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What is the possible no of reflexive relations on a set of 5 elements?
Wiki User
∙ 13y ago
The total no. of reflexive relations on a set A having n elements is 2^n(n-1).
Thus, the required no. is 2^20 = 1 048 576
Wiki User
∙ 13y ago
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What is null set in mathematics?
a set having no elements, or only zeros as elements.
Which is not a subset of the set 1235?
Any set that has elements that are not in that set.
What is cordinality of set?
The cardinality of a set is the number of elements in the set.
How many subsets does set A have if the set A has 3 elements?
In a subset each element of the original may or may not appear - a choice of 2 for each element; thus for 3 elements there are 2 × 2 × 2 = 2³ = 8 possible subsets.
What is the name of the set that contains all elements?
Name of the set which contains all elements is UNIVERSAL SET. It is usually represented by (U)
What is the possible number of reflexive relations on a set of 5 elements?
2 power 20
What is the total number of reflexive and symmetric relations on a set containing n elements?
the total no of reflexive relation on an n- element set is 2^(n^2-n).
How do you Derive The Number of reflexive relation of a Set 's' having n elements?
make a table as I did below for the set {a,b,c} with 3 elements. A table with all n elements will represent all the possible relations on that set of n elements. We can use the table to find all types of relations, transitive, symmetric etc. | a | b | c | --+---+---+---+ a | * | | | b | | * | | c | | | * | The total number of relations is 2^(n^2) because for each a or b we can include or not include it so there are 2 possibilities and there are n^2 elements so 2^(n^2) total relations. A relation is reflexive if contains all pairs of the form {x,x) for any x in the set. So this is the diagonal of your box. THESE ARE FIXED! No, in reflexive relation we still can decide to include or not include any of the other elements. So we have n diagonal elements that are fixed and we subtract that from n^2 so we have 2^(n^2-n) If you do the same thing for symmetric relations you will get 2^(n(n+1)/2). We get this by picking all the squares on the diagonal and all the ones above it too.
What is the possible number of symmetric relations on a set of 5 elements?
The number is 5! = 120
Is the empty set reflexive?
The empty set is both reflexive and irreflexive.
How many equivalence relations are there on a set with four elements?
16
What is the largest equivalence relation on a set A?
An equivalence relation on a set is one that is transitive, reflexive and symmetric. Given a set A with n elements, the largest equivalence relation is AXA since it has n2 elements. Given any element a of the set, the smallest equivalence relation is (a,a) which has n elements.
What is the elements of the finite set?
They are the members of the set. It is not possible to list them without knowing what the set is.
How many symmetric relations are there on a set with eight elements?
2^32 because 2^(n*(n+1)/2) is the no of symmetric relation for n elements in a given set
How do you Derive The Number of symetric relation of a Set 's' having n elements?
2^(n^2+n)/2 is the number of symmetric relations on a set of n elements.
What does reflexive property looks like?
The binary operator ~ is reflexive if a ~ a for every element a in the relevant set.
Is the divides relation on the set of positive integers is reflexive?
No, it is not.
