MATH 289
PROBLEM SET 1: INDUCTION
1. The induction Principle
The following property of the natural numbers is intuitively clear:
Axiom 1. Every nonempty subset of the set of nonnegative integers Z≥0 = {0, 1, 2, 3, . . . }
has a smallest element.
This axiom can be thought of a instance of the extremal principle which will be discussed
later.
Suppose that k is an integer and we want to prove that
“P (n) is true for every positive integer n”,
where P (n) is a proposition (statement) which depends on a positive integer n. Proving
P (1), P (2), P (3) individually, would take an infinite amount of time. Instead, we can use
the Induction Principle:
Theorem 2 (Induction Principle). Assume that k is an integer and P (n) is a proposition
for all n ≥ k.
(1) Suppose that P (k) is true, and
(2) for any integer m ≥ k for which P (m) is true, P (m + 1) is true.
Then P (n) is true for all integers n ≥ k.
For k = 1, the induction principle can be compared to an infinite sequence of dominos
tiles, numbered 1,2,3, etc.
1
2
3
4
5
6
7
If the m-th domino tile falls, it will hit the (m + 1)-th domino tile and the (m + 1)-th
domino tile will fall as well. If the first domino tile falls, then all domino tiles will fall down.
(Here P (n) is the statement:“the n-th domino tile falls down”)
1
2
3
4
5
6
7
Proof of the Induction Principle. Let S to be the set of all nonnegative integers i for which
P (k + i) is false. If S is nonempty, then S has a smallest element, say l by Axiom 1. By
property (1)), P (k) is true and l > 0. The statement P (k + l − 1) is true because l is the
minimal element of S. Therefore, statement P (k + l) is true because of property (2). Hence
l 6∈ S. Contradiction! We conclude that S is empty, and P (n) is true for all n ≥ k.
A typical example of the induction principle is the following:
1
Example 3. Prove that
1 + 2 + 3 + ··· + n =
(1)
n(n + 1)
.
2
for every positive integer n.
Proof. We prove (1) by induction on n. For n = 1 we check that
1=
1 · (1 + 1)
.
2
Suppose that (1) is true for n = m. Then
1 + 2 + · · · + m + (m + 1) = (1 + 2 + · · · + m) + (m + 1) =
m(m + 1)
(m + 1)(m + 2)
+ (m + 1) =
.
2
2
So (1) is true for n = m + 1. Now (1) is true for all positive integers n by the induction
principle.
=
Remark 4. When the German mathematician Carl Friedrich Gauss (1777–1855) was 10
years old, his school teacher gave the class an assignment to add all the numbers from 1 to
100. Gauss gave the answer almost immediately: 5050. This is how (we think) he did it:
Write the numbers from 1 to 100 from left to right. Write under that the numbers from 1
to 100 in reverse order.
1
2
3 · · · 100
100 99 98 · · · 1
101 101 101 · · · 101
|
{z
}
100
Each of the 100 column sums is 101. This shows that
2 · (1 + 2 + · · · + 100) = 100 · 101
and
1 + 2 + · · · + 100 =
100 · 101
= 50 · 101 = 5050.
2
This easily generalizes to a proof of (1).
A formula similar to (1) exists for the sums of squares, namely
(2)
12 + 22 + · · · + n2 =
n(n + 1)(2n + 1)
.
6
Example 5. Give and prove a formula for
13 + 23 + · · · + n3
We have seen similar examples, namely (1) and (2). We can also add the formula
10 + 20 + 30 + · · · + n0 = n.
Let
pk (n) = 1k + 2k + 3k + · · · + nk
where k ∈ Z≥0 . The examples so far suggest that pk (n) is a polynomial of degree k + 1 (and
1
that the leading coefficient is k+1
). Let us assume that p3 (n) is a polynomial of degree 4.
