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logo1 The Principle of Induction Examples Strong Induction Associativity

Induction

Bernd Schr¨

• der

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

Introduction

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

Introduction

• 1. The Principle of Induction from Peano’s Axioms looks a

little funny to people who have some experience with induction.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

Introduction

• 1. The Principle of Induction from Peano’s Axioms looks a

little funny to people who have some experience with induction.

• 2. Induction proofs typically are about statements P(n), not

about sets S.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

Introduction

• 1. The Principle of Induction from Peano’s Axioms looks a

little funny to people who have some experience with induction.

• 2. Induction proofs typically are about statements P(n), not

about sets S. (Nonetheless, the induction proofs so far were similar to “regular” induction.)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

Introduction

• 1. The Principle of Induction from Peano’s Axioms looks a

little funny to people who have some experience with induction.

• 2. Induction proofs typically are about statements P(n), not

about sets S. (Nonetheless, the induction proofs so far were similar to “regular” induction.)

• 3. We will now state induction in its more customary form.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

Theorem.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Let P(n) be a statement

about the natural number n.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Let P(n) be a statement

about the natural number n. If P(1) is true

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Let P(n) be a statement

about the natural number n. If P(1) is true and if, for all n ∈ N, truth of P(n) implies truth of P(n+1)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Let P(n) be a statement

about the natural number n. If P(1) is true and if, for all n ∈ N, truth of P(n) implies truth of P(n+1), then P(n) holds for all natural numbers.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Let P(n) be a statement

about the natural number n. If P(1) is true and if, for all n ∈ N, truth of P(n) implies truth of P(n+1), then P(n) holds for all natural numbers.

• Proof. Consider the set S :=
• n ∈ N : P(n) is true
• .

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Let P(n) be a statement

about the natural number n. If P(1) is true and if, for all n ∈ N, truth of P(n) implies truth of P(n+1), then P(n) holds for all natural numbers.

• Proof. Consider the set S :=
• n ∈ N : P(n) is true
• . Then

1 ∈ S.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Let P(n) be a statement

about the natural number n. If P(1) is true and if, for all n ∈ N, truth of P(n) implies truth of P(n+1), then P(n) holds for all natural numbers.

• Proof. Consider the set S :=
• n ∈ N : P(n) is true
• . Then

1 ∈ S. Moreover, if n ∈ S, then the statement P(n) is true.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Let P(n) be a statement

about the natural number n. If P(1) is true and if, for all n ∈ N, truth of P(n) implies truth of P(n+1), then P(n) holds for all natural numbers.

• Proof. Consider the set S :=
• n ∈ N : P(n) is true
• . Then

1 ∈ S. Moreover, if n ∈ S, then the statement P(n) is true. By assumption, this implies that P(n+1) is true.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Let P(n) be a statement

about the natural number n. If P(1) is true and if, for all n ∈ N, truth of P(n) implies truth of P(n+1), then P(n) holds for all natural numbers.

• Proof. Consider the set S :=
• n ∈ N : P(n) is true
• . Then

1 ∈ S. Moreover, if n ∈ S, then the statement P(n) is true. By assumption, this implies that P(n+1) is true. That is, n+1 ∈ S.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Let P(n) be a statement

about the natural number n. If P(1) is true and if, for all n ∈ N, truth of P(n) implies truth of P(n+1), then P(n) holds for all natural numbers.

• Proof. Consider the set S :=
• n ∈ N : P(n) is true
• . Then

1 ∈ S. Moreover, if n ∈ S, then the statement P(n) is true. By assumption, this implies that P(n+1) is true. That is, n+1 ∈ S. By the Principle of Induction, S = N.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Let P(n) be a statement

about the natural number n. If P(1) is true and if, for all n ∈ N, truth of P(n) implies truth of P(n+1), then P(n) holds for all natural numbers.

• Proof. Consider the set S :=
• n ∈ N : P(n) is true
• . Then

1 ∈ S. Moreover, if n ∈ S, then the statement P(n) is true. By assumption, this implies that P(n+1) is true. That is, n+1 ∈ S. By the Principle of Induction, S = N. Thus for all n ∈ N the statement P(n) is true.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Let P(n) be a statement

about the natural number n. If P(1) is true and if, for all n ∈ N, truth of P(n) implies truth of P(n+1), then P(n) holds for all natural numbers.

• Proof. Consider the set S :=
• n ∈ N : P(n) is true
• . Then

1 ∈ S. Moreover, if n ∈ S, then the statement P(n) is true. By assumption, this implies that P(n+1) is true. That is, n+1 ∈ S. By the Principle of Induction, S = N. Thus for all n ∈ N the statement P(n) is true.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Let P(n) be a statement

about the natural number n. If P(1) is true and if, for all n ∈ N, truth of P(n) implies truth of P(n+1), then P(n) holds for all natural numbers.

