Math 55 - Fall 2001 - Poonen - Practice Midterm

Math 55 - Fall 2001 - Poonen - Practice Midterm - WED 16:43...

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Unformatted text preview: 05/15/2002 WED 16:43 FAX 6434330 MOFFITT LIBRARY 001 Prof. Bjorn Pooncn October 5, 2001 MATH 55 PRACTICE MIDTERM Do not write your answers on this sheet. Instead please write your name, your student ID, your TA’S name, your section time, and all your answers in your blue books. In general, you must show your work to get credit. Total: 100 pts., 75 minutes. (1) For each of (a)~(g) below: If the proposition is true, write TRUE. Tt‘ the propo- sition is false, write FALSE. (Please do not use the abbreviations T and F, since in handwriting they are sometimes indistiguishable.) No explanations are required in this problem. (a) The circle {(mgy) : a: E R, y E R, and 322 + y2 2 1} is a countable set. (b) The propositions p —5 -q and q —> -1p are logically equivalent. (c) If p is a prime number, then there is a prime q satisfying p < q < p + 6. (d) There exists a. positive integer m such that 30, 77, m are pairwise relatively prime. (e) In the RSA cryptosystem, when Bob wants to send Alice a message, it is essential that Bob keeps his encryption method secret. (f) 3405 E 13 (mod 7]. (g) Tt is valid to deduce fiq, if -Ip and p —) q have been proved already. (2) Find a compound proposition involving p, q, 7’ that is true if and only if at least two of p, q, r are false. (3) Let S be the subset ofN defined recursively by 0 7 E S o 107i: 6 5 whenever it E S, and ok—ZESwhenever kESande2. Describe S, and determine whether 107 is in S. (4) For which real numbers r is it true that (filinl3 + 6n2(log n)3)(5/n + 7/ log n) is 0(71”)? Explain. (5) Find the set of integer solutions to the system of congruences a: E 4 (mod 7) and m E 6 (mod 13). (6) Prove by induction that 2“ > 101; holds for sufficiently large positive integers n . (7) Prove that x/é is irrational. (8) Prove or disprove the following statement: if f is an injective function from A to B, and g is a surjective function from B to C, then 9 o f is a bijective function from A to C. This is the end! At this point, you may want to look over this sheet to make sure you have not omitted any problems. 05/15/2002 WED 16:44 FAX 6434330 MOFFITT LIBRARY 002 P4094155) Fall 200i J @ern looms Math 55 Practice Midterm Solutions (prepared by Nick Meyer) 1a. FALSE. There is a one—to—one correspondence between the halfeopen interval H = [0, l) C R and the circle C C 1R2; this correspondence can be demonstrated by the bijective function f : H —> C that sets f(.'c) : (cos 27rm,sin 27m) for :c E H. Since H is a nonempty interval of the real numbers, it is uncountable; since C is in oneto-one correspondence with H, it must also be. lb. TRUE. The two statements are contrapositives of each other, hence logically equivalent. 1c. FALSE. Let p = 23; there is no prime 9 strictly between p and p + 6 : 29. (Note: there is nothing special about 6 here. For any positive 72, we have that 2, 3,4, . . . , n all divide nl; hence 2|(n! + 2), 3l(nl -|— 3), . . .,n|(nE + We thus have constructed a sequence of n — 1 consecutive composite numbers 7 namely, 7;! + 2 through 77.! + 1d. TRUE. 30 = 2 X 3 x 5; 77 : 7 X 11. We must merely find an m which does not have any of 2, 3, 5, 7, 11 as a factor. 13 naturally suggests itself. 1e. FALSE. Indeed, everyone knows how Bob encrypted his message. To encrypt message M, Bob calculates ciphertext C = M 9 mod n and sends C to Alice. 