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Chapter 2: Mathematical Induction
MAT246H1S Lec0101 Burbulla
Chapter 2 Lecture Notes
Mathematical Induction
Winter 2017
Chapter 2 Lecture Notes Mathematical Induction
MAT246H1S Lec0101 Burbulla
Chapter 2: Mathematical Induction
Chapter 2: Mathematical Induction
2.1: The Principle of Mathematical Induction
2.2: The Principle of Complete Mathematical Induction
Chapter 2 Lecture Notes Mathematical Induction
MAT246H1S Lec0101 Burbulla
Chapter 2: Mathematical Induction
2.1: The Principle of Mathematical Induction
2.2: The Principle of Complete Mathematical Induction
The Well-Ordering Principle
Here is an important property of natural numbers which we regard
as ‘obvious:’
The Well-Ordering Principle: every non-empty set of natural
numbers contains a least element. That is, if S ⊆ N and S 6= ∅,
then there is an m ∈ S such that m ≤ n for all n ∈ S.
This principle gets its name from considering the ‘natural’ ordering
of numbers:
1 < 2 < 3 < 4 < ··· < n < n + 1 < ···
Then, if S is any non-empty set of natural numbers, we can ‘order’
them using ‘less than;’ some number m ∈ S must be the least one.
Note: we have not proved the well-ordering principle. Instead we
take it as an axiom, a proposition we regard as true without proof.
Chapter 2 Lecture Notes Mathematical Induction
Chapter 2: Mathematical Induction
MAT246H1S Lec0101 Burbulla
2.1: The Principle of Mathematical Induction
2.2: The Principle of Complete Mathematical Induction
Example 1
Let S be the set of ages, in whole years, of everyone taking this
course. Then S is non-empty, since there is at least one student in
this course, actually many more. Of all the students taking the
course, there must be a student whose age is less than or equal to
the age of all the other students in the course. That student’s age
is the least number in S.
Chapter 2 Lecture Notes Mathematical Induction
MAT246H1S Lec0101 Burbulla
Chapter 2: Mathematical Induction
2.1: The Principle of Mathematical Induction
2.2: The Principle of Complete Mathematical Induction
The Principle of Mathematical Induction
As we have pointed out previously, the set of natural numbers N is
closed under addition. That is, if m, n ∈ N, then m + n ∈ N. In
particular this means that if n ∈ N then n + 1 ∈ N. Consequently it
is possible to calculate every number n ∈ N simply by ‘adding 1’s:’
2 = 1 + 1, 3 = 2 + 1, 4 = 3 + 1, . . . , n = (n − 1) + 1, . . . etc
This observation can be formalized as
The Principle of Mathematical Induction: if S is any set of
natural numbers such that
I
A: 1 is in S, and
I
B: k + 1 is in S whenever k is in S,
then S = N.
Chapter 2 Lecture Notes Mathematical Induction
Chapter 2: Mathematical Induction
MAT246H1S Lec0101 Burbulla
2.1: The Principle of Mathematical Induction
2.2: The Principle of Complete Mathematical Induction
Proof of The Principle of Mathematical Induction
It may seem to you that the principle of mathematical induction is
intuitively obvious, and does not require proof. After all, by
property A, 1 ∈ S; so by property B, 1 + 1 = 2 ∈ S. Then by
property B, 2 + 1 = 3 ∈ S. By property B, again, 3 + 1 = 4 ∈ S.
And so on, ad infinitum. So S = N.
Surprise 1: we can prove the principle of mathematical induction
by using the well-ordering principle. That is, suppose S is a set of
natural numbers such that S satisfies properties A and B. Let T be
the set of natural numbers that are in N but not in S. (Aside:
S = N if and only if T = ∅.) Suppose T 6= ∅. Then T is a
non-empty set of natural numbers. By the well-ordering principle
there is a least number in T , call it t 6= 1. (Why?) Then t − 1 is
not in T ; therefore t − 1 is in S. By property B, (t − 1) + 1 = t is in
S, and we have a contradiction. (Why?) Thus T = ∅, and S = N.
Chapter 2 Lecture Notes Mathematical Induction
MAT246H1S Lec0101 Burbulla
Chapter 2: Mathematical Induction
2.1: The Principle of Mathematical Induction
2.2: The Principle of Complete Mathematical Induction
But wait! We didn’t prove the well-ordering principle, we just
assumed it. So all we are doing is reducing one obvious statement
to another obvious statement, which may not seem all that useful.
