College Mathematics Clep

Front 1

Union.

Back 1

set of elements which are in either set
For example: let A = (1,2,3) and let B = (3,4,5). The Union of A and B, written A ∪ B = (1,2,3,4,5). There is no need to list the "3" twice.

Front 2

Intersection

Back 2

the set of elements which are in both sets

For example: let A = (1,2,3) and B = (3,4,5). The Intersection of A and B, written A ∩ B = (3). 3 is the only element present in both sets.

Front 3

A = { All prime numbers less than 20 }
B = { All whole numbers less than 10 }

Find A - B

Back 3

{11, 13, 17, 19}. We must identify the set whose elements are present in A but are not present in B:

A = { 2, 3, 5, 7, 11, 13, 17, 19 }
B = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

A – B = { x/x Î A and x Ï B }
The above means A - B = what's in set A, but not in set B.

A – B = { 2, 3, 5, 7, 11, 13, 17, 19 } – { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
= {11, 13, 17, 19}

Front 4

S = {1, 2, 3, 4, 5, 6, 7}

This form of representing a set is called

Back 4

Roster form. A set can be represented in two ways:

(i) Roster form, S = {1, 2, 3, 4, 5, 6, 7}.
(ii) Set builder form, S = {x : x ∈ N; x < 8}.

The ∈ symbol means "element of", so "x ∈ N" indicates that x is an element of N (N is a set containing all natural numbers; i.e. all positive integers {1, 2, 3...}).

Front 5

If n(A) = 5, n(B) = 4, n(A ∩ B) = 2 then n(A ∪ B) =

Back 5

7. n(A) = 5 means that set A contains 5 items. Likewise, set B contains 4 items. The intersection of set A and set B contains 2 items, which means there are two items which set A and B have in common. An example which would match these facts is: A = {1,2,3,4,5}, B = {4, 5, 6, 7}. A ∩ B = {4, 5}, so A ∪ B = {1,2,3,4,5,6,7}.

Here are the calculations:

n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
= 5 + 4 - 2
= 7

The number of items in the set containing the union of set A and B would be 7.