Solutions to Practice Problems for Midterm 2 - Discrete Structures | MATH 455, Exams of Discrete Structures and Graph Theory

Material Type: Exam; Class: Int Discrete Strctrs; Subject: Mathematics; University: University of Massachusetts - Amherst; Term: Spring 2005;

Typology: Exams

Pre 2010

Uploaded on 08/19/2009

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SOLUTIONS TO PRACTICE PROBLEMS FOR MIDTERM #2
MATH 455.2 SPRING 2005
(1) Show that the sequence 0,1,3,7,...,2n1,... satisfies the recurrence relation
cn= 2cn1+ 1 for all n1.
Solution: 8.1 #11
(2) Prove that Fn= 3Fn3+2Fn4for all n4, where Fndenotes the n-th Fibonacci
number.
Solution: 8.1 #26
(3) With each step you take when climbing a staircase, you can move up either one stair
or two stairs. As a result, you can climb the entire staircase taking a combination of
steps that are either one- or two-stair increments. Let cnbe the number of different
ways to climb a staircase with nstairs (n1). Find a recurrence relation for cn.
Find the general formula for cn.
Solution: 8.1 #39. For the general formula compare this sequence with the Fi-
bonacci sequence.
(4) Use iteration method to guess an explicit formula for the sequence given recursively
by ak=ak1
2ak11for all k1 and a0= 2 . Prove that your guess is correct using
induction.
Solution: 8.2 #43
(5) Solve the SOLHRRCC: rk= 2rk1rk2for k2 and r0= 1 , r1= 4 .
Solution: 8.3 #13
(6) If f:RRand g:RRare both 1-1, is f+g:RRalso 1-1? If f:RR
and g:RRare both onto, is f+g:RRalso onto? If f:RRis 1-1 and
c > 0, is c·f:RRalso 1-1?
Solution: 7.2 #32, 33, 34
(7) How many integers must you pick to be sure that at least two of them have the
same remainder when divided by 7?
Solution: 7.3 #17
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SOLUTIONS TO PRACTICE PROBLEMS FOR MIDTERM

MATH 455.2 SPRING 2005

(1) Show that the sequence 0, 1 , 3 , 7 ,... , 2 n^ − 1 ,... satisfies the recurrence relation cn = 2cn− 1 + 1 for all n ≥ 1. Solution: 8.1 #

(2) Prove that Fn = 3Fn− 3 +2Fn− 4 for all n ≥ 4, where Fn denotes the n-th Fibonacci number. Solution: 8.1 #

(3) With each step you take when climbing a staircase, you can move up either one stair or two stairs. As a result, you can climb the entire staircase taking a combination of steps that are either one- or two-stair increments. Let cn be the number of different ways to climb a staircase with n stairs (n ≥ 1). Find a recurrence relation for cn. Find the general formula for cn. Solution: 8.1 #39. For the general formula compare this sequence with the Fi- bonacci sequence.

(4) Use iteration method to guess an explicit formula for the sequence given recursively by ak = (^2) aakk−− 11 − 1 for all k ≥ 1 and a 0 = 2. Prove that your guess is correct using induction. Solution: 8.2 #

(5) Solve the SOLHRRCC: rk = 2rk− 1 − rk− 2 for k ≥ 2 and r 0 = 1, r 1 = 4.

Solution: 8.3 #

(6) If f : R → R and g : R → R are both 1-1, is f + g : R → R also 1-1? If f : R → R and g : R → R are both onto, is f + g : R → R also onto? If f : R → R is 1-1 and c > 0, is c · f : R → R also 1-1?

Solution: 7.2 #32, 33, 34

(7) How many integers must you pick to be sure that at least two of them have the same remainder when divided by 7? Solution: 7.3 # 1

2 SOLUTIONS TO PRACTICE PROBLEMS FOR MIDTERM #2 MATH 455.2 SPRING 2005

(8) There are 50 students in your class. All of them are known to be between 18 and 30 years old. You want to make a bet that the class contains at least k students of the same age. How large can you make k and yet be sure to win your bet? Solution: 7.3 # (9) On a square tray with side 1 meter you place 51 dimes. Prove that no matter how you place them there will be three dimes that can be covered with a plate of radius 15 cm. Solution: Divide the square into 25 squares of side 20 cm each. By the PH principle there will be a square which contains 3 dimes. Cover this square with the plate (the diagonal of the square is about 20*1.4=28 cm and the diameter of the plate is 30 cm). (10) Let f : X → Y and g : Y → Z. If g ◦ f is 1-1 must f be 1-1? What about g? Prove or give a counterexample. Solution: 7.4 #18. The function g may not be 1-1. For example: X = { 1 , 2 }, Y = { 1 , 2 , 3 }, Z = X = { 1 , 2 }, f (1) = 1, f (2) = 2, g(1) = 1, g(2) = g(3) = 2. Then the composition g ◦ f is identity on X , but g is not 1-1. Draw an arrow diagram! (11) Let f : X → Y and g : Y → Z. If g ◦ f is onto must f be onto? What about g? Prove or give a counterexample. Solution: 7.4 #19. The function f may not be onto. See the above example. (12) Prove that any infinite set contains a countable infinite subset. Solution: 7.5 #26. (13) Show that the set of all finite sequences of 0’s and 1’s is countable. Solution: Here is how you can enumerate all finite sequences of 0’s and 1’s. Let the empty sequence be number 1, length one sequences (0 and 1) be number 2 and 3, length two sequences (00, 01, 10, 11) be number 4, 5, 6, and 7, and so on (in general, length n sequences will be enumerated by 2n, 2 n^ + 1,... , 2 n+1^ − 1). Notice that this is equivalent to adding 1 in front of the sequence and taking the integer number whose binary representation is this augmented sequence.