Mathematical Induction
The Induction Principle
In the axiomatic construction of the natural numbers ℕ = {1, 2,
3, ...} (Peano, 1899), one of the axioms needed is the Principle of
Mathematical Induction (PMI):
If S is a subset of ℕ such that;
1) 1 ∈ S, and
2) for all n ∈ ℕ, if n ∈ S then n+1 ∈ S,
then S = ℕ.
This principle expresses a fundamental property of the natural
numbers which is often used in proving statements about sets of
natural numbers.
Definition: A set S of natural numbers satisfying 2) above is
called an inductive set.
The Induction Principle
I often like to think of an analogy to the PMI concerning a string
of dominoes standing on end. If you want to know that all the
dominoes will fall, you need to know two things:
1) the dominoes are arranged so that if one falls, it knocks over
the next one for all the dominoes (the dominoes form an inductive
set); and
2) the first dominoe is knocked over.
Regions of a Plane
In a plane, how many regions are formed by n lines, no three of
which meet in a point, and no two of which are parallel?
2
Notice:
4
n
1
2
3
4
7
# regions
1+1
1+1+2
1+1+2+3
1+1+2+3+4
11
Regions of a Plane
If f(n) = # of regions determined by n lines, the table suggests that
n
2
nn1 n n2
f n = 112⋯n = 1∑ i = 1
=
.
2
2
i=1
But how do we prove that this is always true?
Let S = {n ∈ ℕ | f(n) = ½(n2 + n + 2)}.
1. (Base case) Observe that 1 ∈ S, since f(1) = 2 and
½(12 + 1 + 2) = 2.
2. S is an inductive set.
We need to show that if m ∈ S (inductive hypothesis) then
m+1 ∈ S.
Regions of a Plane
We assume that the formula is correct for m lines and examine the
case of m+1 lines.
When we add a new line, it intersects all of the m old lines in a
distinct point. This divides the line into m+1 segments. Each of
these segments breaks an old region into two regions. So, the new
line adds m+1 regions to the previous total. And under the
induction hypothesis, the number of regions for m+1 lines is:
2
2
m m2
m 3 m4
f m1=
m1=
2
2
2
2
m 2 m1m12 m1 m12
=
=
.
2
2
Thus, S is an inductive set and by the PMI, S = ℕ. That is, the
formula is valid for all natural numbers.
The sum of 3 consecutive cubes is
divisible by 9.
Let S = {n ∈ ℕ: 9| n3 + (n+1)3 + (n+2)3}.
Observe that 1 ∈ S since 13 + 23 + 33 = 1 + 8 + 27 = 36 and 9|36.
Now assume that m ∈ S, i.e., 9| m3 + (m+1)3 + (m+2)3.
Consider
3
3
3
3
3
3
2
m1 m2 m3 =m1 m2 m 9 m 27 m27
3
3
3
2
=m m1 m2 9m 3 m3.
Since 9 divides the sum of the first three terms (induction
hypothesis) and 9 divides the last term, 9 divides this final sum.
Thus, m+1 ∈ S, and S is an inductive set. By the PMI, S = ℕ and
the statement is true for all natural numbers.

n
nn12 n1
∑i =
6
i=1
Let S = {n ∈ ℕ:
n
∑ i 2=
i=1
2
n  n1 2 n1
}
6
Observe that 1 ∈ S since 12 = 1 and (1)(2)(3)/6 = 1.
Now suppose that m ∈ S and consider:
2
2
2
2
2
2
2
1 2 ⋯m1 =1 2 ⋯m m1
2
mm12 m1
mm12
m16m1
=
m12=
6
6
2
m1[2 m m6 m6] m1[m22 m3]
=
=
6
6
m1m112m11
=
.
6
Thus, m+1 ∈ S, so S is an inductive set. By the PMI, S = ℕ and the
formula is valid for all natural numbers.