• Proof. Consider the set S :=
• n ∈ N : P(n) is true
• . Then

1 ∈ S. Moreover, if n ∈ S, then the statement P(n) is true. By assumption, this implies that P(n+1) is true. That is, n+1 ∈ S. By the Principle of Induction, S = N. Thus for all n ∈ N the statement P(n) is true. The proof of P(1) is also called the base step of the induction.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Let P(n) be a statement

about the natural number n. If P(1) is true and if, for all n ∈ N, truth of P(n) implies truth of P(n+1), then P(n) holds for all natural numbers.

• Proof. Consider the set S :=
• n ∈ N : P(n) is true
• . Then

1 ∈ S. Moreover, if n ∈ S, then the statement P(n) is true. By assumption, this implies that P(n+1) is true. That is, n+1 ∈ S. By the Principle of Induction, S = N. Thus for all n ∈ N the statement P(n) is true. The proof of P(1) is also called the base step of the induction. The proof that P(n) implies P(n+1) is also called the induction step.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

Example.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. For all n ∈ N we have that

1+2+3+···+n = n(n+1) 2 .

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. For all n ∈ N we have that

1+2+3+···+n = n(n+1) 2 . Base step, n = 1.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. For all n ∈ N we have that

1+2+3+···+n = n(n+1) 2 . Base step, n = 1. 1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. For all n ∈ N we have that

1+2+3+···+n = n(n+1) 2 . Base step, n = 1. 1 = 1(1+1) 2 .

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. For all n ∈ N we have that

1+2+3+···+n = n(n+1) 2 . Base step, n = 1. 1 = 1(1+1) 2 . Induction step n → n+1.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. For all n ∈ N we have that

1+2+3+···+n = n(n+1) 2 . Base step, n = 1. 1 = 1(1+1) 2 . Induction step n → n+1. 1+2+3+···+n+(n+1)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. For all n ∈ N we have that

1+2+3+···+n = n(n+1) 2 . Base step, n = 1. 1 = 1(1+1) 2 . Induction step n → n+1. 1+2+3+···+n+(n+1) = n(n+1) 2 +(n+1)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. For all n ∈ N we have that

1+2+3+···+n = n(n+1) 2 . Base step, n = 1. 1 = 1(1+1) 2 . Induction step n → n+1. 1+2+3+···+n+(n+1) = n(n+1) 2 +(n+1) = n(n+1)+2(n+1) 2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. For all n ∈ N we have that

1+2+3+···+n = n(n+1) 2 . Base step, n = 1. 1 = 1(1+1) 2 . Induction step n → n+1. 1+2+3+···+n+(n+1) = n(n+1) 2 +(n+1) = n(n+1)+2(n+1) 2 = (n+1)

• (n+1)+1
• 2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. For all n ∈ N we have that

1+2+3+···+n = n(n+1) 2 . Base step, n = 1. 1 = 1(1+1) 2 . Induction step n → n+1. 1+2+3+···+n+(n+1) = n(n+1) 2 +(n+1) = n(n+1)+2(n+1) 2 = (n+1)

• (n+1)+1
• 2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Base step

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Base step, n = 3.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Base step, n = 3. (See book.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Base step, n = 3. (See book. We also did quadrilaterals in the presentation on proofs.)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Induction step.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Induction step. Finding the right corner.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Induction step. Finding the right corner.

✑✑✑✑✑✑

• ☞

☞ ☞ ☞

...

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Induction step. Finding the right corner.

✑✑✑✑✑✑

• ☞

☞ ☞ ☞

...

s Ak

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Induction step. Finding the right corner.

✑✑✑✑✑✑

• ☞

☞ ☞ ☞

...

s s Ak−1 Ak

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Induction step. Finding the right corner.

✑✑✑✑✑✑

• ☞

☞ ☞ ☞

...

s s s Ak−1 Ak Ak+1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Induction step. Finding the right corner.

✑✑✑✑✑✑

• ☞

☞ ☞ ☞

...

s s s Ak−1 Ak Ak+1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Induction step. Finding the right corner.

✑✑✑✑✑✑ ✑✑✑

• ☞

☞ ☞ ☞

...

s s s s s s Ak−1 Aj−1 Ak Aj Ak+1 Aj+1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Induction step. Cutting a “convex corner”.

✑✑✑✑✑✑ ✑✑✑

• ☞

☞ ☞ ☞

...

s s s s s s Ak−1 Aj−1 Ak Aj Ak+1 Aj+1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Induction step. Cutting a “convex corner”.

