1. The 3rd Report of Abstract Agebra IMPoME 2012
2.4
The Principle of Mathematical Induction
Assignment from:
Sri Rejeki
Ahmad Wachidul Kohar
2.4.1 The standard PMI
Suppose we consider a set S, which is assumed to be a subset of the natural numbers N, and
suppose S is known to have the following properties:
(I1) 1 ∈ S
(I2) If n ≥ 1 and n ∈ S, then n + 1 ∈ S
Theorem 2.4.1 (PMI). Suppose S ⊆ N has the properties that
(I1) 1 ∈ S
(I2) If n ≥ 1 and n ∈ S, then n + 1 ∈ S. Then S = N.
Proof:
Suppose S ⊆ N satisfies properties I1–I2. Suppose also that S = N. Then either S ⊆ N or N ⊆ S.
But S ⊆ N is assumed, so N ⊆ S. Thus there exists n ∈ N such that n ∈/ S. If we define T = NS
= N ∩ S , then T ⊆ N and n ∈ T . Thus T is a nonempty subset of the natural numbers, which, by
the WOP, contains a smallest element a. Now it is impossible that a = 1 because 1 ∈ S. Thus a >
1, so that a −1 ∈ N. Furthermore, since a is the smallest element of T , it must be that a −1 ∈/ T .
Thus a − 1 ∈ S. But by I2, since a − 1 ∈ S, it follows that a = (a − 1) + 1 ∈ S. But if a ∈ S, then
a ∈/ T . This is a contradiction. Thus there is no smallest element of T , which means that T = ∅.
Therefore, N ⊆ S, from which S = N.
Theorem 2.4.2 (Sample). For all n ∈ N, P (n).
Proof:
We use induction on n ≥ 1.
(I1) P (1).
(I2) Suppose n ≥ 1 and P (n). Then . . . . Thus, P (n + 1).
Therefore, by induction, P (n) is true for all n ∈ N.
Sri Rejeki and Ahmad Wachidul Kohar Page 1

2. The 3rd Report of Abstract Agebra IMPoME 2012
Theorem 2.4.3. For all n ∈ N,
…………………………………………………………………………(2.54)
Proof:
We use induction on n ≥ 1.
(I1) For the case n = 1, we have and 1(1 + 1)/2 = 1,
so that Eq. (2.54) holds true for n = 1.
(I2) Suppose n ≥ 1 and . Then
Thus, Eq. (2.54) holds for n +1.
By the PMI, it follows that Eq. (2.54) holds for all n ∈ N.
Theorem 2.4.4. Suppose A, B1 , B2 ,..., Bn are sets. Then
A ∪ ( B1 ∩ B2 ∩ ··· ∩ Bn ) = ( A ∪ B1 ) ∩ ( A ∪ B2 ) ∩ ··· ∩ ( A ∪ Bn )……………………(2.56)
Proof:
To clean up the notation, we write Eq. (2.56) as
……………………………………………………(2.57)
(I1) If n = 1, then ∩k=1 Bk = B1 , so that both sides of Eq. (2.57) are simply A ∪ B1 .
(I2) Suppose n ≥ 1 and Eq. (2.57) holds for n. Then
=
=
Thus, Eq. (2.57) holds for n + 1.
By the PMI, Eq. (2.57) holds for all n ∈ N.
2.4.2 Variation of the PMI
Supposed that j is any whole number and that S ⊆ W has the following properties.
(J1) j ∈ S
(J2) If n ≥ j and n ∈ S, then n + 1 ∈ S.
By mimicking almost word for word the proof of Theorem 2.4.1, we would have a similar PMI
rooted, if you will, at j ∈ W.
Sri Rejeki and Ahmad Wachidul Kohar Page 2