2
Since p3 (0) is an empty sum, we have that p3 (0) = 0. We can write p3 (n) = an4 + bn3 +
cn2 + dn. for certain real numbers a, b, c, d. We have
(3)
n3 = p3 (n)−p3 (n−1) = a(n4 −(n−1)4 )+b(n3 −(n−1)3 )+c(n2 −(n−1)2 )+d(n−(n−1)) =
= a(4n3 − 6n2 + 4n − 1) + b(3n2 − 3n + 1) + c(2n − 1) + d =
= n3 (4a) + n2 (−6a + 3b) + n(4a − 3b + 2c) + (−a + b − c + d)
Comparing coefficients in (3) gives us the linear equations:
(4)
(5)
(6)
(7)
1
0
0
0
=
=
=
=
4a
−6a + 3b
4a − 3b + 2c
−a + b − c + d
We solve the system of equations and find a = 41 , b = 12 , c =
conjecture the following formula:
1
4
and d = 0. We now should
13 + 23 + · · · + n3 = 41 n4 + 21 n3 + 14 n2 .
Finding this formula was the hard part. It is now not so hard to prove this formula by
induction:
Proof. We will prove that
13 + 23 + · · · + n3 = 41 n4 + 21 n3 + 14 n2 .
(8)
by induction on n. The case n = 0 is clear, because both sides of the equation are equal to
0. If (8) is true for n = m − 1, then
13 + 23 + · · · + (m − 1)3 = 41 (m − 1)4 + 21 (m − 1)3 + 41 (m − 1)2 .
From this follows that
13 + 23 + · · · + (m − 1)3 + m3 = 14 (m − 1)4 + 21 (m − 1)3 + 41 (m − 1)2 + m3 =
= 41 (m4 − 4m3 + 6m2 − 4m + 1) + 21 (m3 − 3m2 + 3m − 1) + 14 (m2 − 2m + 1) + m3 =
= 14 m4 + 12 m3 + 14 m2 ,
so (8) is true for n = m. By induction follows that (8) is true for all n ∈ Z≥0 .
Notice that
1 4
m
4
+ 12 m3 + 14 m2 = ( 12 n(n + 1))2
which leads to the following aesthetic formula:
13 + 23 + · · · + n3 = (1 + 2 + · · · + n)2 .
Example 6. What is the value of
1
1
1
+
+
+ · · ·?
1·2 2·3 3·4
3
Let us compute the partial sums. Perhaps we will find a pattern.
1
2 4
2
1
+
= + = ,
1·2 2·3
6 6
3
1
1
1
2
1
8
1
9
3
+
+
= +
=
+
=
= ,
1·2 2·3 3·4
3 12
12 12
12
4
1
1
1
1
3
1
15
1
16
4
+
+
+
= +
=
+
=
= .
1·2 2·3 3·4 4·5
4 20
20 20
20
5
A pattern emerges. Namely, it seems that
1
1
1
1
+
+ ··· +
=1−
.
1·2 2·3
n(n + 1)
n+1
(9)
Proof. By induction on n we prove:
1
1
1
1
+
+ ··· +
=1−
.
1·2 2·3
n(n + 1)
n+1
(10)
For n = 1 we check
1
1
=1− .
1·2
2
If (10) is true for n = m, then
1
1
1
1
+
+ ··· +
+
=
1·2 2·3
m(m + 1) (m + 1)(m + 2)
1 1
1
1
+(
−
)=1−
.
m+1
m+1 m+2
m+2
Hence (10) is true for n = m + 1. By induction, (10) is true for all integers n ≥ 1. We have
= 1−
1
1
1
1 +
+
+ · · · = lim 1 −
= 1.
n→∞
1·2 2·3 3·4
n+1
Example 7 (UMUMC, 1988). Let Sn be the set of all pairs (x, y) with integral coordinates
such that x ≥ 0, y ≥ 0 and x + y ≤ n. Show that Sn cannot be covered by the union of n
straight lines.