✑✑✑✑✑✑

• ☞

☞ ☞ ☞

...

s s s Ak−1 Ak Ak+1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Induction step. Cutting a “convex corner”.

✑✑✑✑✑✑

• ☞

☞ ☞ ☞

...

s s s Ak−1 Ak Ak+1

αk−1 Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Induction step. Cutting a “convex corner”.

✑✑✑✑✑✑

• ☞

☞ ☞ ☞

...

s s s Ak−1 Ak Ak+1

αk αk−1 Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Induction step. Cutting a “convex corner”.

✑✑✑✑✑✑

• ☞

☞ ☞ ☞

...

s s s Ak−1 Ak Ak+1

αk αk−1 αk+1 Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Induction step.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Induction step. Cutting a “concave corner”.

✑✑✑✑✑✑ s s s Ak−1 Ak Ak+1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Induction step. Cutting a “concave corner”.

✑✑✑✑✑✑ s s s Ak−1 Ak Ak+1 ✆ ✆ ✆ ✆✆

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Induction step. Cutting a “concave corner”.

✑✑✑✑✑✑ s s s Ak−1 Ak Ak+1 ✆ ✆ ✆ ✆✆ ✘✘✘✘✘✘✘✘✘ ✘

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Induction step. Cutting a “concave corner”.

✑✑✑✑✑✑ s s s Ak−1 Ak Ak+1 ✆ ✆ ✆ ✆✆ ✘✘✘✘✘✘✘✘✘ ✘ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Induction step. Cutting a “concave corner”.

✑✑✑✑✑✑

...

s s s Ak−1 Ak Ak+1 ✆ ✆ ✆ ✆✆ ✘✘✘✘✘✘✘✘✘ ✘ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Induction step. Cutting a “concave corner”.

✑✑✑✑✑✑

...

s s s Ak−1 Ak Ak+1 ✆ ✆ ✆ ✆✆ ✘✘✘✘✘✘✘✘✘ ✘ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Induction step. Cutting a “concave corner”.

✑✑✑✑✑✑

...

s s s Ak−1 Ak Ak+1 ✆ ✆ ✆ ✆✆ ✘✘✘✘✘✘✘✘✘ ✘ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Induction step. Cutting a “concave corner”.

✑✑✑✑✑✑

...

s s s Ak−1 Ak Ak+1 ✆ ✆ ✆ ✆✆ ✘✘✘✘✘✘✘✘✘ ✘ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Induction step. Cutting a “concave corner”.

✑✑✑✑✑✑

...

s s s Ak−1 Ak Ak+1 ✆ ✆ ✆ ✆✆ ✘✘✘✘✘✘✘✘✘ ✘ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔

αk−1 Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Induction step. Cutting a “concave corner”.

✑✑✑✑✑✑

...

s s s Ak−1 Ak Ak+1 ✆ ✆ ✆ ✆✆ ✘✘✘✘✘✘✘✘✘ ✘ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔

αk αk−1 Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Example. The sum of the interior angles of a plane n-gon with

n ≥ 3 vertices is (n−2)π. Induction step. Cutting a “concave corner”.

✑✑✑✑✑✑

...

s s s Ak−1 Ak Ak+1 ✆ ✆ ✆ ✆✆ ✘✘✘✘✘✘✘✘✘ ✘ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔

αk αk−1 αk+1 Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

Theorem.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Every nonempty subset A of

N has a smallest element.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Every nonempty subset A of

N has a smallest element. Proof.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Every nonempty subset A of

N has a smallest element.

• Proof. Let /

0 = A

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Every nonempty subset A of

N has a smallest element.

• Proof. Let /

0 = A ⊆ N.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Every nonempty subset A of

N has a smallest element.

• Proof. Let /

0 = A ⊆ N. Suppose for a contradiction that A does not have a smallest element.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Every nonempty subset A of

N has a smallest element.

• Proof. Let /

0 = A ⊆ N. Suppose for a contradiction that A does not have a smallest element. Let P(n) =“{1,...,n}∩A = / 0”.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Every nonempty subset A of

N has a smallest element.

• Proof. Let /

0 = A ⊆ N. Suppose for a contradiction that A does not have a smallest element. Let P(n) =“{1,...,n}∩A = / 0”. We will prove P(n) for all n ∈ N.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Every nonempty subset A of

N has a smallest element.

• Proof. Let /

0 = A ⊆ N. Suppose for a contradiction that A does not have a smallest element. Let P(n) =“{1,...,n}∩A = / 0”. We will prove P(n) for all n ∈ N. Base step, n = 1.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Every nonempty subset A of

N has a smallest element.