Stamps
Show that any postage > 7¢ can be made using only 3¢ and 5¢
stamps.
5¢
3¢
8
3¢
3¢
12
3¢
3¢
5¢
3¢
9
3¢
3¢
5¢
5¢
3¢
10
5¢
3¢
13
5¢
3¢
11
3¢
3¢
14
5¢
3¢
5¢
5¢
5¢
15
Stamps
Let S = {postages > 7 which can be made from 3 and 5}
We have already seen that 8 ∈ S.
Suppose that m ∈ S.
If a 5¢ is used to make up m¢, then remove the 5¢ stamp and
replace it with 2 3¢ stamps. This will give m+1 ∈ S.
If no 5¢ is used, then m¢ is made up entirely by 3¢ stamps.
Since m > 7, there must be at least 3 3¢ stamps used. Remove 3 3¢
stamps and replace them with 2 5¢ stamps. So in this case also m+1
∈ S.
Thus, S is an inductive set and by the PMI all natural numbers >7
are in S.
Size of the Power Set
Theorem: If a set A has n elements then P (A) has 2n elements.
Pf: Let S = {n ∈ ℕ: the statement above is true}
1∈S
If A has 1 element, i.e., A = {x}, then P (A) = {∅, A} which
has 2 = 21 elements.
Suppose m ∈ S.
Consider a set A with m+1 elements.
Pick one element, say x, in A.
There are just two types of subsets of A, those that contain x
and those that don't.
The subsets that do not contain x are subsets of A' = A – {x}
and since A' has m elements, there are 2m of these.
Size of the Power Set
The subsets that contain x are of the form {x} ∪ {a subset of A'}.
Thus, there are also 2m of these.
In total there are 2m + 2m = 2m+1 subsets of A.
So, m+1 ∈ S, and S is an inductive set. By the PMI, S = ℕ and
the statement is true for all natural numbers.

All horses have the same color
Let S = {n ∈ ℕ: Any collection of n horses have the same color}
Clearly 1 ∈ S.
Now suppose that m ∈ S.
Consider a set of m+1 horses.
Cut out one horse from this herd, and you are left with a set of m
horses.
By the induction hypothesis, they all have the same color.
Now, return the horse that was cut out, and cut out a different
horse. By the induction hypothesis again, the remaining horses all
have the same color, so the horse that was cut out the first time
has the same color as all the other horses.
Thus, m+1 ∈ S, S is an inductive set and by the PMI all horses
have the same color.
??????????
Caveats
As this last example shows, small cases can make a big
difference ... making sure that an argument works in the small
cases makes a lot of sense.
Also, it should be emphasized that mathematical induction is a
proof technique ... it is not a problem solving strategy. You can
not find answers by induction, you can only prove that answers
you suspect are valid are in fact valid.
Principle of Complete Induction (PCI)
Theorem: If S is a set of natural numbers satisfying the condition:
for each natural number m, {i ∈ ℕ| i < m} ⊆ S implies m ∈ S;
then S = ℕ.
Pf: (By induction) Let K = {n ∈ ℕ| {1,...,n} ⊆ S}.
1 ∈ K : the property that S satisfies holds for all natural numbers,
so it holds in particular for m = 1. That is, if {i ∈ ℕ| i < 1} ⊆ S then
1 ∈ S. However, since {i ∈ ℕ| i < 1} = ∅ and ∅ ⊆ S for any set S,
we have 1∈ S, and thus, 1 ∈ K since {1} ⊆ S.
Now suppose that k ∈ K. That means that {1, ..., k} ⊆ S. The
property that S satisfies, with m = k+1, says that k+1 ∈ S.
Therefore, {1, ..., k+1} ⊆ S, so k+1 ∈ K. K is thus an inductive set
and by the PMI K = ℕ. Thus, ℕ ⊆ S, but we already know that S ⊆
ℕ, so we have S = ℕ.

