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Mathematical Induction – IntroductionLecture 21Section 5.2
Robb T. Koether
Hampden-Sydney College
Thu, Feb 20, 2014
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 1 / 34
1 The Principle
2 The Method
3 ExamplesSums of IntegersSums of Squares of IntegersSums of Products of IntegersMaking Change
4 Assignment
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 2 / 34
Outline
1 The Principle
2 The Method
3 ExamplesSums of IntegersSums of Squares of IntegersSums of Products of IntegersMaking Change
4 Assignment
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 3 / 34
The Principle of Mathematical Induction
Let P(n) be a predicate defined for all integers n ≥ 0.If the following two statements are true
P(0),For all k ≥ 0, if P(k), then P(k + 1),
then the statementFor all integers n ≥ 0, P(n).
is true.
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 4 / 34
The Principle
The first statement shows that P(0) is true.The second statement shows that P(0)→ P(1).Now that we have P(1), the second statement also shows thatP(1)→ P(2).Now that we have P(2), the second statement also shows thatP(2)→ P(3).And so on.Therefore, we conclude P(n) for all n ≥ 0.
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 5 / 34
The Principle
0 1 2 3 4 5
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 6 / 34
The Principle
0 1 2 3 4 5
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 6 / 34
The Principle
0 1 2 3 4 5
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 6 / 34
The Principle
0 1 2 3 4 5
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 6 / 34
The Principle
0 1 2 3 4 5
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 6 / 34
The Principle
0 1 2 3 4 5
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 6 / 34
The Principle
0 1 2 3 4 5
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 6 / 34
The Principle
0 1 2 3 4 5
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 6 / 34
The Principle
0 1 2 3 4 5
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 6 / 34
Outline
1 The Principle
2 The Method
3 ExamplesSums of IntegersSums of Squares of IntegersSums of Products of IntegersMaking Change
4 Assignment
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 7 / 34
The Method
The basic step.Choose a starting point a, typically 0 or 1.Prove P(n) for that starting point a, e.g., prove P(0) or P(1).
The inductive step.Suppose that P(k) is true for some k ≥ a.Prove that it follows that P(k + 1) must be true.
Conclude that P(n) is true for all n ≥ a.
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 8 / 34
Outline
1 The Principle
2 The Method
3 ExamplesSums of IntegersSums of Squares of IntegersSums of Products of IntegersMaking Change
4 Assignment
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 9 / 34
Outline
1 The Principle
2 The Method
3 ExamplesSums of IntegersSums of Squares of IntegersSums of Products of IntegersMaking Change
4 Assignment
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 10 / 34
Sum of Integers
TheoremLet n ≥ 1. Then
n∑i=1
i =n(n + 1)
2.
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 11 / 34
Sum of Integers
Proof.The basic step:
When n = 1, we have
n∑i=1
i =1∑
i=1
i
= 1,
and
n(n + 1)
2=
1 · 22
= 1.
Therefore, the statement is true when n = 1.
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 12 / 34
Sum of Integers
Proof.The basic step:
When n = 1, we have
n∑i=1
i =1∑
i=1
i
= 1,
and
n(n + 1)
2=
1 · 22
= 1.
Therefore, the statement is true when n = 1.
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 13 / 34
Sum of Integers
Proof.The inductive step:
Suppose that the statement is true when n = k , for some integerk ≥ 1.That is, suppose that
∑ki=1 i = k(k+1)
2 for some integer k ≥ 1.Then
k+1∑i=1
i =
(k∑
i=1
i
)+ (k + 1)
=k(k + 1)
2+ (k + 1)
=k(k + 1)
2+
2(k + 1)
2
=(k + 1)(k + 2)
2.
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 14 / 34
Sum of Integers
Proof.The inductive step:
Suppose that the statement is true when n = k , for some integerk ≥ 1.That is, suppose that
∑ki=1 i = k(k+1)
2 for some integer k ≥ 1.Then
k+1∑i=1
i =
(k∑
i=1
i
)+ (k + 1)
=k(k + 1)
2+ (k + 1)
=k(k + 1)
2+
2(k + 1)
2
=(k + 1)(k + 2)
2.
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 15 / 34
Sum of Integers
Proof.The inductive step continued:
Therefore, the statement is true when n = k + 1.
