Unit 6: Mathematical Induction
1
Subunit 6(a): Ordinary Induction
2
Principle of Mathematical Induction
(Ordinary)
Let 𝑃(𝑛)be a statement about the positive integer
𝑛. To prove that 𝑃 𝑛 is true for all positive
integers 𝑛, it is sufficient to do the following:
• (Basic step): Prove that 𝑃(1) is true
• (Inductive step): Prove that for every positive
integer 𝑘, 𝑃(𝑘) → 𝑃(𝑘 + 1).
If we succeed in doing these two things, it
follows that 𝑃(𝑛) is true for all positive integers
𝑛.
3
Example 1
𝑛(𝑛+1)
2
• Prove that 1 + 2 + ⋯ + 𝑛 =
for every
positive integer 𝑛.
Solution: Let the given statement be 𝑃(𝑛).
Basic step: 𝑃(1) says that 1 =
1(1+1)
2
which is
true.
Inductive step: Let 𝑘 be any positive integer.
We have to show that 𝑃(𝑘) → 𝑃(𝑘 + 1).
Suppose that 𝑃 𝑘 is true. This means that
4
Example 1 (continued)
1 + 2 + ⋯+ 𝑘 =
𝑘(𝑘+1)
.
2
Then,
k (k + 1)
+ k +1
2
k 
= (k + 1) + 1
2 
(k + 1)(k + 2)
=
2
1 + 2 +  k + (k + 1) =
which is 𝑃(𝑘 + 1). We have shown that 𝑃(𝑘) →
𝑃(𝑘 + 1) for every positive integer 𝑘.
It follows that 𝑃(𝑛) is true for every positive integer
𝑛 by the principle of mathematical induction.
5
Principle of Mathematical Induction
(modified)
Let 𝑃(𝑛) be a statement about the integer 𝑛 and
let 𝑏 be a fixed integer. To prove that 𝑃(𝑛) is
true for every integer 𝑛 with 𝑛 ≥ 𝑏, it is
sufficient to do the following:
• (Basic step): Prove that 𝑃(𝑏) is true.
• (Inductive step): Prove that for every integer 𝑘
with 𝑘 ≥ 𝑏, we have 𝑃(𝑘) → 𝑃(𝑘 + 1).
It will follow that 𝑃(𝑛) is true for all integers 𝑛
with 𝑛 ≥ 𝑏.
6
Example 2
Prove that 2𝑛 < 𝑛! for every integer 𝑛 with 𝑛 ≥
4.
Solution: Let the given statement be 𝑃(𝑛).
• Basic step: 𝑃(4) says that 24 < 4! which is
true.
• Inductive step: Let 𝑘 be a positive integer with
𝑘 ≥ 4. We have to show that 𝑃 𝑘 →
𝑃 𝑘 + 1 . Suppose that 𝑃(𝑘) is true.
7
Example 2 (continued)
Thus, 2𝑘 < 𝑘!.
Multiplying both sides by 2 gives
2𝑘+1 < 2 𝑘! . Next, note that 2 < 𝑘 + 1 since
𝑘 ≥ 4. Therefore, 2 𝑘! < 𝑘 + 1 𝑘! . Thus,
2𝑘+1 < 𝑘 + 1 !. We have shown that 𝑃(𝑘) →
𝑃(𝑘 + 1)for every integer 𝑘 with 𝑘 ≥ 4.
It follows that 𝑃(𝑛) is true for every integer 𝑛
with 𝑛 ≥ 4.
8
Divisibility
Let 𝑎 and 𝑏 be integers with 𝑎 ≠ 0. We say that 𝑎
divides 𝑏 and write 𝑎|𝑏 provided that there exists an
integer 𝑐 such that 𝑏 = 𝑎𝑐. Whenever 𝑎|𝑏, we also
say the following
• 𝑎 is a factor of 𝑏
• 𝑏 is a multiple of 𝑎
• 𝑏 is divisible by 𝑎
• 𝑎 is a divisor of 𝑏.
If 𝑎 does not divide 𝑏, we write a |⁄b
For example, 3|6 but 2 |⁄ 5.
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Example 3
• Prove that 7|(32𝑛+1 + 2𝑛+2 ) for all nonnegative integers 𝑛.
Solution: Let the given statement be 𝑃(𝑛). 𝑃(0)
says that 7|(31 +22 ) which is true. Next, let 𝑘 be
a non-negative integer such that 𝑃(𝑘) is true.
Hence, 7|(32𝑘+1 + 2𝑘+2 ) so that there is an
integer 𝑞 with 32𝑘+1 + 2𝑘+2 = 7𝑞. Thus,
32𝑘+1 = 7𝑞 − 2𝑘+2
(1)
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Example 3 (continued)
Note that
32 ( k +1) +1 + 2( k +1) + 2 = 32 k +3 + 2 k +3
= 3232 k +1 + 2 k +3
= 9(7q − 2 k + 2 ) + 2 k +3
= 7(9q) − 9(2 k + 2 ) + 2(2 k + 2 )
= 7(9q) − 7(2 k + 2 )
= 7(9q − 2 k + 2 )
Thus, 7 | (32 ( k +1) +1 + 2( k +1) + 2 ) since 9q − 2 k + 2 is an integer.