First we should try a few small cases, say n = 0, 1, 2, 3, 4:
n=0
n=1
n=2
n=3
n=4
Notice that Sn is a subset of Sn+1 . This will be helpful for our induction proof:
4
Proof. We prove the statement by induction on n, the case n = 0 being trivial. Suppose that
one needs at least n + 1 lines to cover Sn . Define Cn+1 = Sn+1 \ Sn . The set Cn+1 consists
of n + 2 points on the line x + y = n + 1. Suppose that k lines `1 , `2 , . . . , `k cover Sn+1 .
case 1: One of the lines is equal to the line x + y = n + 1. Without loss of generality we
may assume that `k is equal to the line x + y = n + 1. Then `1 , `2 , . . . , `k−1 cover Sn because
`k ∩ Sn = ∅. From the induction hypothesis follows that k − 1 ≥ n + 1, so k ≥ n + 2.
case 2: None of the lines are equal to the line x + y = n +1. Then each of the lines intersects
the line x + y = n + 1 in at most one point, and therefore it intersects the set Cn+1 in at
most one point. Since Cn+1 has n + 2 elements, there must be at least n + 2 lines.
So in both cases we conclude that one needs at least n + 2 lines to cover Sn+1 .
2. Strong Induction
The following example illustrates that sometimes one has to make a statement stronger
in order to be able to prove it by induction.
Example 8. Prove that
999, 999
1
1 3 5
· · ···
<
.
2 4 6
1, 000, 000
1000
Since 1000 =
(11)
√
1, 000, 000 one might suggest that
1 3 5
2n − 1
1
· · ···
<√
2 4 6
2n
2n
for all n ≥ 1. Let us try to prove (11). We can check (11) for small n (which gives some
validity to our conjecture that this inequality holds). Suppose that (11) holds for n = m:
(12)
1 3 5
2m − 1
1
· · ···
<√
2 4 6
2m
2m
We have to prove (11) for n = m + 1:
(13)
2m + 1
1
1 3 5
· · ···
<√
.
2 4 6
2m + 2
2m + 2
If we divide (13) by (12) we obtain
(14)
2m + 1
≤
2m + 2
r
2m
.
2m + 2
If (12) and (14) are true, then (13) is true. By squaring (14) we see that (14) is equivalent
to
2
2m
2m + 1
≤
2m + 2
2m + 2
and to
(15)
(2m + 1)2 ≤ (2m + 2)(2m)
So if (15) is true then our induction proof is complete. Unfortunately (15) is not true and
we are stuck.
Sometimes it is easier to prove a stronger statement by induction:
5
Proof. We prove
2n − 1
1
1 3
· ···
<√
2 4
2n
2n + 1
by induction on n. The case n = 1 is clear because
1
1
<√ .
2
3
Suppose that (16) is true for n = m:
2m − 1
1
1 3
· ···
<√
(17)
2 4
2m
2m + 1
Since
(2m + 1)(2m + 3) = (2m + 2)2 − 1 < (2m + 2)2
we have that
2
2m + 1
2m + 1
<
2m + 2
2m + 3
and
r
2m + 1
2m + 1
(18)
<
.
2m + 2
2m + 3
Multiplying (17) by (18) yields
13
2m + 1
1
(19)
···
<√
,
24
2m + 2
2m + 3
so (16) is true for n = m + 1. This shows that (16) is true for all positive integers n. In
particular, for n = 500, 000 we get
999, 999
1
1
1 3
· ···
<√
<
.
2 4
1, 000, 000
1000
1, 000, 001
(16)
Below is a trickier proof of Example 8.
Proof. Let
A=
1 · 3 · 5 · · · 999, 999
2 · 4 · 6 · · · 1, 000, 000
B=
2 · 4 · 6 · · · 1, 000, 000
.
3 · 5 · 7 · · · 1, 000, 001
and
Clearly A < B because
1
2 3
4
999, 999
1, 000, 000
< , < ,...,
<
.