• Proof. Let /

0 = A ⊆ N. Suppose for a contradiction that A does not have a smallest element. Let P(n) =“{1,...,n}∩A = / 0”. We will prove P(n) for all n ∈ N. Base step, n = 1. 1 is the smallest element of N.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Every nonempty subset A of

N has a smallest element.

• Proof. Let /

0 = A ⊆ N. Suppose for a contradiction that A does not have a smallest element. Let P(n) =“{1,...,n}∩A = / 0”. We will prove P(n) for all n ∈ N. Base step, n = 1. 1 is the smallest element of N. So 1 ∈ A.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Every nonempty subset A of

N has a smallest element.

• Proof. Let /

0 = A ⊆ N. Suppose for a contradiction that A does not have a smallest element. Let P(n) =“{1,...,n}∩A = / 0”. We will prove P(n) for all n ∈ N. Base step, n = 1. 1 is the smallest element of N. So 1 ∈ A. Hence P(1) is true.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Every nonempty subset A of

N has a smallest element.

• Proof. Let /

0 = A ⊆ N. Suppose for a contradiction that A does not have a smallest element. Let P(n) =“{1,...,n}∩A = / 0”. We will prove P(n) for all n ∈ N. Base step, n = 1. 1 is the smallest element of N. So 1 ∈ A. Hence P(1) is true. Induction step, n → n+1.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Every nonempty subset A of

N has a smallest element.

• Proof. Let /

0 = A ⊆ N. Suppose for a contradiction that A does not have a smallest element. Let P(n) =“{1,...,n}∩A = / 0”. We will prove P(n) for all n ∈ N. Base step, n = 1. 1 is the smallest element of N. So 1 ∈ A. Hence P(1) is true. Induction step, n → n+1. Assume {1,...,n}∩A = /

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Every nonempty subset A of

N has a smallest element.

• Proof. Let /

0 = A ⊆ N. Suppose for a contradiction that A does not have a smallest element. Let P(n) =“{1,...,n}∩A = / 0”. We will prove P(n) for all n ∈ N. Base step, n = 1. 1 is the smallest element of N. So 1 ∈ A. Hence P(1) is true. Induction step, n → n+1. Assume {1,...,n}∩A = / 0 and consider the element n+1.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Every nonempty subset A of

N has a smallest element.

• Proof. Let /

0 = A ⊆ N. Suppose for a contradiction that A does not have a smallest element. Let P(n) =“{1,...,n}∩A = / 0”. We will prove P(n) for all n ∈ N. Base step, n = 1. 1 is the smallest element of N. So 1 ∈ A. Hence P(1) is true. Induction step, n → n+1. Assume {1,...,n}∩A = / 0 and consider the element n+1. n+1 is the smallest element of N\{1,...,n}.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Every nonempty subset A of

N has a smallest element.

• Proof. Let /

0 = A ⊆ N. Suppose for a contradiction that A does not have a smallest element. Let P(n) =“{1,...,n}∩A = / 0”. We will prove P(n) for all n ∈ N. Base step, n = 1. 1 is the smallest element of N. So 1 ∈ A. Hence P(1) is true. Induction step, n → n+1. Assume {1,...,n}∩A = / 0 and consider the element n+1. n+1 is the smallest element of N\{1,...,n}. Thus n+1 ∈ A.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Every nonempty subset A of

N has a smallest element.

• Proof. Let /

0 = A ⊆ N. Suppose for a contradiction that A does not have a smallest element. Let P(n) =“{1,...,n}∩A = / 0”. We will prove P(n) for all n ∈ N. Base step, n = 1. 1 is the smallest element of N. So 1 ∈ A. Hence P(1) is true. Induction step, n → n+1. Assume {1,...,n}∩A = / 0 and consider the element n+1. n+1 is the smallest element of N\{1,...,n}. Thus n+1 ∈ A. Hence P(n+1) is true.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Every nonempty subset A of

N has a smallest element.

• Proof. Let /

0 = A ⊆ N. Suppose for a contradiction that A does not have a smallest element. Let P(n) =“{1,...,n}∩A = / 0”. We will prove P(n) for all n ∈ N. Base step, n = 1. 1 is the smallest element of N. So 1 ∈ A. Hence P(1) is true. Induction step, n → n+1. Assume {1,...,n}∩A = / 0 and consider the element n+1. n+1 is the smallest element of N\{1,...,n}. Thus n+1 ∈ A. Hence P(n+1) is true. So P(n) holds for all n ∈ N

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Every nonempty subset A of

N has a smallest element.

