§ 2.4 Mathematical Induction
What Is (Weak) Induction?
The Principle of Mathematical Induction works like this:
What Is (Weak) Induction?
The Principle of Mathematical Induction works like this:
We want to show some statement P(n) is true for all n ≥ n0 .
What Is (Weak) Induction?
The Principle of Mathematical Induction works like this:
We want to show some statement P(n) is true for all n ≥ n0 .
We show some base case P(n0 ) is true.
What Is (Weak) Induction?
The Principle of Mathematical Induction works like this:
We want to show some statement P(n) is true for all n ≥ n0 .
We show some base case P(n0 ) is true.
We assume P(k) is true for some k ≥ n0 .
What Is (Weak) Induction?
The Principle of Mathematical Induction works like this:
We want to show some statement P(n) is true for all n ≥ n0 .
We show some base case P(n0 ) is true.
We assume P(k) is true for some k ≥ n0 .
We show P(k + 1) is true.
What Is (Weak) Induction?
The Principle of Mathematical Induction works like this:
We want to show some statement P(n) is true for all n ≥ n0 .
We show some base case P(n0 ) is true.
We assume P(k) is true for some k ≥ n0 .
We show P(k + 1) is true.
So, basically, we are trying to show P(k) → P(k + 1) is a tautology
for any choice k ≥ n0 .
The Idea Behind Induction
How to think of induction ...
The Idea Behind Induction
How to think of induction ...
If a baby can take the first step ...
The Idea Behind Induction
How to think of induction ...
If a baby can take the first step ...
If you can step on the first rung of a ladder ...
When To Use (Weak) Induction
We use mathematical induction when ...
we have a series and we are trying to prove the general formula.
When To Use (Weak) Induction
We use mathematical induction when ...
we have a series and we are trying to prove the general formula.
we are trying to prove a ‘counting’ problem.
When To Use (Weak) Induction
We use mathematical induction when ...
we have a series and we are trying to prove the general formula.
we are trying to prove a ‘counting’ problem.
there seems to be a pattern that seems to change in a fixed
manner.
First (Famous) Example
Example
Prove that if n is a positive integer, then
1 + 2 + ... + n =
n
X
i=1
i=
n(n + 1)
2
First (Famous) Example
Example
Prove that if n is a positive integer, then
1 + 2 + ... + n =
n
X
i=1
i=
n(n + 1)
2
Proof.
Let P(n) be the proposition that the sum of the first n positive integers
is n(n+1)
2 . We will show this is true by mathematical induction.
First (Famous) Example
Example
Prove that if n is a positive integer, then
1 + 2 + ... + n =
n
X
i=1
i=
n(n + 1)
2
Proof.
Let P(n) be the proposition that the sum of the first n positive integers
is n(n+1)
2 . We will show this is true by mathematical induction.
Base Case:
First (Famous) Example
Example
Prove that if n is a positive integer, then
1 + 2 + ... + n =
n
X
i=1
i=
n(n + 1)
2
Proof.
Let P(n) be the proposition that the sum of the first n positive integers
is n(n+1)
2 . We will show this is true by mathematical induction.
Base Case: P(1)
First (Famous) Example
Example
Prove that if n is a positive integer, then
1 + 2 + ... + n =
n
X
i=1
i=
n(n + 1)
2
Proof.
Let P(n) be the proposition that the sum of the first n positive integers
is n(n+1)
2 . We will show this is true by mathematical induction.
Base Case: P(1)
P(1) is true because 1 =
1(1+1)
2 .
First (Famous) Example
Example
Prove that if n is a positive integer, then
1 + 2 + ... + n =
n
X
i=1
i=
n(n + 1)
2
Proof.
Let P(n) be the proposition that the sum of the first n positive integers
is n(n+1)
2 . We will show this is true by mathematical induction.
Base Case: P(1)
P(1) is true because 1 =
1(1+1)
2 .
First (Famous) Example
Proof.
Inductive Step: We assume that P(k) holds for any positive integer k.
That is, we assume
1 + 2 + ... + k =
k
X
i=1
i=
k(k + 1)
2
First (Famous) Example
Proof.
Inductive Step: We assume that P(k) holds for any positive integer k.
That is, we assume
1 + 2 + ... + k =
k
X
i=1
We want to show P(k + 1) is true.
i=
k(k + 1)
2
First (Famous) Example
Proof.
Inductive Step: We assume that P(k) holds for any positive integer k.
