## Set Representation

- Listing method
- Set builder notation (rules of inclusion)

## Listing Method

Representing a set S using the listing method consists of listing all the elements of S.

S1 = set of all vowels in the English alphabet

S1 = {a, e, i, o, u} (listing method)

S2 = set of all positive integers less than 10

S2 = {1,2,3,4,5,6,7,8,9} (listing method)

S3 = set of all positive even integers less than 10

S3 = {2,4,6,8} (listing method)

## Set Builder Notation

Assume we want to describe the set of all even integers.

-6 | 2*(-3) |

-4 | 2*(-2) |

-2 | 2*(-1) |

0 | 2*0 |

2 | 2*1 |

4 | 2*2 |

6 | 2*3 |

Even\,=\{2n|n\,\in\,Z\}\\set \,builder\,notation

- the set of elements of two times n, such that n is an element of Z

Quick Review: Z = set of integers = {… , -3, -2, -1, 0, 1, 2, 3, …}

Assume we want to describe the set of all odd integers:

-5 | 2*(-3)+1 |

-3 | 2*(-2)+1 |

-1 | 2*(-1)+1 |

1 | 2*0+1 |

3 | 2*1+1 |

5 | 2*2+1 |

7 | 2*3+1 |

Odd\,=\,\{2n+1|n\,\in\,Z\}

Assume we want to describe the set of rational numbers Q.

Q={…,1/1,1/2,1/3,1/4,…}

We can’t really represent this set accurately using the listing method. However, using the set-builder method, we can represent Q more accurately as follow. We can write Q is the set of elements of the form n over m, such that n and m are elements of Z and m is different from zero.

Q\,=\,\{\frac{n}{m}|n,m\,\in\,Z\,and\,m\,\neq\,0\}

Quick Review: | means “such that” or “whereas”, x|x is in my bag means x such that x is in my bag. x|x is an odd number means x whereas x is an odd number.

Bag\,=\,\{pen,book,laptop,...\}\\Bag\,=\,\{x|x\,is\,in\,my\,bag\}

## Examples:

S1\,=\,\{3n|n\,\in\,N\,and\,n<6\}\\S1\,=\,\{3*1,3*2,3*3,3*4,3*5\}\\n\,=\,1,2,3,4,5\\S1\,=\,\{3,6,9,12,15\}

**Quick Review: **

**≤**means less than or equal to- Z is a set of integers

S2\,=\,\{2^n|n\,\in\,Z\,and\,0\,\leq\,n\leq\,4\}

- n is in the set Z (integer)
- n is more then or equal to 0
- n is less than or equal to 4
- n = {0, 1, 2, 3, 4}

S2\,=\,\{2^0,2^1,2^2,2^3,2^4\}\\2^0\,=\,1\\2^1\,=\,2*1\,=\,2\\2^2\,=\,2 *2\,=\,4\\2^3\,=\,2*2*2\,=\,8\\2^4\,=\,2*2*2*2\,=\,16\\Therefore,also:\\S2\,=\,\{1,2,4,8,16\}

**Quick Review:**

**Negative Exponents** –

S3\,=\,\{2^{-n}|n\,\in\,Z\,and\,0\,\leq\,n\,<\,8\}

- n is in set Z (integers)
- n is more than or equal to 0
- n is less than 8
- n(-)
- n = {0, -1, -2, -3, -4, -5, -6, -7}

S3\,=\,\{2^0,2^{-1},2^{-2},2^{-3},2^{-4},2^{-5},2^{-6},2^{-7}\}\\S3\,=\,\{1,(\frac{1}{2}),(\frac{1}{2*2}),(\frac{1}{2*2*2}),(\frac{1}{2*2*2*2}),(\frac{1}{2*2*2*2*2}),(\frac{1}{2*2*2*2*2*2}),(\frac{1}{2*2*2*2*2*2*2})\}\\S3\,=\,\{1,\frac{1}{2},\frac{1}{8},\frac{1}{16},\frac{1}{32},\frac{1}{64},\frac{1}{128}\}

## Sample Exercises

Rewrite the following sets using the set builder method:

**Problem 1:**

S_1\,=\,\{\frac{1}{2},\frac{1}{4},\frac{1}{6},\frac{1}{8},\frac{1}{10}\}

**Solution:**

S_1\,=\,\{\frac{1}{2n}|n\,\in\,Z\,and\,0\,<\,n\,\leq\,5\}

**Problem 2**

S_2\,=\,\{1, 2,4,8,16,32,64,128\}

Solution:

S_2\,=\,\{2^n|n\,\in\,Z\,and\,0\,\leq\,n\,<\,8\}

**Problem 3**

S_3\,=\,\{...,-9,-6,-3,0,3,6,9...\}

**Solution**

S_3\,=\,\{3n|n\,\in\,Z\}\\because\;of\;the\;...\,no\;need\;for:\\and\,{-3}\,\leq\,n\,\leq\,3

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