// Numbas version: exam_results_page_options {"name": "Luis's copy of set5 - Cartesian Products", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"advice": "

a)

$A \\times B$ is the set of all pairs $(a,b)$, where $a \\in A$ and $b \\in B$.

b)

$B \\cap D$ is the set of all elements present in both $B$ and $D$, i.e. $\\var{set2 and set5}$.

$A \\cap C$ is the set of all elements present in both $A$ and $C$, i.e. $\\var{set1 and set4}$.

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

c)

$(A\\cap C)\\times (A\\cap C)\\times (C\\cap D)$ is the set of all triples $(x,y,z)$, where $x \\in A \\cap C$, $y \\in A \\cap C$ and $z \\in C \\cap D$. Note that $x$ and $y$ do not have to be different.

d)

$A-C$ is the set of all elements present in $A$ but not in $C$, i.e. $\\var{set1-set4}$.

$C-A$ is the set of all elements present in $C$ but not in $A$, i.e. $\\var{set4-set1}$.

$(A-C) \\cup (C-A)$ is the set of all elements which are either in $A-C$, or in $C-A$, so $(A-C) \\cup (C-A) = \\var{(set1-set4) or (set4-set1)}$.

e)

$(A \\times D)$ is the set of all pairs of elements $(a,d)$, with $a \\in A$ and $d \\in D$, i.e. $\\var{set(product(list(set1),list(set5)))}$.

$C \\times B)$ is the set of all pairs of elements $(c,b)$, with $c \\in C$ and $b \\in B$, i.e. $\\var{set(product(list(set4),list(set2)))}$.

$(A \\times D) \\cap (C \\times B)$ is the set of all pairs present in both of the previous sets.

f)

$C \\cap D$ is the set of all elements in both $C$ and in $D$, so $C \\cap D = \\var{set(set4 and set5)}$.

$C - D$ is the set of all elements in $C$ and not in $D$, so $C-D = \\var{set(set4 - set5)}$.

$(C \\cap D) \\times (C - D)$ is the set of all pairs of elements $(x,y)$, where $x$ is in $C \\cap D$ and $y$ is in $C - D$, so $C \\cap D) \\times (C-D) = \\var{set(product(list(set4 and set5),list(set4 - set5)))}$.

Similarly, $(D - C) \\times (C \\cap D) = \\var{set(product(list(set5-set4),list(set4 and set5)))}$.

Finally, $[(C \\cap D) \\times (C - D)] \\cup [(D - C) \\times (C \\cap D)]$ is the set of all pairs present in either of the above sets, i.e. $\\var{set16}$.

", "preamble": {"css": "", "js": ""}, "question_groups": [{"questions": [], "pickingStrategy": "all-ordered", "pickQuestions": 0, "name": ""}], "name": "Luis's copy of set5 - Cartesian Products", "variable_groups": [{"name": "Random numbers", "variables": ["a", "b"]}, {"name": "Lists", "variables": ["list1", "list2", "list4", "list5", "list9", "list10", "list_extra"]}], "type": "question", "functions": {"mod_set": {"language": "javascript", "type": "list", "definition": "//returns all integers which are divisible by c betweeen a and b\nvar l=[];\nfor(var i=a;iLet $A=\\var{set1}$, let $B=\\var{set2}$, let $C=\\var{set4}$ and let $D=\\var{set5}$.

List the elements of the following sets.

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().

", "parts": [{"variableReplacements": [], "prompt": "

$A \\times B=\\;$[[0]]

$(B\\cap D)\\times (A\\cap C)=\\;$[[0]]

$(A\\cap C)\\times (A\\cap C)\\times (C\\cap D)=\\;$[[0]]

$[(A - C)\\cup (C-A)]\\times [(B-D)\\cup(D-B)]=\\;$[[0]]

$(A\\times D)\\cap (C\\times B)=\\;$[[0]]

$[(C\\cap D)\\times (C-D)]\\cup [(D-C)\\times (C\\cap D)]=\\;$[[0]]

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