Download Practice with Induction and Solutions

Math 433
Induction Practice
Problem 1 Prove by induction that if A = {1, 2, 3, . . . , n}, then the power set,
P(A), has 2n elements.
Problem 2 Every integer greater than 1 is divisible by a prime.
Problem 3 For every n ∈ N, 2n3 + 3n2 + n is divisible by 6.
Problem 4 Prove by induction that
1 · 2 + 2 · 3 + 3 · 4 + · · · n · (n + 1) =
n(n + 1)(n + 2)
.
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(1)
Problem 5 If fn are the Fibonacci numbers, then for all n ∈ N,
n
X
f2r−1 = f2n .
(2)
r=1
1. Solution. Recall that for a (particular) set, A, the power set, P(A), is
the collection of subsets of A. We’ll use induction.
What is the sequence of statements?
(Sn): The number of elements in P({1, 2, . . . , n}) is 2n .
First we check that (S1) is true. But, the subsets of the set {1} are
{∅, {1}}. That is,
P({1}) = {∅, {1}}.
Hence, there are exactly two elements in P({1}).
Now assume that (S n) is true. That is, assume that the number of
elements in P({1, 2, . . . , n}) is 2n . We have to show that (S n+1) is true.
So, we have to count the number of subsets of {1, 2, . . . , n, n + 1}. But,
a subset of this last set is one of two types. Either the subset contains
n + 1 or it doesn’t. Those subsets which do not contain n + 1 are just
subsets of {1, 2, . . . , n}. And we know how many of these we have: 2n .
Those subsets which contain n + 1 can be described by taking a subset of
{1, 2, . . . , n} and throwing in the element n + 1. For example, if n + 1 = 5
and the subset of {1, 2, . . . , 5} which we would like to describe is {1, 3, 5},
we could say that
{1, 3, 5} = {1, 3} ∪ {5}.
Or, more generally, every set which contains n + 1 can be written as
A ∪ {n + 1}, where A is a subset of {1, 2, . . . , n}. Hence, to count these
sets we only have to count the A’s which are subsets of {1, 2, . . . , n}.
Again, this is just 2n . So, we’ve divided our problem into counting in two
distinct situations and in each case there are 2n elements. Therefore, the
total number of elements is 2n + 2n = 2 · 2n = 2n+1 . Therefore, we’ve
verified that (S n+1) is true. Hence, by PMI we know all (n ≥ 1) the
statements (S n) are true.
2. Solution. We did this in class.
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3. Solution. Let (S n): 2n3 + 3n2 + n is divisible by 6. (S 1) is just that
2 + 3 + 1 is divisible by 6, which is trivial. Assume that (S1), . . . , (Sn) are
true. Prove that (Sn + 1) is true. But,
2(n + 1)3 + 3(n + 1)2 + (n + 1)
= 2(n3 + 3n2 + 3n + 1) + 3(n2 + 2n + 1) + (n + 1)
= (2n3 + 3n2 + n) + (6n2 + 6n + 2 + 6n + 3 + 1)
= (2n3 + 3n2 + n) + 6(n2 + 2n + 1)
The first term is divisible by 6 since (Sn) is true and the second term is
a multiple of 6. Hence, the last quantity is divisible by 6.
4. Solution. (Fill in the details.) Check n = 1. Assume true for a particular
n. That is, (1) is true. But,
1 · 2 + 2 · 3 + 3 · 4 + · · · n · (n + 1) + (n + 1)(n + 2) =
= (n + 1)(n + 2)
n(n + 1)(n + 2)
+ (n + 1)(n + 2)
3
(n + 1)(n + 2)(n + 3)
n
+1 =
,
3
3
which shows that (S n+1) is true.
5. Solution. Recall f1 = 1, f2 = 1 and for n ≥ 2, fn = fn−2 + fn−1 . Check
n = 1. Assume (2) is true. Then,
n+1
X
r=1
f2r−1 =
n
X
f2r−1 + f2n+1 = (by the ind. hyp. )f2n + f2n+1
r=1
= f2(n+1) ( by the def. of Fib. ).
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