• Proof. Let /

0 = A ⊆ N. Suppose for a contradiction that A does not have a smallest element. Let P(n) =“{1,...,n}∩A = / 0”. We will prove P(n) for all n ∈ N. Base step, n = 1. 1 is the smallest element of N. So 1 ∈ A. Hence P(1) is true. Induction step, n → n+1. Assume {1,...,n}∩A = / 0 and consider the element n+1. n+1 is the smallest element of N\{1,...,n}. Thus n+1 ∈ A. Hence P(n+1) is true. So P(n) holds for all n ∈ N and A is empty

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Every nonempty subset A of

N has a smallest element.

• Proof. Let /

0 = A ⊆ N. Suppose for a contradiction that A does not have a smallest element. Let P(n) =“{1,...,n}∩A = / 0”. We will prove P(n) for all n ∈ N. Base step, n = 1. 1 is the smallest element of N. So 1 ∈ A. Hence P(1) is true. Induction step, n → n+1. Assume {1,...,n}∩A = / 0 and consider the element n+1. n+1 is the smallest element of N\{1,...,n}. Thus n+1 ∈ A. Hence P(n+1) is true. So P(n) holds for all n ∈ N and A is empty, contradiction.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Every nonempty subset A of

N has a smallest element.

• Proof. Let /

0 = A ⊆ N. Suppose for a contradiction that A does not have a smallest element. Let P(n) =“{1,...,n}∩A = / 0”. We will prove P(n) for all n ∈ N. Base step, n = 1. 1 is the smallest element of N. So 1 ∈ A. Hence P(1) is true. Induction step, n → n+1. Assume {1,...,n}∩A = / 0 and consider the element n+1. n+1 is the smallest element of N\{1,...,n}. Thus n+1 ∈ A. Hence P(n+1) is true. So P(n) holds for all n ∈ N and A is empty, contradiction. Therefore A must have a smallest element.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction. Every nonempty subset A of

N has a smallest element.

• Proof. Let /

0 = A ⊆ N. Suppose for a contradiction that A does not have a smallest element. Let P(n) =“{1,...,n}∩A = / 0”. We will prove P(n) for all n ∈ N. Base step, n = 1. 1 is the smallest element of N. So 1 ∈ A. Hence P(1) is true. Induction step, n → n+1. Assume {1,...,n}∩A = / 0 and consider the element n+1. n+1 is the smallest element of N\{1,...,n}. Thus n+1 ∈ A. Hence P(n+1) is true. So P(n) holds for all n ∈ N and A is empty, contradiction. Therefore A must have a smallest element.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

Definition.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Definition. A set F is called finite iff F is empty or there is an

n ∈ N and a bijective function f : {1,...,n} → F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Definition. A set F is called finite iff F is empty or there is an

n ∈ N and a bijective function f : {1,...,n} → F. Sets that are not finite are called infinite.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Definition. A set F is called finite iff F is empty or there is an

n ∈ N and a bijective function f : {1,...,n} → F. Sets that are not finite are called infinite. For finite sets F = / 0 we set |F| := n with n as above

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Definition. A set F is called finite iff F is empty or there is an

n ∈ N and a bijective function f : {1,...,n} → F. Sets that are not finite are called infinite. For finite sets F = / 0 we set |F| := n with n as above and we set |/ 0| := 0

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Definition. A set F is called finite iff F is empty or there is an

n ∈ N and a bijective function f : {1,...,n} → F. Sets that are not finite are called infinite. For finite sets F = / 0 we set |F| := n with n as above and we set |/ 0| := 0, where 0 is an element that is not in N.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Definition. A set F is called finite iff F is empty or there is an

n ∈ N and a bijective function f : {1,...,n} → F. Sets that are not finite are called infinite. For finite sets F = / 0 we set |F| := n with n as above and we set |/ 0| := 0, where 0 is an element that is not in N. For infinite sets I we set |I| := ∞, where ∞ is an element that is not in N∪{0}.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Definition. A set F is called finite iff F is empty or there is an

n ∈ N and a bijective function f : {1,...,n} → F. Sets that are not finite are called infinite. For finite sets F = / 0 we set |F| := n with n as above and we set |/ 0| := 0, where 0 is an element that is not in N. For infinite sets I we set |I| := ∞, where ∞ is an element that is not in N∪{0}. Theorem.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Definition. A set F is called finite iff F is empty or there is an

n ∈ N and a bijective function f : {1,...,n} → F. Sets that are not finite are called infinite. For finite sets F = / 0 we set |F| := n with n as above and we set |/ 0| := 0, where 0 is an element that is not in N. For infinite sets I we set |I| := ∞, where ∞ is an element that is not in N∪{0}.

