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Chapter 3
Induction
In this chapter we learn a way to prove a statement of the form (∀n ∈ N)P (n). For
example, suppose we were adding even natural numbers and we detected a pattern.
That is,
2
2+4
=
=
2+4+6
=
..
.
1·2
2·3
3·4
So we conjecture 2 + 4 + · · · + 2n = n · (n + 1) for all n. How do we prove this when
we cannot check the statement for every n ∈ N?
Here is a way to do it. Suppose we prove the statement P (1). Then we prove
the connective statement P (n) → P (n + 1) is true for each n ∈ N. That is, we
assume P (n) is true and we show P (n + 1) follows. Then, since P (1) is true and
P (1) → P (2), P (2) is true, and in the same way P (3) is true and so on. A bit
like dominos falling. This method of proof is called the Principle of Mathematical
Induction. More formally,
Theorem 3.1. (Principle of Mathematical Induction) Suppose P (n) is a statement
for each n ∈ N. Moreover, suppose
(i) Base Step: P (1) is true.
(ii) Inductive Step: for each n ∈ N, P (n) implies P (n + 1).
Then P (n) is true for all n ∈ N.
We will prove the theorem below. The proof will require a further assumption
about the natural numbers.
29
30
3. Induction
Example 3.2. Prove 2 + 4 + · · · + 2n = n · (n + 1) for all n ∈ N.
Proof. Let P (n) be the statement 2 + 4 + · · · + 2n = n · (n + 1).
(i) Base Step: the statement P (1) is 2 = 1 · 2 which is obviously true.
(ii) Inductive Step: suppose 2 + 4 + · · · + 2n = n · n(n + 1). Adding 2(n + 1) to
both sides gives
2 + 4 + · · · + 2n + 2(n + 1) =
=
n(n + 1) + 2(n + 1)
(n + 1)(n + 2).
This is P (n+1). Thus, by the Principle of Mathematical Induction, 2+4+· · ·+2n =
n · (n + 1) for all n ∈ N
A strategy for proving the inductive step is to write down, in the scratch area,
the statement P (n + 1). When P (n) involves an equality or an inequality. Pick a
side (left or right) and manipulate it to look like the same side of P (n + 1). Finally
manipulate the other side to make it look like the corresponding side of P (n + 1).
We illustrate this in the next example.
Example 3.3. Prove 2n−1 ≤ n!, where n! = n(n − 1)(n − 2) · · · 1.
Scratch Work. We can easily check P (1). The difficult part is showing P (n) →
P (n + 1). The statement P (n + 1) is the statement 2n ≤ (n + 1)!, and, again, we
need to derive this from 2n−1 ≤ n!. Following the above suggestion we multiply
P (n) by 2 to make the left side of P (n) look like the left side of P (n + 1). Then we
deal with the right side. We find
2 2n−1 ≤ 2n!.
That is, 2n ≤ 2n!. Since 2 ≤ n + 1 for n ≥ 1, we get 2n ≤ (n + 1)n! or 2n ≤ (n + 1)!.
This is P (n+ 1). In the proof below we work with the right side-just to be different.
Proof. Let P (n) be the statement 2n−1 ≤ n!.
(i) Base Step: P (1) is 20 ≤ 1 or 1 ≤ 1 which is true.
(ii) Inductive Step: Suppose P (n) is true for all n ∈ N. That is, suppose 2n−1 ≤
n!. Multiplying both sides by n + 1, we find (n + 1)2n−1 ≤ (n + 1)n! = (n + 1)!.
Moreover, we have 2 ≤ n + 1 for all n ∈ N. Hence,
2 2n−1 ≤ (n + 1)2n−1 ≤ (n + 1)!.
That is, 2n ≤ (n + 1)!. This is P (n + 1).
The result follows from the Principle of Mathematical Induction.
Example 3.4. Suppose x, y ∈ R and 0 < x ≤ y. Then xn ≤ y n for all n ∈ N.
Proof. Let P (n) be the statement xn ≤ y n .
(i) Base Step: P (1) is x ≤ y which by assumption is true.