Therefore, the statement is true for all n ≥ 1.
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 16 / 34
Sum of Integers
Proof.The inductive step continued:
Therefore, the statement is true when n = k + 1.
Therefore, the statement is true for all n ≥ 1.
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 17 / 34
Outline
1 The Principle
2 The Method
3 ExamplesSums of IntegersSums of Squares of IntegersSums of Products of IntegersMaking Change
4 Assignment
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 18 / 34
Sums of Squares
TheoremLet n ≥ 1. Then
n∑i=1
i2 =n(n + 1)(2n + 1)
6.
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 19 / 34
Inductive Proof
Proof.Let n = 1.Then
1∑i=1
i2 = 12 = 1
and
n(n + 1)(2n + 1)
6=
1 · 2 · 36
= 1.
Therefore, the statement is true when n = 1.
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 20 / 34
Inductive Proof
Proof.Suppose that the statement is true when n = k , for some integerk ≥ 1.That is, suppose that
k∑i=1
i2 =k(k + 1)(2k + 1)
6.
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 21 / 34
Inductive Proof
Proof.Then
k+1∑i=1
i2 =k(k + 1)(2k + 1)
6+ (k + 1)2
=(k + 1)[k(2k + 1) + 6(k + 1)]
6
=(k + 1)(2k2 + 7k + 6)
6
=(k + 1)(k + 2)(2k + 3)
6.
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 22 / 34
Inductive Proof
Proof.Therefore, the statement is true when n = k + 1.Therefore, the statement is true for all n ≥ 1.
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 23 / 34
Outline
1 The Principle
2 The Method
3 ExamplesSums of IntegersSums of Squares of IntegersSums of Products of IntegersMaking Change
4 Assignment
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 24 / 34
Example
TheoremFor all n ≥ 1,
n∑i=1
i(i + 1) =n(n + 1)(n + 2)
3.
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 25 / 34
Non-inductive Proof
Proof.Let n ≥ 1. Then
n∑i=1
i(i + 1) =n∑
i=1
i2 +n∑
i=1
i
=n(n + 1)(2n + 1)
6+
n(n + 1)
2
=n(n + 1)
2
[2n + 1
3+ 1]
=n(n + 1)(n + 2)
3.
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 26 / 34
Outline
1 The Principle
2 The Method
3 ExamplesSums of IntegersSums of Squares of IntegersSums of Products of IntegersMaking Change
4 Assignment
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 27 / 34
Making Change
TheoremLet n ≥ 4 be an integer. Then n¢ can be obtained using only 2¢ and 5¢coins.
The predicate P(n) is “n¢ can be obtained using only 2¢ and 5¢coins.”The starting point is 4.
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 28 / 34
Making Change
Proof.The basic step:
When n = 4, we have 4¢ = 2¢ + 2¢.Therefore, the statement is true when n = 4.
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 29 / 34
Making Change
Proof.The inductive step:
Suppose that the statement is true when n = k , for some integerk ≥ 4.That is, suppose that k¢ can be obtained using only 2¢ and 5¢coins for some integer k ≥ 4.We consider 2 cases:
Case 1: k¢ uses a 5¢ coin.Case 2: k¢ does not use a 5¢ coin.
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 30 / 34
Making Change
Proof.The inductive step continued:
Case 1: Suppose that k¢ uses a 5¢ coin.Then replace it with three 2¢ coins to make (k + 1)¢.
Case 2: Suppose that k¢ does not use a 5¢ coin.Then it must use at least two 2¢ coins.Replace them with one 5¢ coin to make (k + 1)¢.
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 31 / 34
Making Change
Proof.The inductive step continued:
Therefore, the statement is true when n = k + 1.
Therefore, the statement is true for all n ≥ 4.
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 32 / 34
Outline
1 The Principle
2 The Method
3 ExamplesSums of IntegersSums of Squares of IntegersSums of Products of IntegersMaking Change
4 Assignment
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 33 / 34
Assignment
AssignmentRead Section 5.2, pages 244 - 256.Exercises 2, 4, 5, 7, 14, 26, 29, page 256.
Robb T. Koether (Hampden-Sydney College) Mathematical Induction – Introduction Thu, Feb 20, 2014 34 / 34