Discrete Mathematics Exam 2

studied byStudied by 17 people
5.0(1)
Get a hint
Hint

Proof by Contradiction

1 / 65

flashcard set

Earn XP

66 Terms

1

Proof by Contradiction

  1. Suppose the statement to be proved is false. That is, suppose that the negation of the statement is true.

  2. Show that this supposition lead logically to a contradiction.

  3. Conclude that the statement to be proved is true.

New cards
2

Method of Proof by Contraposition

  1. Express the statement to be proved in the form ∀x in D ,if P (x ) then Q (x )

  2. Rewrite this statement in the contrapositive form ∀x in D ,if ∼Q (x ) then ∼P (x )

  3. Prove the contrapositive by direct proof. a. Suppose x is a (particular but arbitrarily chosen) element in D such that Q (x ) is false. b. Show that P (x ) is false.

New cards
3

The Irrationality of √2

By the Pythagorean Theorem, the length of the diagonal equals the square root of 2. So the square root of 2 is irrational!

New cards
4

Are There Infinitely Many Prime Numbers?

The number of primes is infinite

New cards
5

variable

In computing, refers to a storage location in the computer’s memory.

New cards
6

assignment statement

value to a variable: x := e.

New cards
7

x := e.

x is assigned the value e

New cards
8

Conditional statements

using current values of variables to determine which statement is executed next

New cards
9

If then else

If condition is true, then s1 is executed and execution moves to the next algorithm statement If condition is false, then s2 is executed and execution moves to the next algorithm statement

New cards
10

Iterative statements

when a sequence of algorithm statements is to be executed over and over again

New cards
11

While loop

while (condition) [statements that make up the body of the loop] end while

New cards
12

Trace Table

to trace the execution of the following algorithm by finding the values of all the algorithm variables each time they are changed during execution

New cards
13

For-next loop

for variable := initial expression to final expression [statement that make up the body of the loop] next (same) variable

New cards
14

Division Algorithm

or an integer a and a positive integer d , the quotient-remainder theorem guarantees the existence of integers q and r such that a = dq + r and 0 ≤r <d .

New cards
15

gcd(a,b)

greatest common divisor.

  1. d is a common divisor of both a and b. In other words, d |a and d |b.

  2. For every integer c , if c is a common divisor of both a and b, then c is less than or equal to d .

New cards
16

Euclidean Algorithm

a way to find the greatest common divisor of two positive integers, a and b

New cards
17

Sequence

A function whose domain is either all the integers between two given integers or all the integers greater than or equal to the integer

New cards
18

Term

individual element ak (read: “a sub k ”)

New cards
19

Subscript/Index

The k in ak is called

New cards
20

Initial term

m is the subscript of the

New cards
21

Final term

n is the subscript of the

New cards
22

Infinite Sequence

notation am ,am+1,am+2,...

New cards
23

explicit formula or general formula

a rule that shows how the values of ak depend on k

New cards
24

the summation from k equals m to n of a-sub-k

n ∑ak k =m

New cards
25

Telescoping sum

Some sums can be written so that successive cancellation of terms collapses the final result

New cards
26

product from k equals m to n of a-sub-k

n ∏ak k =m

New cards
27

Dummy variable

The variable used to represent the index of summation

New cards
28

factorial

the product of all the integers up to and including a given integer. n!

New cards
29

Zero factorial

denoted 0! is defined to be 1: 0! = 1

New cards
30

n choose r

(n/r) without the /

New cards
31

Proving an equality

  1. Transform the left hand side and the right hand side independently until they are seen to be equal

  2. Transform one side of the equation until it is seen to be the same as the other side of the equation.

New cards
32

Ordinary induction

assume P (k ) is true and show P (k + 1) is true.

New cards
33

Strong induction

Assume P (i ) is true for i ≤k , show P (k + 1) is true.

New cards
34

Well-Ordering Principle for the Integers

Let S be a set of integers containing one or more integers all of which are greater than some fixed integer. Then S has a least element,

New cards
35

Quotient-Remainder Theorem

Given any integer n and any positive integer d , there exist integers q and r such that n = dq + r and 0 ≤r <d .

New cards
36

AXB

Let A and B be sets. A relation from A to B

New cards
37

x is related to y by R

xRy if and only if (x ,y ) ∈R .

