This preview shows page 1. Sign up to view the full content.
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 onetoone 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, 2662, so 2a2, so we must have 2a. But this means that 4u2, which means that 4662, which
implies that 2b2, which means that 2b. 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. ...
View
Full
Document
This test prep was uploaded on 04/01/2008 for the course MATH 55 taught by Professor Strain during the Spring '08 term at Berkeley.
 Spring '08
 STRAIN
 Math

Click to edit the document details