A Probability Course for the Actuaries A Preparation for Exam P/1

16 BASIC OPERATIONS ON SETS
Example 1.11
(a) Use induction to show that if n(A) = n then n(P(A)) = 2 n .
(b) If P(A) has 256 elements, how many elements are there in A
Solution.
(a) We apply induction to prove the claim. If n = 0 then A = ∅ and in this
case P(A) = {∅}. Thus n(P(A)) = 1 = 2 0 . As induction hypothesis, suppose
that if n(A) = n then n(P(A)) = 2 n . Let B = {a 1 , a 2 , · · · , a n , a n+1 }. Then
P(B) consists of all subsets of {a 1 , a 2 , · · · , a n } together with all subsets of
{a 1 , a 2 , · · · , a n } with the element a n+1 added to them. Hence, n(P(B)) =
2 n + 2 n = 2 · 2 n = 2 n+1 .
(b) Since n(P(A)) = 256 = 2 8 then n(A) = 8
Example 1.12
Use induction to show ∑ n
i=1 (2i − 1) = n2 , n ≥ 1.
Solution.
If n = 1 we have 1 2 = 2(1)−1 = ∑ 1
i=1
(2i−1). Suppose that the result is true
for up to n. We will show that it is true for n + 1. Indeed, ∑ n+1
∑ i=1
(2i − 1) =
n
i=1 (2i − 1) + 2(n + 1) − 1 = n2 + 2n + 2 − 1 = (n + 1) 2