New cards
38

arrow diagram for R

  1. Represent the elements of A as points in one region and elements of B as points in another.

  2. For each x ∈A, y ∈B , draw an arrow from x to y if and only if (x ,y ) ∈R .

New cards
39

function F from a set A to a set B

a relation with domain A and co-domain B that satisfies the following two properties:

  1. For every element x in A, there is an element y in B such that (x ,y ) ∈F .

  2. For all elements x in A and y in B , if (x ,y ) ∈F and (x ,z ) ∈F ,then y = z

New cards
40

congruent modulo 2

mEn if, and only if, m mod 2 = n mod 2

New cards
41

reflexive

referring back to itself. or every x ∈A, xRx

New cards
42

symmetric

for every x , y ∈A, if xRy then yRx

New cards
43

transitive

or every x, y, z ∈A, if xRy and yRz then xRz

New cards
44

transitive closure

R is the relation R ton A that satisfies the following three properties:

  1. R tis transitive.

  2. R ⊆R t.

  3. If S is any other transitive relation that contains R , then R t⊆S .

New cards
45

Union

A ∪B , is the set of all elements that are in at least one of A and B .

New cards
46

Intersection

A ∩B , is the set of all elements that are common to both A and B .

New cards
47

partition

a finite or infinite collection of nonempty, mutually disjoint subsets whose union is A.

New cards
48

m ≡ n(mod d )

m is congruent to n modulo d , d |(m −n)

New cards
49

encryption

the method by which information is converted into secret code that hides the information's true meaning

New cards
50

Plaintext

A message before conversion

New cards
51

ciphertext

The converted plaintext

New cards
52

Decryption

To convert the ciphertext back into plaintext

New cards
53

Caeser Cipher

Encrypts messages by changing each letter of the alphabet to the one three places farther along with wrapping X to A, Y to B, and Z to C.

New cards
54

public key cryptography

potential recipient of encrypted messages openly distributes a public key containing encryption information

New cards
55

RSA Cryptography

M = C^d mod pq, where d >0 and d is an inverse to e modulo (p −1)(q −1). p=5, q=11

New cards
56

Subset

A ⊆B ⇔∀x ,if x ∈A then x ∈B .

New cards
57

proper subset

  1. A ⊆B , and

  2. there is at least one element in B that is not in A.

New cards
58

venn diagram

a diagram that uses circles to represent mathematical or logical sets pictorially inside a rectangle

New cards
59

Union, denoted A ∪B

is the set of all elements that are in at least one of A or B .

New cards
60

intersection, denoted A ∩B

is the set of all elements that are common to both A and B

New cards
61

difference, denoted B −A

the set of all elements that are in B and not A

New cards
62

complement, denoted A^c

the set of all elements in U that are not in A

New cards
63

empty set, ∅

the set with no elements

New cards
64

disjoint

A ∩B = ∅

New cards
65

mutually disjoint

Ai ∩Aj = ∅ whenever i /= j

New cards
66

partition

a. A is the union of all the Ai b. the sets A1,A,2A3,...are mutually disjoint.

New cards

Explore top notes

note Note
studied byStudied by 62 people
... ago
4.0(1)
note Note
studied byStudied by 48 people
... ago
5.0(1)
note Note
studied byStudied by 3 people
... ago
5.0(1)
note Note
studied byStudied by 33 people
... ago
5.0(2)
note Note
studied byStudied by 7 people
... ago
5.0(1)
note Note
studied byStudied by 58 people
... ago
5.0(1)
note Note
studied byStudied by 58 people
... ago
5.0(1)

Explore top flashcards

flashcards Flashcard (35)
studied byStudied by 833 people
... ago
5.0(1)
flashcards Flashcard (27)
studied byStudied by 133 people
... ago
5.0(5)
flashcards Flashcard (40)
studied byStudied by 1 person
... ago
5.0(1)
flashcards Flashcard (74)
studied byStudied by 31 people
... ago
5.0(2)
flashcards Flashcard (65)
studied byStudied by 30 people
... ago
5.0(1)
flashcards Flashcard (41)
studied byStudied by 20 people
... ago
5.0(1)
flashcards Flashcard (40)
studied byStudied by 18 people
... ago
5.0(1)
flashcards Flashcard (25)
studied byStudied by 9 people
... ago
5.0(1)
robot