That is, we assume
1 + 2 + ... + k =
k
X
i=1
i=
k(k + 1)
2
We want to show P(k + 1) is true.
Aside: P(k + 1) is the same thing as the formula we have but we
replace k with k + 1. So our goal is to arrive at
k+1
X
i=1
=
(k + 1)(k + 2)
2
Proof.
Consider
k+1
X
i=1
i=
k
X
i=1
i + (k + 1)
Proof.
Consider
k+1
X
i=1
i=
k
X
i + (k + 1)
i=1
k(k + 1)
=
+k+1
2
Proof.
Consider
k+1
X
i=1
i=
k
X
i + (k + 1)
i=1
k(k + 1)
=
+k+1
2
k(k + 1) 2(k + 1)
=
+
2
2
Proof.
Consider
k+1
X
i=1
i=
k
X
i + (k + 1)
i=1
k(k + 1)
=
+k+1
2
k(k + 1) 2(k + 1)
=
+
2
2
k2 + 3k + 2
=
2
Proof.
Consider
k+1
X
i=1
i=
k
X
i + (k + 1)
i=1
k(k + 1)
=
+k+1
2
k(k + 1) 2(k + 1)
=
+
2
2
k2 + 3k + 2
=
2
(k + 1)(k + 2)
=
2
Proof.
Consider
k+1
X
i=1
i=
k
X
i + (k + 1)
i=1
k(k + 1)
=
+k+1
2
k(k + 1) 2(k + 1)
=
+
2
2
k2 + 3k + 2
=
2
(k + 1)(k + 2)
=
2
This last equation shows that P(k + 1) is true under the assumption
that P(k) is true. So, by mathematical induction, we have shown that
that
n
X
n(n + 1)
i=
∀n ∈ Z +
2
i=1
Union/Intersection Example
Example
Let A1 , A2 , . . . , An be any n sets. Show that
n
[
i=1
!
Ai
=
n
\
i=1
Ai
Union/Intersection Example
Example
Let A1 , A2 , . . . , An be any n sets. Show that
n
[
i=1
Note: Does this look familiar?
!
Ai
=
n
\
i=1
Ai
Union/Intersection Example
Example
Let A1 , A2 , . . . , An be any n sets. Show that
n
[
i=1
!
Ai
=
n
\
Ai
i=1
Note: Does this look familiar? This is just an extension of
DeMorgan’s law.
Union/Intersection Example
Example
Let A1 , A2 , . . . , An be any n sets. Show that
n
[
i=1
!
Ai
=
n
\
Ai
i=1
Note: Does this look familiar? This is just an extension of
DeMorgan’s law.
Proof.
Let P(n) be the predicate that the equality holds for any n sets. We
will prove by mathematical induction that for all n ≥ 1, P(n) is true.
Union/Intersection Example
Proof.
Base Case:
Union/Intersection Example
Proof.
Base Case: P(1):
Union/Intersection Example
Proof.
Base Case: P(1):
Certainly A1 = A1 is true.
Union/Intersection Example
Proof.
Base Case: P(1):
Certainly A1 = A1 is true.
Inductive Step: We assume P(k)
k
[
i=1
!
Ai
=
k
\
i=1
is true. We want to show P(k + 1) is true.
Ai
Union/Intersection Example
Proof.
Consider
k+1
[
i=1
!
Ai
= A1 ∪ A2 ∪ . . . ∪ Ak ∪ Ak+1
Union/Intersection Example
Proof.
Consider
k+1
[
!
Ai
= A1 ∪ A2 ∪ . . . ∪ Ak ∪ Ak+1
i=1
= (A1 ∪ A2 ∪ . . . ∪ Ak ) ∪ Ak+1
Union/Intersection Example
Proof.
Consider
k+1
[
!
Ai
= A1 ∪ A2 ∪ . . . ∪ Ak ∪ Ak+1
i=1
= (A1 ∪ A2 ∪ . . . ∪ Ak ) ∪ Ak+1
= (A1 ∪ A2 ∪ . . . ∪ Ak ) ∩ Ak+1
Union/Intersection Example
Proof.
Consider
k+1
[
!
Ai
= A1 ∪ A2 ∪ . . . ∪ Ak ∪ Ak+1
i=1
= (A1 ∪ A2 ∪ . . . ∪ Ak ) ∪ Ak+1
= (A1 ∪ A2 ∪ . . . ∪ Ak ) ∩ Ak+1
!
k
\
=
Ai ∩ Ak+1
i=1
Union/Intersection Example
Proof.