7?. and e comprise Alice’s “public key”; as the name suggests, they are public information. 1f. TRUE. By Fermat’s Little Theorem, 36 E 1 (mod 7). Hence 3405 = ((367)6)(33) (mod 7). (The last congruence follows because 7|(27 -— l 3); equivalently, 27 mod 7 6. 1g. FALSE. pr is false and q is true, then both —up and p —) q are true, but fig is not. 03 33227:1 13 mod 7: 2. It is easy to construct such a compound proposition in disjunctive normal form (see Section 1.2, Exercise 26). (p /\ -Iq /\ -14) V (10 /\ q A fir) V (op /\ -Iq /\ r) V (-‘p A sq /\ or) will do nicely. 3. 5 consists of all positive even integers, and also the numbers 1, 3, 5,7. Hence 107 is not in 3. Proof: Rules 1 and 3 together tell us that l, 3, 5,7 are all in 5. Rule 2 says that a. * lUk is in S, for a = 1, 3, 5, 7 and Is: any positive integer. These numbers can become arbitrarily large, and are all even. Rule 3 now implies that all positive even numbers are in S. Multiplying an even number by ten yields an even number, so applying Rule 2 gives us nothing new; we’re done. 4. The given expression is 0(71’") for all real 7' 2 3. To prove this, we first recall that log n is 0(n‘l) for any 16 > O, and equivalently that (log n.)l is 0(n) for any I > 0. Now we note that 4((nl)3 < 4(n+l)3, so is certainly 0(n3); however, since it’s equal to 4723 for integer it, it’s certainly not 0(n3). By our above recollection, 6n.2(log n)3 is 0(n3). Thus the first term in the given expression is 0(n3) (though not 0013)). Let us examine the second term. 5/77. is 0(n'1; 7/ logo is 0(n0), but — again, by our above recollection — it isn’t 0(nk) for any k < 0. Thus the second term is 001°). The entire expression is 0(n3 * no) 2 0(n3), but not 0(nk) for any It < 3. 5. Using the notation of Theorem 4, Section 2.5, we have: m = 91, m1 = 7, m2 = 13, a1 = 4, e2 : 6. Thus M1 = m/m1 = 13 and M3 = m/mg = 7. 13y, E 1 (mod 7) 4—} iyl E 1 (mod 7), so yl can be —1. 7:92 E 1 (mod 13); y; = 2 is an obvious solution. Thus in : aiMiyi +egng2 = -—52+84 : 32 is a solution to the system of congruences; the general solution is all integers m such that :29 E 32 (mod 91). Alternatively, we can find M1 and M2 as above, and then posit that r. = (51141 + ng for some coefficients a, b. Plugging this back into the original system of congruences allows us to find suitable a, b and leads us quickly to the same solution we got above. 6. Claim: 2" > 1073 for all n 3 6. We prove this by induction on n. The base case is n = 6, and indeed 26 = 64 > 10 x 6 = 60. Now, if 2" > 1071, then 2”+1 : 2(2”) > 2(10n) > 10(n + 1), with the last inequality holding as long as n > 1 (so it certainly holds when n 2 6). This suflices to prove the inductive step, so we’re done. 7. Claim: x/E is irrational. 05/15/2002 WED 16:44 FAX 6434330 MOFFITT LIBRARY 003 We prove this by contradiction. Assume 6 is rational; then it can be expressed as (1/1), Where a and b are integers sharing no common factor larger than one. Then 6 = 02/52, so a2 = 652. Well, 2|662, so 2|a2, so we must have 2|a. But this means that 4|u2, which means that 4|662, which implies that 2|b2, which means that 2|b. Thus a and b have the common factor of 2; we have a contradiction! Thus our initial assumption is false, and we’re done. 8. This statement is false. Let A,B, (3' all be Z, the set of integers. Let f : Z —> Z be defined by 2 2:3 (this is injective, but not surjective); let 9 : Z r >— Z be defined by g(m) : m (this is bijective, so is certainly surjective). Then (g o : 23:, so 9 o f is not surjective. ...
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