= b for all a in A and all b in B, prove: a) sup A <= b for all b in B. b) sup A <= inf B. Proof of a): Let z be any upper bound of A and let u be sup A. From the definition of Supremum, u <= z. Also by the definition of Supremum, for given b in B, a <= b for all a in A, so b is an upper bound of the set A for all b in B. Thus by the Completeness Axiom, sup A exists. This means u <= b which means to sup A <= b, for all b in B. Proof of b): Since sup A <= b for all b in B (from above), sup A is a lower bound of set B. By the Completeness Axiom, inf B must exist. All lower bounds of B are less than or equal to inf B. This means sup A, which is one of the lower bounds, is also less than or equals to inf B. Thus, sup A <= inf B. Sent by "Soyun Chang on Tue, 4 Mar 2003 17:50:12

 6. Prove that if lim(x_n) = x and if x >, then there is a positive integer M such that x_n > 0 for all n gt;=M. Since lim(x_n) = x for all epsilon > 0, there exists a K(epsilon) that belongs to N such that if n >= K(epsilon), then |x_n-x| epsilon. Since x > 0, let x = epsilon and let M = K(epsilon). Then for n >= M, |x_n-x|

 8. Suppose (a_n) and (b_n) are any sequences of real numbers with the following properties: (a_n) converges, and lim(a_n) is 0. (b_n) is bounded above by 3 and below -3: so |b_n|<=3 for all n in N. Prove that the sequence (c_n) with c_n defined by c_n := a_n · b_n converges, and that lim(c_n) = 0. We know that (a_n) converges to 0 so 0 <= abs(a_n*b_n) = abs(a_n)abs(b_n). We also know that abs(b_n) <= 3 and since abs(a_n) is nonnegative then abs(a_n)abs(b_n)<= 3abs(a_n) hence: 0 <= abs(a_n)abs(b_n)<= 3 abs(a_n). Since 3 abs(a_n)-->0 the Squeeze Theorem applies, and therefore |a_n*b_n| converges, with limit 0. Sent by Jackie Greenbaum on Tue, 4 Mar 2003 20:20:36 Further comment by the management Since |&nbps;|"a number"|-0 |=|"a number"-0| it seems apparent that a sequence (w_n) converges to 0 if and only if the sequence (|w_n|) converges to 0. This result is being used implicitly in the solution above, I believe. Another different solution to the problem can be written ("... not that there's anything wrong with" the one that is given). Since (a_n) converges to 0, given blah>0 there is K(blah) in N so that if n>=K(blah), then |a_n-0|&*lt;0. But we need to show that if epsilon>0, there is some element of N so that when n is bigger than it, |c_n-0|=K(epsilon/3), |c_n-0|=|a_n·b_n|<(epsilon/3)·3=epsilon, so we are done.

 10. Let (x_n) be an infinite sequence. Assume lim(x_n)=L. Let M be a positive integer and suppose S={x_n:n>=M}. Prove that inf S<=L. Let's prove this via contradiction; assume that inf S > L. Firstly, inf S exists because S in non-empty (x_M is a member of S) and all convergent sequences are bounded, so S is bounded. Because (inf S - L) > 0, let's take epsilon = (inf S - L). Then for n >= K(inf S - L), we have |x_n - L| < (inf S - L). Unrolling this inequality yields: -(inf S - L) < x_n - L < (inf S - L). Now if we add L to all sides we get (2L - inf S) < x_n < inf S. But it is not the case that x_n < inf S: contradiction! Therefore, inf S <= L. Sent by Joe Hedberg on Sat, 1 Mar 2003 13:45:51

 14. Let a, b, c, d be real numbers such that a < b and c < d. Prove the inequality bd + ac > ad + bc. If a < b and c < d, then by the definition of greater than, b - a is an element of P, the set of positive numbers, and d - c is an element of P. By the order properties, the product of two positive numbers is also positive so (b - a)(d - c) is also an element of P. Multiplying out those terms gives bd - bc - ad + ac as an element of P. By the commutative property, bd - bc - ad + ac = bd + ac - ad - bc and by the distributive and associative properties, bd + ac - ad - bc = (bd + ac) - (ad + bc), which is an element of P. If (bd + ac) - (ad + bc) is an element of P then by the definition of greater than, bd + ac > ad + bc. Sent by John Oleynick on Fri, 28 Feb 2003 16:23:51