3. The 3rd Report of Abstract Agebra IMPoME 2012
Theorem 2.4.6. Suppose a ∈ R{0} and m, n ∈ W. Then am · an = am+n .
Proof:
We prove by induction on n ≥ 0 (thinking of m as fixed).
(J1) For the case n = 0, we have am · a0 = am · 1 = am = am+0 . Thus the result is true for
n = 0.
(J2) Suppose n ≥ 0 and am · an = am+n . We show that am · an+1 = am+(n+1) .
am · an+1 = am · (an · a) = (am · an ) · a
= am+n · a = a(m+n)+1 = am+(n+1) .
Thus by induction, am · an = am+n for all a ∈ R{0} and m, n ∈ W.
Another example of a recursive definition is the factorial.
0! = 1,
(n + 1)! = (n + 1) · n! for n ≥ 0.
2.4.3 Strong Induction
Suppose has the following properties.
K1.
K2. If and , then .
Theorem 2.4.7 Suppose has the properties K1 and K2 above. Then
Theorem 2.4.8 (Fundamental Theorem of Arithmetic). Every natural number n ≥ 2 can be
written as the product of primes, and this factorization is unique, except perhaps for the order in
which the factors are written.
Proof:
We’ll prove only the existence part now, and save uniqueness for Section 2.7. Since 2 is prime,
the result is true for n = 2. Thus let n ≥ 3 be given, and suppose all of 2, 3,..., n − 1 can be
written as the product of primes. Now if n is itself prime, then its prime factorization is trivial,
and the result is true. On the other hand, if n is composite, then there exist a, b ∈ N such that 1
< a, b < n and n = ab. By the inductive assumption, since 2 ≤ a, b ≤ n − 1, both a and b can
be written as the product of primes. That is,
a = p1 p2 ··· ps
b = q1 q2 ··· qt
,
Sri Rejeki and Ahmad Wachidul Kohar Page 3

4. The 3rd Report of Abstract Agebra IMPoME 2012
where all pk and qk are prime. Thus n = p1 ··· ps q1 ··· qt , and we have found a prime
factorization for n.
EXERCISES
1. Show that the smallest element of a nonempty subset of W is unique.
Answer:
Suppose that U⊂W and U≠Ø. Let a, b are smallest elements in U, where a b so
c U , c a and c b , Since a b , c > b and c > a , there are possibilities a < b or a > b.
It is impossible if a > b because a is smallest element in U.
It is impossible if b > a because b is also smallest element in U.
Therefore a = b. Thus, the smallest number in U is unique
2. Prove the following sum formulas:
n
a. k 1
k2 n n 1 2n 1 / 6
Proof:
We use induction on n ≥ 1.
1
(I1) For the case n = 1, we have k 1
k2 1 and 1 1 1 2 1 1 / 6 1 2 3 / 6 1 , so that
n
k 1
k2 n n 1 2n 1 / 6 holds true for n = 1.
n
(I2) Suppose n ≥ 1 and k 1
k2 n n 1 2n 1 / 6 . Then
n 1 n
n n 1 2n 1
k2 k2 n 12 n 12
k 1 k 1 6
n n 1 2n 1 6 n 1 2 n 1 n 2n 1 6n 1
6 6
n 1 2n 2 7 n 6 n 1 n 2 2n 3
6 6
n
Thus k 1
k2 n n 1 2n 1 / 6 holds for n + 1.
n
By the PMI, It follows that k 1
k2 n n 1 2n 1 / 6 holds for all n ∈ N.
n
b. k 1
1 k k2 1 nn n 1 /2
Proof:
We use induction on n ≥ 1.
1
(I1) For the case n = 1, we have k 1
1 k k2 1 and 1 11 1 1 / 2 1, so that
n
k 1
1 k k2 1 n n n 1 / 2 holds true for n = 1.
Sri Rejeki and Ahmad Wachidul Kohar Page 4

5. The 3rd Report of Abstract Agebra IMPoME 2012
n
(I2) Suppose n ≥ 1 and k 1
1 k k2 1 n n n 1 / 2 . Then
n 1 n 1 nn n 1
k 1
1 k k2 k 1
1 k k2 1n 1n 1
n 12 1n 1
n 12
2
1 nn n 1 2 1n 1
n 12 1n 1
n n 1 2 1n 1
n 12
2 2
n 1 n 1
1 n 1 n 2n 1 1 n 1 n 2
2 2
n
Thus k 1
1 k k2 1 n n n 1 / 2 holds for n + 1.
n
By the PMI, It follows that k 1
1 k k2 1 n n n 1 / 2 holds for all n ∈ N.
n n 2
c. k 1
k3 k 1
k
Proof:
We use induction on n ≥ 1.
1 1 2
(I1) For the case n = 1, we have k 1
k3 1 and k 1
k 1,
n n 2
so that k 1
k3 k 1
k holds true for n = 1.
n n 2
(I2) Suppose n ≥ 1 and k 1
k3 k 1
k . Then
2
n 1 3 n 2
3 2 3 nn 1
k 1
k k 1
k n 1 1 2 ... n n 1 n 13
2
2
n2 n2 4n 1 n 2
n 12 n 1 n 12 n 12
4 4 4
2
n 12 n 2 2
n 1 n 2
4 2
n n 2
Thus k 1
k3 k 1
k holds for n + 1.
n n 2
By the PMI, It follows that k 1
k3 k 1
k holds for all n ∈ N.
n
d. k 1
2k 1 n2
Proof:
We use induction on n ≥ 1.
1 n
(I3) For the case n = 1, we have k 1
2k 1 1 and 12 1 , so that k 1
2k 1 n2
holds true for n = 1.
Sri Rejeki and Ahmad Wachidul Kohar Page 5