2
3 4
5
1, 000, 000
1, 000, 001
It follows that
A2 < AB =
1
1
<
1, 000, 001
1, 000, 000
and A < 1000−1 .
Example 9. Prove that every integer n ≥ 2 is a product of prime numbers.
6
Proof. Let Q(n) be the statement:
“every integer r with 2 ≤ r ≤ n is a product of prime numbers.”
We use induction on n to prove that Q(n) holds for all integers n ≥ 2.
For n = 2 the statement is true because 2 is a prime number. Suppose that Q(m) is true.
We will prove Q(m + 1). Suppose that 2 ≤ r ≤ m + 1. If r ≤ m then r is a product of prime
numbers because Q(m) is true. Suppose that r = m + 1. If m + 1 is a prime number, then
m + 1 is a product of prime numbers and we are done. Otherwise, m + 1 can be written as
a product ab with 1 ≤ a, b ≤ m. Because Q(m) is true, both a and b are products of prime
numbers. Hence m + 1 = ab is a product of prime numbers.
We have shown that Q(n) holds for all n ≥ 2. In particular, every integer r ≥ 2 is a
product of prime numbers because Q(r) is true.
3. Induction in definitions
We can also use induction in a definition. For example, the Fibonacci numbers is a
sequence of numbers F0 , F1 , F2 , . . . defined by F0 = 0, F1 = 1 and
n ≥ 1.
Fn+1 = Fn + Fn−1 ,
By (strong) induction on n we can prove that Fn is well-defined for all integers n ≥ 0. The
first few Fibonacci numbers are:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, . . .
The sum notation is an example of a recursive definition. Suppose that f (n) is some
function. If a, b are integers and a ≤ b + 1 then we define
b
X
f (n)
n=a
as follows.
a−1
X
f (n) = 0
n=a
and
(20)
b
X
f (n) = f (b) +
n=a
b−1
X
f (n)
n=a
if b ≥ a.
One can then formally prove by induction that
c
X
n=a
f (n) =
b
X
c
X
f (n) +
n=a
f (n).
n=b+1
if a, b, c ∈ Z and a − 1 ≤ b ≤ c. (Induction on c. Start with c = b.)
Similarly we have the product notation.
a−1
Y
f (n) = 1
n=a
7
and
b
Y
f (n) = f (b)
n=a
b−1
Y
f (n).
n=a
if b ≥ a.
For nonnegative integers m and n with m ≤ n we define a binomial coefficient by
n
1
if m = 0 or m = n .
=
n−1
n−1
+
if 0 < m < n
m
m−1
m
If we arrange the binomial coefficients in a triangular shape, we get Pascal’s triangle:
0
0
1
1
0
1
2
2
2
0
1
2
3
3
3
3
0
1
2
3
4
4
4
4
4
0
1
..
.
2
..
.
3
..
.
4
..
.
..
.
After evaluating the binomial coefficients we get:
1
1
1
1
1
..
.
1
2
1
3
4
..
.
3
6
..
.
1
4
..
.
1
..
.
4. Exercises
Exercise 1. * Prove that
12 + 22 + · · · + n2 =
n(n + 1)(2n + 1)
6
for all positive integers n.
Exercise 2. * Show that
1 1
1
1
1
1
1
1 − + − ··· +
−
=
+
+ ··· +
2 3
2n − 1 2n
n+1 n+2
2n
for all n ∈ N.
Exercise 3. * Prove that
n X
m+i
i=0
m
=
m+n+1
.
m+1
for all nonnegative integers m and n.
Exercise 4. * Prove that
n n
n n−1
n n−2 2
n n
(a + b) =
a +
a b+
a b + ··· +
b .
0
1
2
n
n
Exercise 5. *
8
(a) Prove that
xn+1 − 1
.
x−1
for every real number x and every positive integer n.