• Theorem. Let A and B be finite sets so that A ⊆ B and so that

|A| = |B|.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Definition. A set F is called finite iff F is empty or there is an

n ∈ N and a bijective function f : {1,...,n} → F. Sets that are not finite are called infinite. For finite sets F = / 0 we set |F| := n with n as above and we set |/ 0| := 0, where 0 is an element that is not in N. For infinite sets I we set |I| := ∞, where ∞ is an element that is not in N∪{0}.

• Theorem. Let A and B be finite sets so that A ⊆ B and so that

|A| = |B|. Then A = B.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

Proof.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /

0.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N. Suppose for a contradiction that there are finite sets A and B so that

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N. Suppose for a contradiction that there are finite sets A and B so that A ⊆ B

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N. Suppose for a contradiction that there are finite sets A and B so that A ⊆ B, |A| = |B| ∈ N

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N. Suppose for a contradiction that there are finite sets A and B so that A ⊆ B, |A| = |B| ∈ N and A = B.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N. Suppose for a contradiction that there are finite sets A and B so that A ⊆ B, |A| = |B| ∈ N and A = B. Let M ⊆ N be the set of all sizes |A| = |B| ∈ N of such sets.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N. Suppose for a contradiction that there are finite sets A and B so that A ⊆ B, |A| = |B| ∈ N and A = B. Let M ⊆ N be the set of all sizes |A| = |B| ∈ N of such sets. Then M has a smallest element m.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N. Suppose for a contradiction that there are finite sets A and B so that A ⊆ B, |A| = |B| ∈ N and A = B. Let M ⊆ N be the set of all sizes |A| = |B| ∈ N of such sets. Then M has a smallest element m. Suppose for a contradiction that m was greater than 1.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N. Suppose for a contradiction that there are finite sets A and B so that A ⊆ B, |A| = |B| ∈ N and A = B. Let M ⊆ N be the set of all sizes |A| = |B| ∈ N of such sets. Then M has a smallest element m. Suppose for a contradiction that m was greater than 1. Let A and B be so that A ⊆ B, |A| = |B| = m and A = B.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N. Suppose for a contradiction that there are finite sets A and B so that A ⊆ B, |A| = |B| ∈ N and A = B. Let M ⊆ N be the set of all sizes |A| = |B| ∈ N of such sets. Then M has a smallest element m. Suppose for a contradiction that m was greater than 1. Let A and B be so that A ⊆ B, |A| = |B| = m and A = B. Choose a ∈ A

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N. Suppose for a contradiction that there are finite sets A and B so that A ⊆ B, |A| = |B| ∈ N and A = B. Let M ⊆ N be the set of all sizes |A| = |B| ∈ N of such sets. Then M has a smallest element m. Suppose for a contradiction that m was greater than 1. Let A and B be so that A ⊆ B, |A| = |B| = m and A = B. Choose a ∈ A, let ˜ A := A\{a}

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N. Suppose for a contradiction that there are finite sets A and B so that A ⊆ B, |A| = |B| ∈ N and A = B. Let M ⊆ N be the set of all sizes |A| = |B| ∈ N of such sets. Then M has a smallest element m. Suppose for a contradiction that m was greater than 1. Let A and B be so that A ⊆ B, |A| = |B| = m and A = B. Choose a ∈ A, let ˜ A := A\{a} and let ˜ B := B\{a}.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N. Suppose for a contradiction that there are finite sets A and B so that A ⊆ B, |A| = |B| ∈ N and A = B. Let M ⊆ N be the set of all sizes |A| = |B| ∈ N of such sets. Then M has a smallest element m. Suppose for a contradiction that m was greater than 1. Let A and B be so that A ⊆ B, |A| = |B| = m and A = B. Choose a ∈ A, let ˜ A := A\{a} and let ˜ B := B\{a}. Then |˜ A| = |˜ B| = m−1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N. Suppose for a contradiction that there are finite sets A and B so that A ⊆ B, |A| = |B| ∈ N and A = B. Let M ⊆ N be the set of all sizes |A| = |B| ∈ N of such sets. Then M has a smallest element m. Suppose for a contradiction that m was greater than 1. Let A and B be so that A ⊆ B, |A| = |B| = m and A = B. Choose a ∈ A, let ˜ A := A\{a} and let ˜ B := B\{a}. Then |˜ A| = |˜ B| = m−1, ˜ A = A\{a} ⊆ B\{a} = ˜ B

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N. Suppose for a contradiction that there are finite sets A and B so that A ⊆ B, |A| = |B| ∈ N and A = B. Let M ⊆ N be the set of all sizes |A| = |B| ∈ N of such sets. Then M has a smallest element m. Suppose for a contradiction that m was greater than 1. Let A and B be so that A ⊆ B, |A| = |B| = m and A = B. Choose a ∈ A, let ˜ A := A\{a} and let ˜ B := B\{a}. Then |˜ A| = |˜ B| = m−1, ˜ A = A\{a} ⊆ B\{a} = ˜ B and ˜ A = ˜ B