PCI
The PCI is sometimes called strong induction (and the PMI weak
induction) because its hypothesis is a stronger statement ...
but some would say that it is in fact a weaker result, because it
requires so much more to be assumed in order to prove the same
consequence.
We can however ignore this semantical quibble because these
statements are logically equivalent. We have just proved the PCI
assuming the PMI. We will now prove the PMI assuming the PCI
(that is, taking the PCI as an axiom, we will show that the PMI is a
theorem). This then shows that these two forms of induction are
equally "powerful", and either one may be taken as an axiom in the
development of the natural numbers.
PMI – the theorem
Theorem: Assume the validity of the PCI. If S is an inductive set of
natural numbers that contains 1, then S = ℕ.
Pf: We must show that the set S satisfies the PCI condition:
(∀m ∈ℕ)({1,...,m-1}⊆ S ⇒ m ∈ S)
Case 1: m = 1
1 ∈ S by assumption.
Case 2: m > 1
{1, ..., m-1} is non-empty and contains the integer m-1.
{1, ... , m-1}⊆ S ⇒ m-1 ∈ S.
Since S is an inductive set, m-1 ∈ S ⇒ m ∈ S.
Thus, the condition is satisfied and by the PCI, S = ℕ.

Strong Induction
The PCI can be used in many cases where the PMI would be
awkward to use.
Prop: Every natural number greater than 1 has a prime divisor.
Pf: Let S = {n ∈ ℕ| n > 1 and n has a prime divisor}. Note that
while 1 ∉ S, 2 ∈ S.
For any integer m > 1, assume that {2, ... , m} ⊆ S.
Consider m+1.
If m+1 is prime, then it has itself as a prime divisor, so m+1 ∈ S.
If m+1 is not prime, then m+1 = ab for integers a, b with 1 < a <
m+1 (and the same for b, but we don't need that). Thus, a ∈ {2, ...,
m} and so a ∈ S. Since a has a prime divisor, m+1 has a prime
divisor, i.e., m+1 ∈ S.
Thus, by the PCI, S = {n ∈ ℕ| n > 1}.

Inductive Definitions
In some situations an infinite set of objects indexed by the natural
numbers (i.e., there is a 1st object, a 2nd object, etc.), can be defined
in an inductive manner (also called a recursive definition).
Example: Factorials
n! = n(n-1)(n-2) ... (3)(2)(1)
Can be defined as follows:
1! := 1;
n! := n(n-1!)
Thus, 4! = 4(3!) and 3! = 3(2!) and 2! = 2(1!) = 2(1) so
4! = 4(3!) = 4(3(2!)) = 4(3(2(1))) = 4(3)(2)(1)
To prove that a recursive definition is valid for all n, use PMI.
Example
Let x1 = 1, x2 = 1 and for n ≥ 2, xn+1 := xn + 2xn-1. Prove that xn is
divisible by 3 iff n is divisible by 3.
We observe that xn+3 = xn+2+ 2xn+1= (xn+1+ 2xn) + 2xn+1 = 3xn+1+ 2xn.
We first prove the necessity (⇐): Assume that 3 | n, i.e., n = 3k.
Let S = {k : 3 | x3k}.
Since x3 = 3, we have 1 ∈ S.
Suppose m ∈ S. That is, 3 | x3m. By the observation, x3m+3 = 3x3m+1
+ 2x3m. So, 3 | x3m+3 = x3(m+1). Thus, S is inductive and by the PMI, S
= ℕ, i.e., if 3 | n then 3 | xn.
Example (cont.)
Let x1 = 1, x2 = 1 and for n ≥ 2, xn+1 := xn + 2xn-1. Prove that xn is
divisible by 3 iff n is divisible by 3.
We observe that xn+3 = xn+2+ 2xn+1= (xn+1+ 2xn) + 2xn+1 = 3xn+1+ 2xn.
Now we show the sufficiency (⇒):
Let S = {k : 3 ∤ x3k-1 and 3 ∤ x3k-2}.
Since x1 = 1 and x2 = 1, we have 1 ∈ S.
Suppose m ∈ S. That is, 3 ∤ x3m-1 and 3 ∤ x3m-2. By the
observation, x3m+2 = 3x3m + 2x3m-1, and x3m+1 = 3x3m-1 + 2x3m-2. So,
3 ∤ x3m+2 = x3(m+1)-1 and 3 ∤ x3m+1 = x3(m+1)-2 . Thus, S is inductive and
by the PMI, S = ℕ, i.e., if 3 ∤ n then 3 ∤ xn.

