Sets chapter 2
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Sets Chapter 2. 1. Sets: A Problem Solving Tool 2.1. 2. Sets. A set is a well-defined collection of objects, called elements or members of the set. Let A = {1 , 2, 3, 4 }, which means that “ A is the set containing the elements 1 , 2, 3 , and 4.”

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Sets Chapter 2

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Sets chapter 2

SetsChapter 2

1


Sets a problem solving tool 2 1

Sets: A Problem Solving Tool2.1

2


Sets chapter 2

Sets

A set is a well-defined collection of objects, called elements or members of the set.

Let A = {1, 2, 3, 4}, which means that “A is the set containing the elements 1, 2, 3, and 4.”

4  A “4 is in A” or “4 is an element of the set A”

6  A“6 is not an element of A”

3


Sets chapter 2

  • Let X = {a, b, x, y}. Fill in the blank with or to make

    each statement correct.

  • a___X

  • x___X

  • A___X

4


Sets chapter 2

Sets of Numbers Commonly used in Mathematics

N = {x | x is a natural number} = {1, 2, 3, …}

W = {x | x is a whole number} = {0,1, 2, 3, …}

I = {x | x is an integer} = {…, 2, 1, 0,1, 2, 3, …}

Q = {x | x is a rational number} = {x | x, is of the form

a/b where a, b are integers and b  0}

Q = {x | x is a rational number} = {x | x has a terminating or repeating decimal pattern}

S = {x | x is a irrational number} = {x | x is not rational}

R = {x | x is a real number} = {x | x is the union of the rational and irrational numbers}

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Sets chapter 2

Describing Sets

  • 1. Verbal or written description

  • Roster Method or Listing

  • Set-builder notation

The Empty Set

The symbol { } or Ø represents the empty, or null set.

NOTE:

{Ø} is not the null set. It is the set containing the null set.

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Sets chapter 2

Describe the following sets using the roster method and set builder notation.

  • The set of counting numbers less than 8.

  • The set of counting numbers between 3 and 8.

  • The set of counting numbers between 6 and 7.

  • The set of counting numbers greater than 3.

a. {1, 2, 3, 4, 5, 6, 7}

b. {4, 5, 6, 7}

c. { } or φ

d. {4, 5, 6,…}

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Sets chapter 2

Equality of Sets

Two sets A and B are equal, A = B, if they have the same elements not necessarily listed in the same order.

State whether the sets A and B are equal.

  • A={2n + 1|n is a counting number} B={2n – 1|n is a counting number}

  • A = {1,1,2,2,3} , B = {2,1,3}

ANSWER

1. A ≠ B , not equal

2. A = B , equal

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Sets chapter 2

Subsets

The set A is a subset of B, A ⊆ B, if every element of A is also an element of B.

Proper Subsets

A set A is said to be a proper subset of B, A ⊂ B, if

A ⊆ B but A  B.

The Universal Set

The universal set,U, is the set of all elements under discussion.

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Sets chapter 2

List all the subsets and indicate which are the proper subsets of the given set.

  • U = {a, b}

  • U = {1, 2, 3}

ANSWER

  • Ø, {a}, {b}, and {a, b}. The first three are proper subsets.

Ø

{1}, {2}, {3}

{1, 2}, {1, 3}, {2, 3}

{1, 2, 3}

The first seven are proper subsets.

Note: by definition Ø is a subset of any set.

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Sets chapter 2

Number of Subsets of a Set

A set of n elements has 2n subsets.

Refer to the problems on the previous slide:

22 = 4 and 23 = 8.

Number of Proper Subsets of a Set

A set of n elements has 2n – 1 proper subsets.

Refer to the problems on the previous slide:

22 1 = 4  1 = 3 and 23 1 = 8  1 = 7.

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Example

Example:

Given: A = { 12, 19, 26,…, 68} Find

the number of subsets

the number of proper subsets

To solve this problem you need to determine how many elements are in the set.

Notice the elements in this set are evenly spaced and form what is called an arithmetic sequence. The next term is found by adding a common difference of seven in this example. The total number of elements can be found without having to list all of the elements in this set using the formula An = A1 + (n – 1)d.

An is the last term in the sequence, A1 is the first term, d is the common difference and n is the number of terms in the sequence. Substituting the numbers into the equation and solving for n yields the number of terms in this set.

a. Number of subsets = 29 = 512

b. Number of proper subsets = 29– 1 = 512 – 1 = 511

68 = 12 + (n – 1)(7)

–12 –12

56 = (n – 1)(7)

7 7

8 = n – 1

+1 + 1

9 = n

12

END


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