6. The 3rd Report of Abstract Agebra IMPoME 2012
n
(I4) Suppose n ≥ 1 and k 1
2k 1 n 2 . Then
n 1 n
2k 1 2k 1 2(n 1) 1 n2 2n 1
k 1 k 1
n2 2n 1
n 12
n
Thus k 1
2k 1 n 2 holds for n + 1.
n
By the PMI, It follows that k 1
2k 1 n 2 holds for all n ∈ N.
n
e. k 0
2k 2n 1
1
Proof:
We use induction on n ≥ 0.
0
(I1) For the case n = 0, we have k 0
2k 1and 2 0 1
1 2 1 1,
n
so that k 0
2k 2n 1
1 holds true for n = 0.
n
(I2) Suppose n ≥ 1 and k 0
2k 2n 1
1 . Then
n 1 n
k 0
2k k 0
2k 2n 1
2n 1
1 2n 1
2 2n 1
1
2n 2
1
n
Thus k 0
2k 2n 1
1 holds for n + 1.
n
By the PMI, It follows that k 0
2k 2n 1
1 holds for all n ∈ N.
n
f. k 1
1/ k k 1 n/ n 1
Proof:
We use induction on n ≥ 1.
1
(I1) For the case n = 1, we have k 1
1/ k k 1 1 / 2 and 1 / 1 1 1 / 2 , so that
n
k 1
1/ k k 1 n / n 1 holds true for n = 1.
n
(I2) Suppose n ≥ 1 and k 1
1/ k k 1 n / n 1 . Then
n 1 n n 1
k 1
1/ k k 1 k 1
1/ k k 1 1/ n 1 n 2
n 1 n 1 n 2
nn 2 1 n 2 2n 1 n 12 n 1
n 1 n 2 n 1 n 2 n 1 n 2 n 2
n
Thus k 1
1/ k k 1 n / n 1 holds for n + 1.
n
By the PMI, It follows that k 1
1/ k k 1 n / n 1 holds for all n ∈ N.
Sri Rejeki and Ahmad Wachidul Kohar Page 6

7. The 3rd Report of Abstract Agebra IMPoME 2012
3. Suppose n ∈ N and a, b1, b2, …,bn ∈ R. Show that
n n
a b
k 1 k k 1
abk
Answer:
n
a b
k 1 k
a b1 b2 ... bn
ab1 ab2 .... abn
n
k 1
abk
4. Suppose m, n ∈ N and a1, a2, …, am, b1, b2, …,bn ∈ R. Show that
m n m n
j 1
aj b
k 1 k j 1 k 1
a j bk
Answer:
m n n
j 1
aj b
k 1 k
a1 a2 ... a m b
k 1 k
n n n
a1 b
k 1 k
a2 b
k 1 k
... a m b
k 1 k
n n n
ab
k 1 1 1
a b
k 1 2 k
... a b
k 1 m k
m n
j 1 k 1
a j bk
n n
5. Suppose a1, a2, …, an ∈ R. Show that k
a
1 k k 1
ak
Proof:
We use induction on n ≥ 1.
1 a1 , if a1 0
(I1) For the case n = 1, we have a
k 1 k
a1 and
a1 , if a1 0
1 a1 , if a1 0
k 1
ak a1
a1 , if a1 0
n n
so that a
k 1 k k 1
ak holds true for n = 1.
n n
(I2) Suppose n ≥ 1 and a
k 1 k k 1
ak . Then
n 1 n
a
k 1 k
a
k 1 k
a k 1 , by theorem2.3.22
n n
a
k 1 k
ak 1 k 1
ak ak 1
n 1
k 1
ak
Sri Rejeki and Ahmad Wachidul Kohar Page 7

