In Class Problem Set #12
CSE 1400 and MTH 2051
Fall 2012
Induction
1. Follow the mathematical induction steps below to prove there are
2n truth assignments on n ≥ 1 Boolean variables.
(a) Establish a basis: Show there are 21 = 2 truth assignments on
n = 1 variables.
Answer: Let p be that variable. There are 2 truth assignments
on p:
p = 0 (False)
and
p = 1 (True)
You could argue there is 20 = 1 truth assignments on n = 0 (no)
variables: The empty assignment that maps no variables to no
truth value.
(b) Make a hypothesis: For some n ∈ N, there are 2n truth assignments on n variables.
Answer: Hypothesis: For some n ∈ N, there are 2n truth assignments on n variables.
(c) Use the hypothesis: Show there are 2n+1 truth assignments on
n + 1 variables.
Answer: Consider n + 1 Boolean variables. To talk about them,
give them names:
p 0 , p 1 , . . . , p n −1 , p n
By the induction hypothesis there are 2n truth assignments
on the first n variables. There are two (and only two) ways to
extend each of these truth assignments to n + 1 variable:
p0 , p1 , . . . , p n−1 , False
and
p0 , p1 , . . . , p n−1 , True
Therefore there are 2n + 2n = 2 · 2n = 2n+1 truth assignments on
n + 1 variables.
2. Follow the mathematical induction steps below to prove that a set
X with n ≥ 0 elements has 2n subsets.
(a) Establish a basis: Show that if X is the empty set (∅) then X
has 1 subset.
Answer: The empty set is a subset of itself (∅ ⊆ ∅) and ∅ is the
only subset of ∅.
You can check n = 2 too by noting there
are 4 truth assignments on two Boolean
variables: p and q.
in class problem set #12
(b) Make a hypothesis: For some n ≥ 0, if X has n elements1 then
X has 2n subsets.
Answer: Hypothesis: For some n ≥ 0, if |X| = n, then X has
2n subsets.
2
The 50¢ way to say this is “the cardinality |X| of X is equal to n.
1
(c) Use the hypothesis: Show a set X with n + 1 elements has 2n+1
subsets.
Answer: To talk about the elements of X, give them names
X = { x 0 , x 1 , . . . , x n −1 , x n }
If A is a subset of X (A ⊆ X), then either
xn 6∈ A
xn ∈ A
or
• If xn 6∈ A, then A is a subset of the first n variables and
(by the induction hypothesis) there are 2n subsets like A. If
xn ∈ A, then A = A0 ∪ { xn } where A0 is a subset of the first
n variables. Therefore there are 2n subsets A with xn ∈ A,
• If xn ∈ A, then A = A0 ∪ { xn } where A0 is a subset of
the first n variables. Therefore there are 2n subsets A with
xn ∈ A.
Therefore there are
2n + 2n = 2 · 2n = 2n +1
subsets of a set with n + 1 variables.
3. Follow the mathematical induction steps below to prove there are
n
22 Boolean functions of n ≥ 0 Boolean variables.
0
(a) Establish a basis: Show there are 22 = 21 = 2 Boolean functions on 0 (no) variables.
Answer: There are two Boolean constant (functions) True and
False.
n
(b) Make a hypothesis: For some n ∈ N, there are 22 Boolean
functions on n variables.
You can check n = 1 too by noting there
are 4 Boolean functions on one Boolean
variables:
B0 ( p ) = False, B1 ( p ) = p
B2 ( p ) = ¬ p, B3 ( p ) = True
n
Answer: Hypothesis: For some n ∈ N, there are 22 Boolean
functions on n variables.
(c) Use the hypothesis: Show there are 22
n + 1 variables.
n +1
Boolean functions on
Answer: Look what’s happening in an example. There are 8 =
23 truth assignments on 3 variables. To define a function, we
must map each of these 8 assignments to one of two values:
True or False. So there are 28 different Boolean functions on
3 variables. By problem 1, there are 2n+1 truth assignments on
n + 1 variables. A Boolean function is defined by mapping each
These four functions can be described
by columns in a truth table. Each row
in the table is a truth assignment. Each
column is a function.
Input
p
0
1
Four functions on a single input
B0 ( p ) B1 ( p ) B2 ( p )
B3 ( p )
q
q
q
q
False
p
¬p
True
0
0
0
1
1
0
1
1
in class problem set #12
3
truth assignment to one of the values: True or False. Therefore
n +1
there are 22
Boolean functions on n + 1 variables.
4. Follow the mathematical induction steps below to prove 2n > n
for n ∈ N.
(a) Establish a basis: Show 2n > n for n = 0.
Answer: For n = 0, 20 = 1 > 0. Note also, for n = 1, 21 = 2 > 1.
(b) Make a hypothesis: For some n ∈ N,2n > n.
Answer: Hypothesis: For some n ∈ N, 2n > n.
(c) Use the hypothesis: Show 2n+1 > n + 1.
Answer: By the induction hypothesis, 2n > n and for every n,
2n > 1. Therefore
2 n + 1 = 2 n 21 = 2 · 2 n = 2 n + 2 n > n + 1
5. Follow the steps of mathematical induction to prove the sum of
−1
2
the first n odd natural numbers ∑nk=
0 (2k + 1) is equal to n .
Answer:
• Basis: When n = 0, n2 = 0 and the sum of no terms is empty
and defined to have the value 0.
• Hypothesis: For some n ∈ N,
n −1
∑ (2k + 1) = n2
k =0
• Use: Show 2n+1 > n + 1.
n −1
∑ (2k + 1) = n2
k =0
By the induction hypothesis, the sum of the first n odd natural
numbers is n2 , that is,
n −1
∑ (2k + 1) = n2
k =0
The (n + 1)st odd natural number is 2n + 1. Therefore, the sum
of the first n + 1 odd natural numbers is
n
n −1
k =0
k =0
∑ (2k + 1) = ∑ (2k + 1) + (2n + 1) = n2 + (2n + 1) = (n + 1)2
Don’t like empty arguments? When
n = 1, n2 = 1 and the sum has one
term,
1−1
∑ (2k + 1) = (2 · 0 + 1) = 1 = 12
k =0
That is, the value of the sum equals the
value of the function.
in class problem set #12
6. Pretend you have a fortune controlled by a trustee. The instructions to the trustee are to give you one-half of the fortune at the
start of the first year, ole-half of what’s left at the start of the
second year, and repeating, at the start of each year, one-half of
what’s left will be given to you. At the start of the nth year, what
percentage of the fortune will have been given to you?
k
Answer: To answer the question sum the powers 12 from k = 1 to
k = n.
n 2
1 1 1
1
1
∑ 2 = 2 + 4 + 8 + · · · + 2n
k =1
n +1
1
−1
2
=
−1
1
2 −1
n+1
1
= 2 1−
−1
2
n +1
1
= 1−2
2
n
1
= 1−
2
4