Using induction, show that the cardinality of the powerset of A is 2" where n is the number of elements in A. In other words, given a set A with |A| = n then |P(A)| helpful to see the practice problem on induction to see what is required of your proof). = 2". (You may find it

College Algebra (MindTap Course List)
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ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter8: Sequences, Series, And Probability
Section8.5: Mathematical Induction
Problem 42E
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Using induction, show that the cardinality of the powerset of A is 2" where n is the number of
elements in A. In other words, given a set A with |A| = n then |P(A)| = 2". (You may find it
helpful to see the practice problem on induction to see what is required of your proof).
Transcribed Image Text:Using induction, show that the cardinality of the powerset of A is 2" where n is the number of elements in A. In other words, given a set A with |A| = n then |P(A)| = 2". (You may find it helpful to see the practice problem on induction to see what is required of your proof).
Using induction, show that n! < n" for all n > 1. (You may find it helpful to see the practice
problem on induction to see what is required of your proof).
Solution:
We use induction on a sequence of statements:
• 2! < 22 (statement 2)
• 3! < 33 (statement 3)
• n! < n" (statement n)
Base Case (statement 2):
2! = 2 * 1 = 2 < 4 = 2²
so that 2! < 22
Inductive Step: Assume statement n is true, that is n! < n". Then:
(n +1)! = n! * (n + 1)
< п" * (п+1)
< (n +1)" * (п + 1)
= (n +1)"+1
Where we use statement n in the second line and the fact that n" < (n +1)" in the third (since n
is positive). This statement shows that (n +1)! < (n + 1)"n+1.
By induction, n! < n" for all n > 1.
Transcribed Image Text:Using induction, show that n! < n" for all n > 1. (You may find it helpful to see the practice problem on induction to see what is required of your proof). Solution: We use induction on a sequence of statements: • 2! < 22 (statement 2) • 3! < 33 (statement 3) • n! < n" (statement n) Base Case (statement 2): 2! = 2 * 1 = 2 < 4 = 2² so that 2! < 22 Inductive Step: Assume statement n is true, that is n! < n". Then: (n +1)! = n! * (n + 1) < п" * (п+1) < (n +1)" * (п + 1) = (n +1)"+1 Where we use statement n in the second line and the fact that n" < (n +1)" in the third (since n is positive). This statement shows that (n +1)! < (n + 1)"n+1. By induction, n! < n" for all n > 1.
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