Solutions to Homework 9, Math 55
2. The generating function is 1 + 4x + 16x2 + 64x3 + 256x4 =
8. (a) Since (x2 + 1)3 = (1 + x2 )3 =
an = 0.
(b) Similarly, (3x 1)3 =
otherwise an = 0.
(c) We have 12x2 =
then an = 0.
Solutions to Midterm II. Discrete Mathematics 55
Instructor: Zvezdelina Stankova
Problem 1 (20pts). True or False? To discourage guessing, the problem will be graded as follows:
2 pts for each correct answer. 0 pts for a blank. -2 pts for each incorrect
Rob Bayer Instructions
Math 55 Worksheet
June 22, 2009
Introduce yourselves! Despite popular belief, math is in fact a team sport! Find some blackboard space, a piece of chalk, and decide who will be your first scribe. Do the problems below, having a dif
Solutions to Midterm I. Discrete Mathematics 55
Instructor: Zvezdelina Stankova
Problem 1 (20pts). True or False?
To discourage guessing, the problem will be graded as follows:
2 pts for each correct answer.
0 pts for a blank.
-2 pts for each incorrect an
MATH 55 1/29/14
(a) Use a direct proof to show that the product of two rational numbers is rational.
(b) Prove or disprove that the product of two irrational numbers is irrational.
Solution: (a) Left to you. (b) The product of two irrational numbers is
Math 55: Discrete Mathematics
UC Berkeley, Fall 2011
Homework # 7, due Wedneday, March 14
Happy Pi Day!
(If any errors are spotted, please email them to morrison at math dot
berkeley dot edu.)
6.5.10 A croissant shop has plain croissants, cherry croissant
Problem 1. Show that the set of real numbers that are solutions of quadratic equations
ax2 + bx + c = 0, where a, b, c are integers, are countable.
Goal: I want show the set of quadra
MATH55 HOMEWORK #10
I can be reached at [email protected]
1. An Introduction to Discrete Probability
7.1.8 What is the probability that a ve-card poker hand contains the ace of hearts?
Solution: There are 52 dierent hands of ve cards
HW 2 Solutions, MATH 55, Spring 2016
1.7, 1 Let 2k + 1 and 2j + 1 be two odd integers. Then 2k + 1 + 2j + 1 = 2(j + k) + 2 = 2(j + k + 1),
which two divides, and so is even.
1.7, 8 Let n = a2 and suppose n+2 = b2 . We can assume that a, b are posiiv
Math 55 Homework 9 Solutions
(For odd-numbered problems, see the back of the book.)
20. There is no way to pay negative pesos, so xn = 0 for n < 0. There is one way to pay zero pesos (no coins), so
x0 = 1. For n > 0, we just subtr
Math 55 Homework 8 Solutions
2. We have p(3) = 2p(n) for each n 6= 3, and p(3) + 5p(n) = 1 since each outcome besides 3 is equally likely and
there are five such other outcomes. Solving, we get p(n) = 17 for each n 6= 3 and p(
Math 55 Homework 10 Solutions
Due April 23
22. Every asymmetric relation is antisymmetric.
Reason: A relation is asymmetric if
(a, b) R (b, a)
so for an asymmetric relation, (a, b) R (b, a) R is always false. A
relation is antisymmetric if
Math 55 Homework 11 Solutions
May 2, 2014
For odd numbered problems, see the back of the book for more details.
12. Since G is undirected, if there is an edge between u and v then there is an edge between v and u and so uRv
implies vRu. As
Math 55: Discrete Mathematics,
UC Berkeley, Spring 2014
Solutions to Homework #1 (due January 29)
January 26, 2014
1.1 # 12
a. If you have the flu, then you miss the final examination.
b. You do not miss the final examination, if and only if you pass the
Math 55 Homework 3 Solutions
20. Give an example of a function from N to N that is:
1. one-to-one, but not onto. f (n) = 2n.
2. onto, but not one-to-one. f (n) = d n2 e.
3. both onto and one-to-one, but not the identity funct
Math 55, Spring 2014
Functions and Sets Worksheet
Let f : A B be a function, S, T A and U, V, B subsets. For each of the following statements
either prove it with a proof or disprove it with a counter-example. For each false statement, see
if you ca
Math 55 Homework 5 Solutions
due March 5
4. Let P (n) be the statement
i3 = ( n(n1)
(a) What is the statement P (1)?
P (1) says that 13 = 1(1+1)
(b) Show that P (1) is true.
We have 13 = 1, whereas
1(1 + 1)
Math 55 Homework 7 Solutions
(c) We start by simply taking 2 of each
Weve taken a total of 12, so we still need to choose where to
pick our other 12 from: 12+61
(d) We c
Math 55 Homework 4 Solutions
33. Find each of these values.
(a) (992 mod 32)3 mod 15.
99 3 (mod 32), so 992 9 (mod 32). Then 92 = 81 6 (mod 15), and 6 9 = 54 9 (mod 15). So
(992 mod 32)3 mod 15 = 9.
(b) (34 mod 17)2 mod 11.
Math 55 Homework 2 Solutions
due Feb 5, 2014
a. Let class(x), convert(x), speed(x) denote x is in this class, x owns
a red convertible, and x has gotten a speeding ticket. Using universal instantiation we get convert(Linda) speed(Linda). Using
October 2, 2016
1. This is an example of an Euler circuit. Consider the banks of the river as being
just another two big islands. Suppose we start at vertex A. In a circuit, we
must eventually come back to A, meaning we need 2 bridges (one to
Some basic proofs
September 27, 2016
1. Prove the following:
Let a, b, and c be any integers. If a | b, then a | bc.
Assuming that a | b is true, the next follows:
b = ak
where k is an integer. Then bc = akc = a(kc). Since kc is an integer, it follows
Equivalence Relations and Equivalence Classes
September 22, 2016
1. Which of the relations , |, and are reflexive?
All three are reflexive. In other words:
Every number is less than or equal to itself: a a.
Every number is divisible
MATH 55 - HOMEWORK 5 SOLUTIONS
Exercise 0.1. 24
Proof. We have n = 43 59 = 2537 and clearly 2525 < 2537 but 252525 > 2537 so our blocks will consist of 4 digits.
The numbers associated to the letters in ATTACK are A = 00, C = 02, K = 10, and T = 19. Since
HW 3 Solutions, MATH 55, Spring 2016
2.3, 28 If an inverse g existed, then we would have f g(x) = x. Take x = 1; then f (g(1) =
eg(1) 0, so f (g(1) 6= 1. The problem here is that f is not surjective as a function to
the real numbers. If the codomain