set6 - Cartesian Products and Complements - Numbas at mathcentre.ac.uk

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Newcastle University Mathematics and Statistics

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Newcastle University Mathematics and Statistics 9 years, 2 months ago

Created this.

Name	Status	Author	Last Modified
set6 - Cartesian Products and Complements	draft	Newcastle University Mathematics and Statistics	20/11/2019 14:51
Lois's copy of set6 - Cartesian Products and Complements	draft	Lois Rollings	04/04/2020 15:00
Luis's copy of set6 - Cartesian Products and Complements	draft	Luis Hernandez	30/11/2018 17:37
Listar los elementos de los siguientes conjuntos...	Ready to use	Luis Hernandez	14/12/2020 19:22

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Question statement

In this question the universal set is $U=\var{univ}$ .

Let $A=\var{set1}$ and let $B=\var{set2}$ .

For subsets $A$ and $B$ of $U$ , the universal set for the Cartesian product $A\times B$ is $U\times U$ .

List the elements of the following sets.

Note that you input sets in the form set(a,b,c,d) .

For example set(1,2,3)gives the set $\{1,2,3\}$ .

Element $(a,b)$ of a Cartesian product is entered, and represented as $[a,b]$ .

For example set([1,1],[1,2],[2,3])gives the set $\{[1,1], [1,2], [2,3]\}$ .

The empty set is input as set().

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	3
b	integer	2
ext	set	set(56,76)
int	set	set(28,26)
list1	list	[ 18, 7, 4 ]
list2	list	[ 85, 55 ]
set1	set	set(18,7,4,28,26)
set2	set	set(85,55,28,26)
set3	set	Set of 4 items
set4	set	Set of 4 items
set4_1	set	Set of 20 items
set4_2	set	Set of 20 items
set5	set	Set of 6 items
set5_1	set	set(18,7,4)
set5_2	set	set(56,76)
set6	set	Set of 4 items
set6_1	set	Set of 18 items
set6_2	set	Set of 18 items
set7	set	Set of 4 items
univ	set	set(18,7,4,28,26,85,55,56,76)
univ_2	set	Set of 81 items

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(2..3)

JME function reference

Description

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `integer`

→ Used by:

b
list1

This variable doesn't seem to be used anywhere.

Parts

Part a) Gap-fill
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Part b) Gap-fill
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Part c) Gap-fill
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Part d) Gap-fill
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Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

$(A^c\cap B^c) \times (A\cap B)=\;$ Unnamed gap

Advice

a)

$A^c$ is the set of $U-A$ of all elements in $U$ and not in $A$ , so $A^c = \var{univ - set1}$ .

$B^c$ is the set of $U-A$ of all elements in $U$ and not in $B$ , so $B^c = \var{univ - set1}$ .

$A^c \cap B^c$ is the set of all elements present in both $A^c$ and $B^c$ . This is equivalent to the set of all elements in neither $A$ nor $B$ , i.e. $\var{(univ-set1) and (univ-set2)}$ .

$A \cap B$ is the set of all elements present in both $A$ and $B$ , i.e $\var{set1 and set2}$ .

So $(A^c \cap B^c) \times (A \cap B)$ is the set of all pairs $(x,y)$ , where $x$ is in $A^c \cap B^c$ , and $y$ is in $A \cap B$ .

b)

$(U \times A)^c$ is the set of all pairs $(x,y)$ in $U \times U$ which are not in $U \times A$ . Since $U$ is the universal set, this is equivalent to $U \times (A^c)$ , the product of $U$ with the set of elements not in $A$ .

Similarly, $(U \times B)^c$ is equivalent to $U \times (B^c)$ .

Again because $U$ is the universal set, $(U \times A)^c \cap (U \times B)^c = U \times (A^c \cap B^c)$ .

By a similar argument, $(A \times U)^c \cap (B \times U)^c = (A^c \cap B^c) \times U$ .

So $(U\times A)^c\cap (U\times B)^c\cap (A\times U)^c\cap (B\times U)^c$ is equivalent to $(A^c \cap B^c) \times (A^c \cap B^c)$ . That is, the set of all pairs of two elements that are in neither $A$ nor $B$ .

c)

$A-B$ is the set of elements which are in $A$ but not $B$ , i.e. $\var{set1-set2}$ .

$(A \cup B)^c$ is the set of elements $U - (A \cup B)$ which are in $U$ and not in $A \cup B$ , so $(A \cup B)^c = \var{univ-(set1 or set2)}$ .

d)

$[(A \cup B) \times U]^c$ is equivalent to $(A \cup B)^c \times U$ .

So $[(A \cup B) \times U]^c \cap [U \times (A \cap B)] = (A \cup B)^c \times (A \cap B)$ .

e)

$A^c-B$ is the set of all elements which are in $A^c$ but not $B$ . That's equivalent to the set of elements which are in neither $A$ nor $B$ , i.e. $(A \cup B)^c = \var{univ-(set1 or set2)}$ .

Similarly, $B^c - A = (B \cup A)^c = (A \cup B)^c = \var{univ-(set1 or set2)}$ .

Give a worked solution to the whole question.

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set6 - Cartesian Products and Complements

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Taxonomy: mathcentre

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Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

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Warning

Generated value: `integer`

Parts

Gap-fill

a)

b)

c)

d)

e)

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set6 - Cartesian Products and Complements

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

a)

b)

c)

d)

e)

Generated value: `integer`