0
Still curious? Ask our experts.
Chat with our AI personalities
Steve
Beau
ProfessorAdd your answer:
Earn +20 pts
What is the name of the eight grouping symbols?
The eight (8) grouping symbols related to set theory include the following: ∈ "is an element (member) of" ∉ "is not an element (member) of" ⊂ "is a proper subset of" ⊆ "is a subset of" ⊄ "is not a subset of" ∅ the empty set; a set with no elements ∩ intersection ∪ union
What is the largest number in a data set?
It's the maximum.Probably C, the continuum.The cardinality (count) of the infinite set of integers is Aleph-null. Then C = 2^(Aleph-null).
Prove that A contains N elements and the different subsets of A is equal to 2?
Assuming the question is: Prove that a set A which contains n elements has 2n different subsets.Proof by induction on n:Base case (n = 0): If A contains no elements then the only subset of A is the empty set. So A has 1 = 20 different subsets.Induction step (n > 0): We assume the induction hypothesis for all n smaller than some arbitrary number k (k > 0) and show that if the claim holds for sets containing k - 1 elements, then the claim also holds for a set containing k elements.Given a set A which contains k elements, let A = A' u {.} (where u denotes set union, and {.} is some arbitrary subset of A containing a single element no in A'). Then A' has k - 1 elements and it follows by the induction hypothesis that (1) A' has 2k-1 different subsets (which also are subsets of A). (2) For each of these subsets we can create a new set which is a subset of A, but not of A', by adding . to it, that is we obtain an additional 2k-1 subsets of A. (*)So by assuming the induction hypothesis (for all n < k) we have shown that a set A containing kelements has 2k-1 + 2k-1 = 2k different subsets. QED.(*): We see that the sets are clearly subsets of A, but have we covered all subsets of A? Yes. Assume we haven't and there is some subset S of A not covered by this method: if S contains ., then S \ {.} is a subset of A' and has been included in step (2); otherwise if . is not in S, then S is a subset of A' and has been included in step (1). So assuming there is a subset of A which is not described by this process leads to a contradiction.
How can you show that a set is a subset of another?
If you want to show that A is a subset of B, you need to show that every element of A belongs to B. In other words, show that every object of A is also an object of B.
This kind of proof required you to prove that all but one of a set of possible alternatives are false?
elementary proof