However, this is a common process in mathematics: how does one
proposition depend on another? Logically, which propositions are
more basic than other propositions? What is the least we must
assume to prove something else is true? What we have shown is
that if we assume the well-ordering principle, then the principle of
mathematical induction follows as a logical consequence. So in
some sense, the well-ordering principle is more basic than the
principle of mathematical induction. Or is it?
Surprise 2: it turns out that if you assume the principle of
mathematical induction you can prove the well-ordering principle.
(See the Challenging Problems at the end of Section 2.2.) So the
two principles are logically equivalent, which may not surprise you
at all, since you might consider them both ‘equally obvious.’
Chapter 2 Lecture Notes Mathematical Induction
Chapter 2: Mathematical Induction
MAT246H1S Lec0101 Burbulla
2.1: The Principle of Mathematical Induction
2.2: The Principle of Complete Mathematical Induction
Proof by Induction
The principle of mathematical induction can be used to prove
statements about whole numbers. Such a proof is known as proof
by induction; it is a very useful technique that you should become
comfortable with. Perhaps you have seen it before, in high school,
or in a previous math course. Here is an example:
Theorem: for every natural number n,
n
X
n(n + 1)
n(n + 1)
(†) 1 + 2 + 3 + · · · + n =
⇔
i=
.
2
2
i=1
Proof: let S be the set of natural numbers
n ∈ N for which (†)
(
)
n
X
n(n + 1)
i=
.
holds. In set theory notation: S = n ∈ N |
2
i=1
Chapter 2 Lecture Notes Mathematical Induction
MAT246H1S Lec0101 Burbulla
2.1: The Principle of Mathematical Induction
2.2: The Principle of Complete Mathematical Induction
Chapter 2: Mathematical Induction
We shall prove that the formula (†) is true for all natural numbers
by using the principle of mathematical induction to show that
S =N:
I
A: if n = 1 the left side of the formula (†) is simply 1 and the
1 · (1 + 1)
= 1. Hence 1 ∈ S.
2
B: now suppose k ∈ S and check if k + 1 ∈ S :
right side of the formula (†) is
I
1 + 2 + 3 + · · · + k + (k + 1)
k(k + 1)
+ (k + 1), (Why?)
2
k(k + 1) 2(k + 1)
+
2
2
(k + 1)(k + 2)
.
2
=
=
=
Thus (†) holds for n = k + 1, and so k + 1 ∈ S. By the principle of
mathematical induction, S = N.
Chapter 2 Lecture Notes Mathematical Induction
Chapter 2: Mathematical Induction
MAT246H1S Lec0101 Burbulla
2.1: The Principle of Mathematical Induction
2.2: The Principle of Complete Mathematical Induction
Example 2: A Less Formal Way to do Induction Proofs
Prove that for all natural numbers n
12 + 22 + 32 + · · · + n2 =
n(n + 1)(2n + 1)
.
6
If n = 1 the formula is true as the right side becomes 1 · 2 · 3/6 = 1.
If the formula is true for n = k, then
12 + 22 + 32 + · · · + k 2 + (k + 1)2 =
k(k + 1)(2k + 1)
+ (k + 1)2
6
k(k + 1)(2k + 1) 6(k + 1)2
(k + 1)(2k 2 + k + 6k + 6)
=
+
=
6
6
6
(k + 1)(k + 2)(2(k + 1) + 1)
=
, and the formula is true for
6
n = k + 1. By induction the formula is true for all n ∈ N.
Chapter 2 Lecture Notes Mathematical Induction
MAT246H1S Lec0101 Burbulla
Chapter 2: Mathematical Induction
2.1: The Principle of Mathematical Induction
2.2: The Principle of Complete Mathematical Induction
Example 3: A Proof by Induction in Linear Algebra
Theorem: Eigenvectors corresponding to distinct eigenvalues of
the n × n matrix A are linearly independent.
Proof: by induction on the number of distinct eigenvalues of A.
Suppose λ1 6= λ2 and A ~u1 = λ1 ~u1 , A ~u2 = λ2 ~u2 , ~ui 6= ~0. We claim
{~u1 , ~u2 } is independent. Suppose a1 ~u1 + a2 ~u2 = ~0. Then
λ1 (a1~u1 + a2~u2 ) = ~0
λ1 a1~u1 + λ1 a2~u2 = ~0
⇔
A(a1~u1 + a2~u2 ) = ~0
a1 λ1~u1 + a2 λ2~u2 = ~0
Subtract the last two equations:
(λ1 − λ2 )a2~u2 = ~0 ⇒ a2 = 0.