Consider
k+1
[
!
Ai
= A1 ∪ A2 ∪ . . . ∪ Ak ∪ Ak+1
i=1
= (A1 ∪ A2 ∪ . . . ∪ Ak ) ∪ Ak+1
= (A1 ∪ A2 ∪ . . . ∪ Ak ) ∩ Ak+1
!
k
\
=
Ai ∩ Ak+1
i=1
=
k+1
\
i=1
!
Ai
Union/Intersection Example
Proof.
Consider
k+1
[
!
Ai
= A1 ∪ A2 ∪ . . . ∪ Ak ∪ Ak+1
i=1
= (A1 ∪ A2 ∪ . . . ∪ Ak ) ∪ Ak+1
= (A1 ∪ A2 ∪ . . . ∪ Ak ) ∩ Ak+1
!
k
\
=
Ai ∩ Ak+1
i=1
=
k+1
\
!
Ai
i=1
So, we have shown that P(k + 1) is true when we assume P(k) is true,
and so by mathematical induction we have proven our statement.
Inequalities
Example
Use mathematical induction to prove the inequality
n < 2n
for all positive integers n.
Inequalities
Example
Use mathematical induction to prove the inequality
n < 2n
for all positive integers n.
Proof.
Let P(n) be the proposition that n < 2n .
Inequalities
Example
Use mathematical induction to prove the inequality
n < 2n
for all positive integers n.
Proof.
Let P(n) be the proposition that n < 2n .
Base Case:
Inequalities
Example
Use mathematical induction to prove the inequality
n < 2n
for all positive integers n.
Proof.
Let P(n) be the proposition that n < 2n .
Base Case: P(1)
Inequalities
Example
Use mathematical induction to prove the inequality
n < 2n
for all positive integers n.
Proof.
Let P(n) be the proposition that n < 2n .
Base Case: P(1)
Since 1 < 21 = 2, P(1) is true.
Inequalities
Example
Use mathematical induction to prove the inequality
n < 2n
for all positive integers n.
Proof.
Let P(n) be the proposition that n < 2n .
Base Case: P(1)
Since 1 < 21 = 2, P(1) is true.
Inductive Step: Assume P(k) is true. That is, assume k < 2k for
k ∈ Z + . We want to show that P(k + 1) is true.
Inequalities
Proof.
Consider
k + 1 < 2k + 1
Inequalities
Proof.
Consider
k + 1 < 2k + 1
≤ 2k + 2k
Inequalities
Proof.
Consider
k + 1 < 2k + 1
≤ 2k + 2k
= 2 · 2k
Inequalities
Proof.
Consider
k + 1 < 2k + 1
≤ 2k + 2k
= 2 · 2k
= 2k+1
Inequalities
Proof.
Consider
k + 1 < 2k + 1
≤ 2k + 2k
= 2 · 2k
= 2k+1
This shows that P(k + 1) is true, based on the assumption that P(k) is
true. Therefore, by mathematical induction, we have shown that for
all n ∈ Z + , n < 2n .
Harmonic Series Example
Definition
The harmonic series, Hj , is the series
j
X
1
i=1
i
=1+
1 1
+ + ...
2 3
Harmonic Series Example
Definition
The harmonic series, Hj , is the series
j
X
1
i=1
i
=1+
1 1
+ + ...
2 3
Example
Prove for any nonnegative integer n, H2n ≥ 1 +
n
2
Harmonic Series Example
Proof.
Let P(n) be the proposition that H2n ≥ 1 + n2 .
Harmonic Series Example
Proof.
Let P(n) be the proposition that H2n ≥ 1 + n2 .
Base Case:
Harmonic Series Example
Proof.
Let P(n) be the proposition that H2n ≥ 1 + n2 .
Base Case: P(0)
Harmonic Series Example
Proof.
Let P(n) be the proposition that H2n ≥ 1 + n2 .
Base Case: P(0)
P(0) is true because H20 = H1 = 1 ≥ 1 + 20 .
Harmonic Series Example
Proof.
Let P(n) be the proposition that H2n ≥ 1 + n2 .
Base Case: P(0)
P(0) is true because H20 = H1 = 1 ≥ 1 + 20 .
Inductive Step: Assume P(k) is true. That is, assume H2k ≥ 1 + 2k ,
whenever k is a nonnegative integer.
Harmonic Series Example
Proof.