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N. Suppose for a contradiction that there are finite sets A and B so that A ⊆ B, |A| = |B| ∈ N and A = B. Let M ⊆ N be the set of all sizes |A| = |B| ∈ N of such sets. Then M has a smallest element m. Suppose for a contradiction that m was greater than 1. Let A and B be so that A ⊆ B, |A| = |B| = m and A = B. Choose a ∈ A, let ˜ A := A\{a} and let ˜ B := B\{a}. Then |˜ A| = |˜ B| = m−1, ˜ A = A\{a} ⊆ B\{a} = ˜ B and ˜ A = ˜ B (otherwise A

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N. Suppose for a contradiction that there are finite sets A and B so that A ⊆ B, |A| = |B| ∈ N and A = B. Let M ⊆ N be the set of all sizes |A| = |B| ∈ N of such sets. Then M has a smallest element m. Suppose for a contradiction that m was greater than 1. Let A and B be so that A ⊆ B, |A| = |B| = m and A = B. Choose a ∈ A, let ˜ A := A\{a} and let ˜ B := B\{a}. Then |˜ A| = |˜ B| = m−1, ˜ A = A\{a} ⊆ B\{a} = ˜ B and ˜ A = ˜ B (otherwise A = ˜ A∪{a}

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N. Suppose for a contradiction that there are finite sets A and B so that A ⊆ B, |A| = |B| ∈ N and A = B. Let M ⊆ N be the set of all sizes |A| = |B| ∈ N of such sets. Then M has a smallest element m. Suppose for a contradiction that m was greater than 1. Let A and B be so that A ⊆ B, |A| = |B| = m and A = B. Choose a ∈ A, let ˜ A := A\{a} and let ˜ B := B\{a}. Then |˜ A| = |˜ B| = m−1, ˜ A = A\{a} ⊆ B\{a} = ˜ B and ˜ A = ˜ B (otherwise A = ˜ A∪{a} = ˜ B∪{a}

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N. Suppose for a contradiction that there are finite sets A and B so that A ⊆ B, |A| = |B| ∈ N and A = B. Let M ⊆ N be the set of all sizes |A| = |B| ∈ N of such sets. Then M has a smallest element m. Suppose for a contradiction that m was greater than 1. Let A and B be so that A ⊆ B, |A| = |B| = m and A = B. Choose a ∈ A, let ˜ A := A\{a} and let ˜ B := B\{a}. Then |˜ A| = |˜ B| = m−1, ˜ A = A\{a} ⊆ B\{a} = ˜ B and ˜ A = ˜ B (otherwise A = ˜ A∪{a} = ˜ B∪{a} = B).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N. Suppose for a contradiction that there are finite sets A and B so that A ⊆ B, |A| = |B| ∈ N and A = B. Let M ⊆ N be the set of all sizes |A| = |B| ∈ N of such sets. Then M has a smallest element m. Suppose for a contradiction that m was greater than 1. Let A and B be so that A ⊆ B, |A| = |B| = m and A = B. Choose a ∈ A, let ˜ A := A\{a} and let ˜ B := B\{a}. Then |˜ A| = |˜ B| = m−1, ˜ A = A\{a} ⊆ B\{a} = ˜ B and ˜ A = ˜ B (otherwise A = ˜ A∪{a} = ˜ B∪{a} = B). But then ˜ A and ˜ B are a counterexample of size m−1 < m, contradicting the choice of m.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N. Suppose for a contradiction that there are finite sets A and B so that A ⊆ B, |A| = |B| ∈ N and A = B. Let M ⊆ N be the set of all sizes |A| = |B| ∈ N of such sets. Then M has a smallest element m. Suppose for a contradiction that m was greater than 1. Let A and B be so that A ⊆ B, |A| = |B| = m and A = B. Choose a ∈ A, let ˜ A := A\{a} and let ˜ B := B\{a}. Then |˜ A| = |˜ B| = m−1, ˜ A = A\{a} ⊆ B\{a} = ˜ B and ˜ A = ˜ B (otherwise A = ˜ A∪{a} = ˜ B∪{a} = B). But then ˜ A and ˜ B are a counterexample of size m−1 < m, contradicting the choice of

• m. Therefore m > 1.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N. Suppose for a contradiction that there are finite sets A and B so that A ⊆ B, |A| = |B| ∈ N and A = B. Let M ⊆ N be the set of all sizes |A| = |B| ∈ N of such sets. Then M has a smallest element m. Suppose for a contradiction that m was greater than 1. Let A and B be so that A ⊆ B, |A| = |B| = m and A = B. Choose a ∈ A, let ˜ A := A\{a} and let ˜ B := B\{a}. Then |˜ A| = |˜ B| = m−1, ˜ A = A\{a} ⊆ B\{a} = ˜ B and ˜ A = ˜ B (otherwise A = ˜ A∪{a} = ˜ B∪{a} = B). But then ˜ A and ˜ B are a counterexample of size m−1 < m, contradicting the choice of

• m. Therefore m > 1.