Then a1~u1 = ~0 ⇒ a1 = 0. So ~u1 , ~u2 are linearly independent. Now
suppose that ~u1 , ~u2 , . . . , ~uk , ~uk+1 are eigenvectors of A
Chapter 2 Lecture Notes Mathematical Induction
Chapter 2: Mathematical Induction
MAT246H1S Lec0101 Burbulla
2.1: The Principle of Mathematical Induction
2.2: The Principle of Complete Mathematical Induction
corresponding to distinct eigenvalues λ1 , λ2 , . . . , λk , λk+1 of A.
Suppose {~u1 , ~u2 , . . . , ~uk , ~uk+1 } is dependent. By induction we can
assume {~u1 , ~u2 , . . . , ~uk } is independent; thus
~uk+1 = a1~u1 + · · · + ak ~uk ,
for some unique scalars ai , 1 ≤ i ≤ k. Multiply both sides by A :
λk+1~uk+1 = a1 λ1~u1 + · · · + ak λk ~uk .
If λk+1 = 0 then λ1 , . . . , λk 6= 0, (Why?) and
~0 = a1 λ1~u1 + · · · + ak λk ~uk ⇒ ai = 0 ⇒ ~uk+1 = ~0.
a1 λ1
a λ
~u1 + · · · + k k ~uk , and so in
λk+1
λk+1
particular λ1 = λk+1 . Either way, we obtain a contradiction. ¶
Does this proof really use the principle of mathematical induction?
If λk+1 6= 0, then ~uk+1 =
Chapter 2 Lecture Notes Mathematical Induction
MAT246H1S Lec0101 Burbulla
Chapter 2: Mathematical Induction
2.1: The Principle of Mathematical Induction
2.2: The Principle of Complete Mathematical Induction
Generalized Principle of Mathematical Induction
One thing illustrated by Example 3 is that we don’t always want to
start an induction proof at n = 1. Some times we wish to start
with a whole number m > 1 and then show some formula or
statement is true for all natural numbers n ≥ m. This can be
formalized by the generalized principle of mathematical induction:
If S is a set of natural numbers with the properties that
I
A: m is in S, and
I
B: k + 1 is in S whenever k is in S and is greater than or
equal to m,
then S = {n ∈ N | n ≥ m}.
Chapter 2 Lecture Notes Mathematical Induction
Chapter 2: Mathematical Induction
MAT246H1S Lec0101 Burbulla
2.1: The Principle of Mathematical Induction
2.2: The Principle of Complete Mathematical Induction
Example 4
Theorem 2.1.5: n! > 2n for all n ≥ 4.
Proof: Aside: 1! = 1 < 2, 2! = 2 < 22 = 4, and 3! = 6 < 23 = 8,
so the formula is not true if n < 4. Let S = {n ∈ N | n! > 2n }.
I
A: 4 ∈ S since 4! = 24 > 24 = 16.
I
B: if k ∈ S then k! > 2k , and so
(k + 1)! = (k + 1)k! > (k + 1)2k > 2 · 2k = 2k+1 ,
so k + 1 is in S too.
Thus by the generalized principle of mathematical induction,
S = {n ∈ N | n ≥ 4}.
Chapter 2 Lecture Notes Mathematical Induction
MAT246H1S Lec0101 Burbulla
Chapter 2: Mathematical Induction
2.1: The Principle of Mathematical Induction
2.2: The Principle of Complete Mathematical Induction
The Principle of Complete Mathematical Induction
If S is any set of natural numbers with the properties that
I
A: 1 is in S, and
I
B: k + 1 is in S whenever k is a natural number and all the
natural numbers from 1 through k are in S,
then S = N.
Compared to the principle of mathematical induction, the principle
of complete mathematical induction assumes more in Step B: it
assumes that n = 1, 2, 3, . . . , k are all in S.
Note: the principle of complete mathematical induction is also
known as the principle of strong mathematical induction.