(b) If x is a real number with |x| < 1 then
1
1 + x + x2 + · · · =
.
1−x
1 + x + x2 + · · · + xn =
Exercise 6. ** Show that the sum of the squares of two consecutive Fibonacci numbers is
again a Fibonacci number.
Exercise 7. ** Cut out a 1 × 1 corner of a 2n × 2n chess board (n ≥ 1). Show that the
remainder of the chess board can be covered with L-shaped tiles (see picture).
2
1
2
1
1
1
The case n = 2 is shown below.
Exercise 8. ** Find and prove a formula for
14 + 24 + · · · + n4 .
Exercise 9. ** Show that
n
n!
=
m
m! (n − m)!
as follows: Define
n
n
n n
+
x + ··· +
x .
f (x) = (1 + x) =
0
1
n
n
and consider f (k) (0) (f (k) (x) is the k-th derivative of f (x)).
Exercise 10. ** Define a sequence a1 , a2 , . . . by a1 =
an explicit formula for an and prove it.
5
2
and an+1 = a2n − 2 for n ≥ 1. Give
Exercise 11 (Division with remainder). ** Suppose that n is a nonnegative integer, and m
is a positive integer. Prove that there exist integers q and r with n = qm + r and 0 ≤ r < m.
Exercise 12. ** Give and prove a formula for
1
1
1
+
+ ··· +
.
1·2·3 2·3·4
n · (n + 1) · (n + 2)
9
Exercise 13 (Expansion in base b). *** Suppose that n is a positive integer, and b is an
integer ≥ 2. Show that there exist an nonnegative integer m, and integers a0 , a1 , . . . , am ∈
{0, 1, 2, . . . , b − 1} such that
(21)
n = am bm + am−1 bm−1 + · · · + a0
and am 6= 0. Moreover, show that m and a0 , a1 , . . . , am are uniquely determined by n. (We
will write (am am−1 · · · a0 )b for the right-hand side in (21)).
Exercise 14 (Zeckendorf’s Theorem). *** Suppose that n is a positive integer. Show that
we can write
n = F i1 + F i2 + · · · + F ik
where k is a positive integer, i1 ≥ 2 and ij ≥ ij−1 + 2 for j = 2, 3, . . . , k. Also show that k
and i1 , . . . , ik are uniquely determined by n.
Exercise 15 (Putnam 1985, B2). *** Define polynomials fn (x) for n ≥ 0 by f0 (x) = 1,
fn (0) = 0 for n ≥ 1, and
d
(fn+1 (x)) = (n + 1)fn (x + 1)
dx
for n ≥ 0. Find, with proof, the explicit factorization of f100 (1) into powers of distinct
primes.
Exercise 16 (Putnam 1987, B2). *** Let r, s and t be integers with 0 ≤ r, 0 ≤ s and
r + s ≤ t. Prove that
s
s
s
t+1
0
1 s + ··· + t =
t +
t
t−s .
(t
+
1
−
s)
r
r+1
r+s
r
Exercise 17. **** Let n = 2k . Prove that we can select n integers from any (2n − 1)
integers such that their sum is divisible by n.
Exercise 18. *** Find the sum of all fractions 1/xy such that gcd(x, y) = 1, x ≤ n, y ≤ n,
x + y > n.
Exercise 19. **** Find and prove a formula for
Z π
2
sinn (x) dx.
0
Exercise 20. **** Suppose that a1 , a2 , . . . , an are positive integers such that a1 ≤ a2 ≤
· · · · · · ≤ an . Prove that
1
1
1
+
+ ··· +
=1
a1 a2
an
implies that an < 2n! .
Exercise 21. *** Generalize example 7: Let Sn,d ⊆ Rd be the set of all integer vectors
(x1 , . . . , xd ) such that xi ≥ 0 for all i and x1 + x2 + · · · + xd ≤ n. Show that Sn,d cannot be
covered with n hyperplanes in Rd .
10