We have shown that P(k ) → P(k + 1) for every
non - negative integer k . Hence, P(n) is true for all
non - negative integers n.
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Subunit 6(b): Strong Induction
12
Example 4
• The Fibonacci sequence is defined recursively
as follows: 𝑓0 = 0, 𝑓1 = 1, 𝑓𝑛 = 𝑓𝑛−1 + 𝑓𝑛−2
for all integers 𝑛 with 𝑛 ≥ 2. The first few
terms are 0, 1, 1, 2, 3, 5, 8, 13. Prove that
𝑓𝑛 =
1 1+ 5
5
2
𝑛
−
1 1− 5
5
2
𝑛
for every non-
negative integer 𝑛.
13
Example 4 (continued)
Solution: Let 𝑃(𝑛) be the statement that has to
1+ 5
2
2
1− 5
.
2
be proved. Let 𝛼 =
𝑎𝑛𝑑 𝛽 =
Note
that 𝛼 2 = 𝛼 + 1 and 𝛽 = 𝛽 + 1. Also, note that
1 𝑛
1 𝑛
𝑃(𝑛) says that 𝑓𝑛 = 𝛼 − 𝛽 .
5
5
Basic Step: It is easily checked that 𝑃(0) and
𝑃(1) are true.
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Example 4 (continued)
• Inductive step: Let 𝑘 be an integer with 𝑘 ≥
2. We will show that if 𝑃 𝑗 is true for every
integer 𝑗 with 0 ≤ 𝑗 < 𝑘 then 𝑃(𝑘) must be
true. So, suppose that 𝑃(𝑗) is true for every
integer 𝑗 with 0 ≤ 𝑗 < 𝑘. This means that
𝑃(𝑘 − 2) and 𝑃(𝑘 − 1) are true. Hence,
𝑓𝑘−2 =
𝑓𝑘−1 =
1 𝑘−2
1 𝑘−2
𝛼
− 𝛽
and
5
5
1 𝑘−1
1 𝑘−1
𝛼
− 𝛽 .
5
5
15
Example 4 (continued)
We have
f k = f k −1 + f k − 2
=
=
=
=
=
1 k −1 1 k −1 1 k − 2 1 k − 2
 −
 +
 −

5
5
5
5
1
1
 k −1 +  k − 2 −
 k −1 +  k − 2
5
5
1 k −2
1 k −2
 ( + 1) −
 (  + 1)
5
5
1 k −2 2 1 k −2 2
  −
  (since  2 =  + 1 and  2 =  + 1)
5
5
1 k 1 k
 −

5
5
(
)
(
)
16
Example 4 (continued)
We have shown that for every integer 𝑘 with 𝑘 ≥
2, if 𝑃(𝑗) is true for every integer 𝑗 with 0 ≤ 𝑗 <
𝑘 then 𝑃(𝑘) is true. Together with the basic step
where it was shown that 𝑃(0)and 𝑃(1) are true,
it follows that 𝑃(𝑛) is true for every nonnegative integer 𝑛 by the principle of strong
induction.
17
Example 5
Consider the sequence defined recursively as
follows: 𝑎1 = 1, 𝑎2 = 2, 𝑎3 = 3, and
𝑎𝑛 = 𝑎𝑛−1 + 𝑎𝑛−2 + 𝑎𝑛−3 for all integers 𝑛
with 𝑛 ≥ 4. Prove that 𝑎𝑛 < 2𝑛 for all positive
integers 𝑛.
Solution: Let 𝑃(𝑛) be the statement that has to
be proved.
Basic step: It is easily seen that 𝑃 1 , 𝑃(2) and
𝑃(3) are true.
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Example 5 (continued)
• Inductive step: Let 𝑘 be an integer with 𝑘 ≥ 4.
Suppose that 𝑃 𝑗 is true for every integer 𝑗
with 1 ≤ 𝑗 < 𝑘. We will show that 𝑃(𝑘) is
true. Note that by our assumption, 𝑃(𝑘 −
3), 𝑃(𝑘 − 2)and 𝑃(𝑘 − 1) are all true. Hence,
𝑎𝑘−3 < 2𝑘−3 , 𝑎𝑘−2 < 2𝑘−2 and 𝑎𝑘−1 < 2𝑘−1 .
Hence, we have
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Example 5 (continued)
ak = ak −1 + ak − 2 + ak −3
 2 k −1 + 2 k − 2 + 2 k −3
= 2 k −3 (2 2 + 2 + 1)
= 2 k −3 (7)  2 k −3 (8) = 2 k .
We have shown that for every integer k with k  4,
if P( j ) is true for every integer j with 1  j  k then
P(k ) is true. Together with the fact that P(1), P(2), and
P(3) are true, it follows that P(n) is true for all positive
integers n.
20