(ii) Inductive Step: Suppose 0 < xn ≤ y n . Multiplying both sides by x, we find
xn+1 ≤ xy n . Since x ≤ y, we have xn+1 ≤ xy n ≤ y y n = y n+1 . That is,
0 < xn+1 ≤ y n+1 . This is P (n + 1).
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3. Induction
The result follows from the Principle of Mathematical Induction.
Example 3.5. Suppose x, y ∈ R and x, y ≥ 0. Then
n ∈ N.
Proof. Let P (n) be the statement
(i) Base Step: P (1) is
x+y
2
≤
≤
n+1
xn +y n
2
for all
xn +y n
.
2
x+y
2 , which is true.
n
n
x+y n
. Multiplying
≤ x +y
2
2
≤
(ii) Inductive Step: Suppose
x+y
2
x+y n
2
x+y n
2
this by
x+y
2 ,
we find
xn + y n x + y
2
2
n+1
n
x
+ x y + xy n + y n+1
4
≤
=
We are hoping
(3.1)
xn+1 + y n+1
xn+1 + xn y + xy n + y n+1
≤
.
4
2
This would require
Factoring we see
0 ≤ xn+1 − xn y − y n x + y n+1 .
xn+1 − xn y − xy n x + y n+1 = (x − y)(xn − y n ).
Thus we see inequality (3.1) is satisfied if
0 ≤ (x − y)(xn − y n ).
Example 3.4 shows xn ≤ y n for all n if x ≤ y. This implies both terms on the
right side of the last inequality are either both positive or both negative and
their product is alway non negative. It follows (3.1) is true for all x, y ≥ 0,
and so
n+1
xn+1 + y n+1
x+y
≤
.
2
2
This is P (n + 1).
The result follows from the Principle of Mathematical Induction.
Example 3.6. For all n ∈ N n3 − n is divisible by 3. We sometimes express this
as 3|(n3 − n).
Proof. Let P (n) be the statement 3|(n3 − n).
(i) Base Step: P (1) is 3|0. That is, 0 is divisible by 3, which is true.
(ii) Inductive Step: Suppose 3|(n3 − n). That is,
3
n3 −n
3
∈ N. We need to show
(n + 1) − (n + 1)
∈ N.
3
Notice
(n + 1)3 − (n + 1) =
=
(n3 + 3n2 + 3n + 1) − (n + 1)
(n3 − n) + 3n(n + 1).
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3. Induction
The first term is divisible by 3 by assumption; the second is also divisible by
3. Thus 3|((n + 1)3 − (n + 1)). This is P (n + 1).
The result follows from the Principle of Mathematical Induction.
Theorem 3.7. Every positive integer is either even or odd.
Proof. Let P (n) be the statement every natural number is either even or odd.
(i) Base Step: The smallest positive integer is 1, and 1 = 0 · 2 + 1 and so 1 is odd.
(ii) Inductive step: Suppose n is either even or odd. If n is even, them n = 2k for
some k ∈ N. Then n + 1 = 2k + 1 and n + 1 is odd. If n is odd, n = 2k + 1
for some k ∈ N. Then n + 1 = 2k + 1 + 1 or n + 1 = 2(k + 1). Thus n + 1 is
even. This implies n + 1 is either even or odd if n is.
The result follows from the Principle of Mathematical Induction.
3.1. Strong Induction
Sometimes the Principle of Mathematical Induction is not adequate to show a
statement P (n) is true for all n ∈ N. There is another form of induction called
Strong Induction. The basic difference is that we assume all P (1), P (2), . . . P (n) to
prove P (n + 1) instead of only P (n).
Theorem 3.8. (Strong Induction) Suppose P (n) is a statement for each n ∈ N.
Moreover, suppose
(i) Base Step: P (1) is true.
(ii) Inductive Step: for each n ∈ N, P (k), 1 ≤ k ≤ n implies P (n + 1).
Then P (n) is true for all n ∈ N.
The next example uses both forms of induction.
Example 3.9. Every amount of postage of 12� or more can be formed using just
4� and 5� stamps.
Proof. We use the Principle of Mathematical Induction. Let P (n) be the statement
n is a combination of 4� and 5� stamps.
(i) Base Step: Note that 12 = 3 · 4 and so P (12) is true.