8. The 3rd Report of Abstract Agebra IMPoME 2012
n n
Thus a
k 1 k k 1
ak holds for n + 1.
6. Prove Eqs. (2.61) and (2.62) for a, b, ∈ R{0} and m, n ∈ W.
Proof:
n
am a mn
Proof:
We use induction on n ≥ 0 (thinking of m as fixed).
n 0 n
(I1) For the case n = 0, we have a m am 1 and a m.0 1 , so that a m a mn
holds true for n = 0.
n n 1
(I2) Suppose n ≥ 0 and a m a mn . We have show that a m a m( n 1)
.
n 1 n
am am am a mn a m a mn m
a m( n 1)
n
Thus a m a mn holds for n + 1.
n
By the PMI, It follows that a m a mn holds for all n ∈ N.
n
ab a nb n
Proof:
We use induction on n ≥ 0 (thinking of m as fixed).
0 n
(I1) For the case n = 0, we have ab 1 and a 0b 0 1 1 1 , so that ab a nb n holds
true for n = 0.
n n 1
(I2) Suppose n ≥ 0 and ab a nb n . We have show that ab a n 1b n 1 .
n 1 n
ab ab ab a n b n ab a n a bn b a n 1b n 1
n
Thus ab a nb n holds for n + 1.
a nb n holds for all n ∈ N.
n
By the PMI, It follows that ab
n
7. For x ∈ R{0} and n ∈ N, define x x1 n
This exercise will show that the rules for exponents in Eqs. (2.60) – (2.62) hold for all a, b ∈
R{0} and m,n ∈ Z.
n 1
a. Show that a an for all n W .
Answer:
1 1 1
(I1) For the case n = 1, we have a a1 a 1 , so that a n
an holds true
for n = 1.
n 1 1
(I2) Suppose n ≥ 1 and a an . We have show that a ( n 1)
an 1
..
( n 1) 1 n 1 1
a a an 1 1
an 1
n 1
Thus a an holds for n + 1.
Sri Rejeki and Ahmad Wachidul Kohar Page 8

9. The 3rd Report of Abstract Agebra IMPoME 2012
1
By the PMI, It follows that a n
an holds for all n ∈ N.
b. Show that a m
a n
a m n
for all m,n ∈ W.
Answer:
m n m ( n) m n
a a a a
c. Show that a m a n
am n
for all m,n ∈ W.
Answer:
ama n
am ( n)
am n
n
d. Show that a m
a mn for all m,n ∈ W.
Answer:
m n
Using 2.61 a a( m)( n )
, based on the second point of theorem 2.3.6
a mn
n
e. Show that a m a mn
for all m,n ∈ W.
Answer:
n
Using 2.61 a m a ( m)( n)
, based on the theorem 2.3.7
a ( m) 1 (n)
, using A11 (commutative property of multiplication)
1 mn
a , based on the theorem 2.3.7
m n
f. Show that a a mn
for all m,n ∈ W.
Answer:
m n
Using 2.61 a a( m)( n )
, based on the first point of theorem 2.3.6
mn
a , based on the theorem 2.3.7
a( 1) mn
mn
a
n
g. Show that ab a nb n
for all m,n ∈ W.
Answer:
n
Using 2.62 ab a nb n
8. Suppose and . Prove the following.
a)
Answer:
For , . This result is true for
Suppose and . We show that
Sri Rejeki and Ahmad Wachidul Kohar Page 9

10. The 3rd Report of Abstract Agebra IMPoME 2012
, Thus by induction for , the equation is true
b)
:
For , . This result is true for , and
Suppose and . We show that .
,
Thus by induction for , the equation is true
9. Prove the following for .
If , then
Answer:
For , we have (It is true based on the given)
For , suppose
Then,
it gives . Thus by induction for , the inequality is true
10. Polynomial multiplication will reveal that
If ., we may divide both sides through by to have
Sri Rejeki and Ahmad Wachidul Kohar Page 10

11. The 3rd Report of Abstract Agebra IMPoME 2012
This algebraic observation suggest a general formula for the sum of powers of a real
number . If , then
Prove the equation above with induction argument rooted at
Answer:
For we obtain
, his result is true for ,
Suppose and , then we have
,
Thus by induction for , the equation is true
11. Use exercise no 10 to prove the following factorization formula:
Answer:
Using the result no. 10 we have,
Then,
Sri Rejeki and Ahmad Wachidul Kohar Page 11

12. The 3rd Report of Abstract Agebra IMPoME 2012
Thus, we obtain
12. If , then . Use this and induction to prove
for all .
Answer:
For , we obtain
Suppose and , then
The last inequality holds since which holds
for
13. Prove the following for
1.
2.
Sri Rejeki and Ahmad Wachidul Kohar Page 12

13. The 3rd Report of Abstract Agebra IMPoME 2012
Answer:
Proofs
1) For , we have
For , and suppose , then
Thus, by induction for , the equation is true.
2) For we have
For and suppose , then we have
Thus, by induction for , the equation is true.
Sri Rejeki and Ahmad Wachidul Kohar Page 13