Let P(n) be the proposition that H2n ≥ 1 + n2 .
Base Case: P(0)
P(0) is true because H20 = H1 = 1 ≥ 1 + 20 .
Inductive Step: Assume P(k) is true. That is, assume H2k ≥ 1 + 2k ,
whenever k is a nonnegative integer. We want to show that if P(k) is
true, then P(k + 1), which states that H2k+1 ≥ 1 + k+1
2 , is also true.
Harmonic Series Example
Proof.
Consider
H2k+1 = 1 +
1
1
1
1 1
+ + ... + k + k
+ . . . + k+1
2 3
2
2 +1
2
Harmonic Series Example
Proof.
Consider
H2k+1 = 1 +
1
1
1
1 1
+ + ... + k + k
+ . . . + k+1
2 3
2
2 +1
2
1
1
= H2k + k
+ . . . + k+1
2 +1
2
Harmonic Series Example
Proof.
Consider
H2k+1 = 1 +
1
1
1
1 1
+ + ... + k + k
+ . . . + k+1
2 3
2
2 +1
2
1
1
= H2k + k
+ . . . + k+1
2
+
1
2
k
1
1
≥ 1+
+ . . . + k+1
+ k
2
2 +1
2
Harmonic Series Example
Proof.
Consider
H2k+1 = 1 +
1
1
1 1
+ + ... + k + k
+ ... +
2 3
2
2 +1
1
= H2k + k
+ ... +
2
+1
k
1
≥ 1+
+ ... +
+ k
2
2 +1
k
≥ 1+
+ 2k ·
2
1
2k+1
1
2k+1
1
k+1
2
1
k+1
2
Harmonic Series Example
Proof.
≥
1
k
+
1+
2
2
Harmonic Series Example
Proof.
≥
k
+
1+
2
3
= +
2
1
2
k
2
Harmonic Series Example
Proof.
≥
1
k
+
1+
2
2
3 k
= +
2 2
2+k+1
=
2
Harmonic Series Example
Proof.
≥
1
k
+
1+
2
2
3 k
= +
2 2
2+k+1
=
2
k+1
=1+
2
Harmonic Series Example
Proof.
≥
This establishes the proof.
1
k
+
1+
2
2
3 k
= +
2 2
2+k+1
=
2
k+1
=1+
2
Divisibility Example
Example
Prove that n3 − n is divisible by 3 whenever n ∈ Z + .
Divisibility Example
Example
Prove that n3 − n is divisible by 3 whenever n ∈ Z + .
Proof.
Let P(n) denote the proposition ‘n3 − n is divisible by 3’.
Divisibility Example
Example
Prove that n3 − n is divisible by 3 whenever n ∈ Z + .
Proof.
Let P(n) denote the proposition ‘n3 − n is divisible by 3’.
Base Case:
Divisibility Example
Example
Prove that n3 − n is divisible by 3 whenever n ∈ Z + .
Proof.
Let P(n) denote the proposition ‘n3 − n is divisible by 3’.
Base Case: P(1)
P(1) is true because 13 − 1 = 0 and 0 is divisible by 3.
Divisibility Example
Example
Prove that n3 − n is divisible by 3 whenever n ∈ Z + .
Proof.
Let P(n) denote the proposition ‘n3 − n is divisible by 3’.
Base Case: P(1)
P(1) is true because 13 − 1 = 0 and 0 is divisible by 3.
Inductive Step: Assume P(k) is true, which is to say that if k ∈ Z + .
k3 − k is divisible by 3. We want to show that P(k + 1) is true, where
P(k + 1) is (k + 1)3 − (k + 1).
Divisibility Example
Proof.
Now,
(k + 1)3 − (k + 1) =
Divisibility Example
Proof.
Now,
(k + 1)3 − (k + 1) = k3 + 3k2 + 3k + 1 − k − 1
Divisibility Example
Proof.
Now,
(k + 1)3 − (k + 1) = k3 + 3k2 + 3k + 1 − k − 1
(k3 − k) + 3(k2 + k)
Divisibility Example
Proof.
Now,
(k + 1)3 − (k + 1) = k3 + 3k2 + 3k + 1 − k − 1
(k3 − k) + 3(k2 + k)
k3 − k is divisible by 3 by our inductive hypothesis. And, since k2 + k
is an integer for any integer k, we have that 3(k2 + k) is also divisible
by 3. It follows that (k3 − k) + 3(k2 + k) is divisible by 3, completing
the proof.