(ii) Inductive Step: Suppose P (n) is true. Then n = a · 4 + b · 5 for some a, b ∈
N ∪ {0}. If a = 0, then n + 1 = (a − 1) · 4 + (b + 1)5. If a = 0, then n = k · 5
for k ≥ 3. Then n + 1 = (k − 3) · 5 + 4 · 4. Thus P (n + 1) follows.
The result follows from the Principle of Mathematical Induction.
Alternatively we may use strong induction. To do this we check P (12), . . . P (15)
first. We have 12 = 3 · 4, 13 = 2 · 4 + 1 · 5, 14 = 1 · 4 + 2 · 5, and 15 = 3 · 5. This is
the base step.
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3.1. Strong Induction
To do the inductive step suppose n ≥ 15 and suppose k = a · 4 + b · 5 for some
a, b ∈ N ∪ {0} and for all 12 ≤ k ≤ n. Then
n+1
= 1 · 4 + [(n + 1) − 4]
= 1 · 4 + (n − 3).
Since we assume P (k) for all 12 ≤ k ≤ n, we have n − 3 = a · 4 + b · 4 for some
a, b ∈ N ∪ {0}. Thus
n+1
=
=
1·4+a·4+b·4
(1 + a) · 4 + b · 5,
and hence P (12) . . . P (n) implies P (n+1). The result follows from strong induction.
We have used the Fundamental Theorem of Arithetic on several occasions. It
is now easily justified.
Definition 3.10. A natural number n > 1 is prime if it cannot be written as the
product of two smaller numbers.
Theorem 3.11. (Fundamental Theorem of Aritmetic) Every natural number n ≥ 3
is either prime or is the product of primes.
Proof. We use strong induction. Let P (n) be the statement in the theorem.
(i) Base Step: Since 2 is prime, P (2) is true.
(ii) Inductive Step: Suppose 1, ≤ k ≤ n is prime or the product of primes. If n + 1
is prime, P (n + 1) follows. If n + 1 is not prime, then a, b ∈ N exists so that
n + 1 = ab. Moreover, a < n + 1 = ab and b < n + 1 = ab. It follows a > 1 and
b > 1. Thus, by assumption, both a and b are either prime or the product of
primes. This means ab = n + 1 is the product of primes, and P (n + 1) follows.
Another application of the Fundamental Theorem of Arithmetic is that there
is no shortage of prime numbers.
Theorem 3.12. There are infinitely many prime numbers.
Proof. Suppose only a funite number of primes exist, say, {p1 , p2 , . . . , pn }. Set
m = p1 p2 . . . , pn + 1. If m is prime m > pi for all 1 ≤ i ≤ n - a contradiction. If m
is not prime, m cannot be factored into primes since m/pi ∈ N for all 1 ≤ i ≤ n.
This contradicts the fundamental theorem of arithmetic.
Remark 3.13. Note that, for there to be an infinite number of primes, we need
the fundamental theorem of arithmetic. This requires induction which requires the
well-order principle - which is an assumption about the natural numbers.
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3. Induction
3.2. Justification of Induction
We need to justify the Principle of Mathematical Induction by proving Theorem
3.1. To do this we need to make a new assumption about the natural numbers. It
turns out that the Principle of Mathematical Induction is equivalent to this new
assumption.
Theorem 3.14. Let P (n) be a statement for each n ∈ N. The following are
equivalent.
(a) The Well-Ordering Property: Every nonempty subset of the natural numbers
has a least element.
(b) The Principle of Mathematical Induction:
(i) Base Step: P (1) is true.
(ii) Inductive Step: for each n ∈ N, P (n) implies P (n + 1).
Then P (n) is true for all n ∈ N.
(c) Strong Induction:
(i) Base Step: P (1) is true.
(ii) Inductive Step: for each n ∈ N, P (k), 1 ≤ k ≤ n implies P (n + 1).
Then P (n) is true for all n ∈ N.
Proof. Suppose N enjoys the well-ordering property and suppose P (1) and P (n) →
P (n + 1) for all n. We prove (b) by contradiction. Suppose P (n) is not true for
all n ∈ N. Set E = {n | P (n) is false}. Then E = ∅ and is a subset of the natural
numbers. By the well-ordering property there is a least such n which we label n0 .