Chapter 2 Lecture Notes Mathematical Induction
Chapter 2: Mathematical Induction
MAT246H1S Lec0101 Burbulla
2.1: The Principle of Mathematical Induction
2.2: The Principle of Complete Mathematical Induction
Proof of the Principle of Complete Mathematical Induction
The principle of complete mathematical induction is also a
consequence of the well-ordering principle. That is, suppose S is a
set of natural numbers that satisfies the two properties:
I
A: 1 is in S, and
I
B: k + 1 is in S whenever k is a natural number and all the
natural numbers from 1 through k are in S.
Let T be the set of natural numbers that are not in S. Suppose
T 6= ∅. Then by the well-ordering principle, there is a least element
in T . Say it is t 6= 1. Then all the numbers 1, 2, 3, . . . , t − 1 must
be in S. By property B, (t − 1) + 1 = t must be in S. But this
contradicts our assumption that t ∈
/ S. Thus we must have T = ∅
and S = N.
Chapter 2 Lecture Notes Mathematical Induction
MAT246H1S Lec0101 Burbulla
Chapter 2: Mathematical Induction
2.1: The Principle of Mathematical Induction
2.2: The Principle of Complete Mathematical Induction
Example 1: Exercise 15 of Section 2.2
Consider the sequence of numbers defined by F1 = 1, F2 = 1, and
Fn = Fn−1 + Fn−2 ,
for n ≥ 3. This is a recursive definition of the sequence of numbers
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, . . .
known as the Fibonacci numbers. We claim
"
√ !n
√ !n #
1+ 5
1
1− 5
Fn = √
−
,
2
2
5
√
1+ 5
2 ,b
√
1− 5
2 ,
which is far from obvious! Let a =
=
and observe
2
that a and b are the roots of the quadratic x = x + 1.
Chapter 2 Lecture Notes Mathematical Induction
Chapter 2: Mathematical Induction
MAT246H1S Lec0101 Burbulla
2.1: The Principle of Mathematical Induction
2.2: The Principle of Complete Mathematical Induction
an − b n
Let S = n ∈ N | Fn = √
.
5
√
5
a−b
A: 1 is in S, since √ = √ = 1 = F1 .
5
5
B: assume all the natural numbers from 1 through k are in S.
In particular k and k − 1 are in S, so
ak − b k
ak−1 − b k−1
√
Fk+1 = Fk + Fk−1 = √
+
=
5
5
I
I
ak−1 (a + 1) − (b k−1 (b + 1))
ak + ak−1 − (b k + b k−1 )
√
√
=
5
5
ak−1 · a2 − b k−1 · b 2
ak+1 − b k+1
√
√
=
=
.
5
5
Thus k + 1 ∈ S; and by complete mathematical induction, S = N.
Chapter 2 Lecture Notes Mathematical Induction
MAT246H1S Lec0101 Burbulla
Chapter 2: Mathematical Induction
2.1: The Principle of Mathematical Induction
2.2: The Principle of Complete Mathematical Induction
Generalized Principle of Complete Mathematical Induction
If S is any set of natural numbers with the properties that
I
A: m is in S, and
I
B: k + 1 is in S whenever k is a natural number and all the
natural numbers from m through k are in S,
then S = {n ∈ N | n ≥ m}. Here is an example that uses the
generalized principle of complete mathematical induction:
Theorem 2.2.4: every natural number other than 1 is a product
of prime numbers.
Proof: by generalized complete induction. Let S be the set of all
natural numbers n that are products of primes. Note: if n is itself
a prime, we consider it to be a “product” of a single prime.
I
A: 2 ∈ S, since 2 is a prime.
Chapter 2 Lecture Notes Mathematical Induction
Chapter 2: Mathematical Induction
I
MAT246H1S Lec0101 Burbulla
2.1: The Principle of Mathematical Induction
2.2: The Principle of Complete Mathematical Induction
B: now suppose that every natural number from 2 to k is in
S. Consider k + 1. If k + 1 is prime, then k + 1 is in S, and
we are done. If k + 1 is not prime, then it is composite. So
there are natural numbers x, y such that
k + 1 = x · y and 1 < x, y < k + 1.
By the inductive hypothesis, both x and y are in S. Thus x
and y are both products of prime numbers, and so x · y is also
a product of primes: all the primes comprising x times all the
primes comprising y . Then by the generalized principle of
complete mathematical induction,
S = {n ∈ N | n ≥ 2}.
Chapter 2 Lecture Notes Mathematical Induction
MAT246H1S Lec0101 Burbulla