Fibonacci Numbers
In 1202, Leonardo Fibonacci (1170-1250) posed the following
famous problem: Suppose a particular breed of rabbit breeds one
new pair of rabbits each month, except that a 1-month old pair is
too young to breed. Suppose further that no rabbit breeds with any
other except its paired mate and that rabbits live forever. Starting
with 1 pair of newborn rabbits, how many pairs of rabbits exist
after n months?
month
# rabbit pairs
1
1
2
1
3
2 (originals + newborns)
4
3 (originals + 1 mo. + new)
5
5 ( ... + 2 new)
6
8 ( ... + 3 new)
Fibonacci Numbers
Let fn be the number of pairs of rabbits alive after n months.
Then,
f1 = f2 = 1 and
fn = fn-1 + fn-2 for n > 2.
The numbers fn are known as the Fibonacci numbers.
Problems involving Fibonacci numbers usually are proved by
induction due to the recursive definitions, and the PCI is often
more useful than the PMI for them.
A Recursion
Let F1= F2 = 1, and Fn+2 = Fn+1 + Fn + Fn+1Fn.
Prove that Fn = 2f(n) - 1 where f(n) is the nth Fibonacci number.
Pf: Let S = {n ∈ ℕ| Fn = 2f(n) – 1}. Note that 1∈ S.
Suppose that {1,...,m} ⊆ S for m > 1. Consider Fm+1.
Fm+1 = Fm + Fm-1 + FmFm-1
= (2f(m)-1) + (2f(m-1)-1) + (2f(m)-1)(2f(m-1)-1)
= 2f(m) + 2f(m-1) -2 + 2f(m)+f(m-1) – 2f(m) -2f(m-1) + 1
= 2f(m)+f(m-1) -1
= 2f(m+1) -1.
Thus m+1 ∈ S. So, by the PCI S = ℕ.

Well-Ordering Principle
This simply stated property of the natural numbers turns out to be
quite powerful and equivalent to the PMI.
Theorem (WOP): Every non-empty subset of ℕ has a smallest
element.
Pf: Let T be a subset of ℕ.
Suppose that T has no smallest element. Let S = ℕ – T.
Assume that {1,...,m} ⊆ S. If m+1 ∉ S, then m+1 ∈ T and would
be the smallest element in T since all smaller integers are not in T.
→← Thus, m+1 ∈ S and by the PCI, S = ℕ. But then, T = ∅.
Thus, if T ≠ ∅ then T has a smallest element.

Using the WOP
Prop: Every natural number greater than 1 has a prime divisor.
Pf: If n is prime, then it is a prime divisor of itself. If n is
composite, then n has a factor other than 1 or n. Let s be the
smallest of these factors (s exists by WOP).
Suppose s is composite. Then s has a divisor d, with 1 < d < s.
Since d divides s, d divides n and this contradicts the definition of
s. Thus, s is prime.

Division Algorithm for ℕ
Theorem: Let a and b be natural numbers with b ≤ a. Then there
exist q ∈ ℕ and r ∈ ℕ ∪ {0} such that:
a = qb + r with 0 ≤ r < b.
Pf: Let T = {s ∈ ℕ | a < sb}. T is non-empty (for instance, a+1 is in
T). By the WOP, T contains a smallest element, w.
Let q = w-1 and r = a – qb.
Since a ≥ b, w > 1, so q ∈ ℕ.
Since q < w, a ≥ qb, so r ≥ 0.
By the definition of r, a = qb + r.
BWOC, suppose r ≥ b.
Then r = a – (w-1)b ≥ b, so a ≥ wb →←
Therefore, r < b.

Odds and Evens
Theorem: Every natural number is either odd or even.
Pf: 1 is odd ( = 2(0) + 1).
Let n be a natural number > 1.
By the division algorithm for natural numbers, there exist
integers q and r so that
n = 2q + r with 0 ≤ r < 2.
Thus, r = 0 or r = 1.
When r = 0, n = 2q is even.
When r = 1, n = 2q +1 is odd.