Since P (1) is true, n0 > 1. By construction, P (n0 ) false while is P (n0 − 1) is true.
Since P (n − 1) → P (n) for all n, we have P (n0 ) is true. This is a contradiction and
the set E must be empty. This proves (b).
Next we show (b) implies (c). Suppose P (1) is true and for each n ∈ N, P (k),
1 ≤ k ≤ n implies P (n + 1). Set Q(n) to be the statement P (k) is true for all
k ≤ n. We prove Q(n) is true for all n ∈ N by using Part (a).
(i) Base Step: Q(1) is the statement P (1) is true, and so Q(1) is true.
(ii) Inductive Step: Suppose Q(n). Then P (k) is true for all k ≤ n. By assumption
P (n + 1) follows. Thus P (k) is true for all k ≤ n + 1. That is, Q(n + 1) is
true.
By the Principle of Mathematical Induction, (b), Q(n) is true for all n ∈ N. That
is, P (n) is true for all n ∈ N and (c) follows.
Finally, suppose (c). We need to show the well-ordering property follows. Let
P (n) be the statement “every subset of N containing n has a least element.” We
use (c), strong induction, to show P (n) is true for all n ∈ N.
(i) Base Step: P (1) means 1 is in the subset. Since 1 is the smallest natural
number, any set containing 1 has a least element. Thus P (1) is true.
(ii) Inductive Step: Suppose P (k) for 1 ≤ k ≤ n. We need to show P (n + 1)
follows. The statement P (n + 1) means every set containing n + 1 has a
minimum element. If n + 1 is the minimum, then P (n + 1) is true. If n + 1is
not the minimum, then there exists i < n + 1 in the set. Such sets have a
3.2. Justification of Induction
35
minimum (since P (1), . . . , P (n) are assumed true), and therefore P (n + 1)
follows. The well-order property follows.
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3. Induction
3.3. Homework
Exercise 3.1. Give a sequence of statements, P (n), for n a natural number such
that P (1), P (2), . . . , P (99) are all true but P (100) is false. Make it simple.
Exercise 3.2. Suppose a1 = 1, a2 = 8, and an = an−1 + 2an−2 for n ≥ 3. Prove
that an = 3 · 2n−1 + 2(−1)n for all n ∈ N.
Exercise 3.3. Suppose x1 = 1 and xn =
1 ≤ xn ≤ 3/2, and xn ≤ xn+1 for all n ∈ N.
1
4 (2xn−1
+ 3) for all n ≥ 2. Prove
Exercise 3.4. Suppose x1 = 3 and xn = 2 − 1/xn−1 for all n ≥ 2. Prove 1 ≤ xn ≤
3, and xn ≥ xn+1 for all n ∈ N.
Exercise 3.5. We say a statement P (n) is eventually true if there exists a N ∈ N
such that P (n) is true for all n ≥ N . For each positive real number x determine if
xn + x < xn+1 is true or false for all n ∈ N or is eventually true.
Exercise 3.6. Prove that
1+
1
1
1 1
+ + ...+ 2 < 2 −
4 9
n
n
for all n ≥ 2.
Exercise 3.7. Let f : R → R satisfy f (x + y) = f (x) + f (y) for all x, y ∈ R. Prove
that
a) Prove f (0) = 0.
b) Prove f (n) = nf (1) for all n ∈ N.
Exercise 3.8. Recall a set S of real numbers is convex if, for all x, y ∈ S and all
0 ≤ t ≤ 1, tx + (1 − t)y ∈ S. For every natural number n ≥ 2 suppose S1 , S2 , . . . , Sn
are convex, prove S1 ∪ S2 ∪ · · · ∪ Sn is a convex set.
Exercise 3.9. Prove that every finite subset of the real numbers has a maximum
element.
Exercise 3.10. Prove, for all n ≥ 1
1
1
1
1
1
1
1 1 1
−
=
+
−
+ ···+
.
1 − + − + ··· +
2 3 4
2n − 1 2n
n+1 n+2 n+3
2n
Exercise 3.11. Prove
n
n(n + 1)
i=
2
i=1
for all n ∈ N by using induction and without using induction.
Exercise 3.12. Prove 6|(n3 − n) for all n ∈ N.