Hence m = 1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N. Suppose for a contradiction that there are finite sets A and B so that A ⊆ B, |A| = |B| ∈ N and A = B. Let M ⊆ N be the set of all sizes |A| = |B| ∈ N of such sets. Then M has a smallest element m. Suppose for a contradiction that m was greater than 1. Let A and B be so that A ⊆ B, |A| = |B| = m and A = B. Choose a ∈ A, let ˜ A := A\{a} and let ˜ B := B\{a}. Then |˜ A| = |˜ B| = m−1, ˜ A = A\{a} ⊆ B\{a} = ˜ B and ˜ A = ˜ B (otherwise A = ˜ A∪{a} = ˜ B∪{a} = B). But then ˜ A and ˜ B are a counterexample of size m−1 < m, contradicting the choice of

• m. Therefore m > 1.

Hence m = 1, A = {a} and B = {b}.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N. Suppose for a contradiction that there are finite sets A and B so that A ⊆ B, |A| = |B| ∈ N and A = B. Let M ⊆ N be the set of all sizes |A| = |B| ∈ N of such sets. Then M has a smallest element m. Suppose for a contradiction that m was greater than 1. Let A and B be so that A ⊆ B, |A| = |B| = m and A = B. Choose a ∈ A, let ˜ A := A\{a} and let ˜ B := B\{a}. Then |˜ A| = |˜ B| = m−1, ˜ A = A\{a} ⊆ B\{a} = ˜ B and ˜ A = ˜ B (otherwise A = ˜ A∪{a} = ˜ B∪{a} = B). But then ˜ A and ˜ B are a counterexample of size m−1 < m, contradicting the choice of

• m. Therefore m > 1.

Hence m = 1, A = {a} and B = {b}. Then a ∈ B = {b}

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N. Suppose for a contradiction that there are finite sets A and B so that A ⊆ B, |A| = |B| ∈ N and A = B. Let M ⊆ N be the set of all sizes |A| = |B| ∈ N of such sets. Then M has a smallest element m. Suppose for a contradiction that m was greater than 1. Let A and B be so that A ⊆ B, |A| = |B| = m and A = B. Choose a ∈ A, let ˜ A := A\{a} and let ˜ B := B\{a}. Then |˜ A| = |˜ B| = m−1, ˜ A = A\{a} ⊆ B\{a} = ˜ B and ˜ A = ˜ B (otherwise A = ˜ A∪{a} = ˜ B∪{a} = B). But then ˜ A and ˜ B are a counterexample of size m−1 < m, contradicting the choice of

• m. Therefore m > 1.

Hence m = 1, A = {a} and B = {b}. Then a ∈ B = {b}, so a = b

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N. Suppose for a contradiction that there are finite sets A and B so that A ⊆ B, |A| = |B| ∈ N and A = B. Let M ⊆ N be the set of all sizes |A| = |B| ∈ N of such sets. Then M has a smallest element m. Suppose for a contradiction that m was greater than 1. Let A and B be so that A ⊆ B, |A| = |B| = m and A = B. Choose a ∈ A, let ˜ A := A\{a} and let ˜ B := B\{a}. Then |˜ A| = |˜ B| = m−1, ˜ A = A\{a} ⊆ B\{a} = ˜ B and ˜ A = ˜ B (otherwise A = ˜ A∪{a} = ˜ B∪{a} = B). But then ˜ A and ˜ B are a counterexample of size m−1 < m, contradicting the choice of

• m. Therefore m > 1.

Hence m = 1, A = {a} and B = {b}. Then a ∈ B = {b}, so a = b and then A = B

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N. Suppose for a contradiction that there are finite sets A and B so that A ⊆ B, |A| = |B| ∈ N and A = B. Let M ⊆ N be the set of all sizes |A| = |B| ∈ N of such sets. Then M has a smallest element m. Suppose for a contradiction that m was greater than 1. Let A and B be so that A ⊆ B, |A| = |B| = m and A = B. Choose a ∈ A, let ˜ A := A\{a} and let ˜ B := B\{a}. Then |˜ A| = |˜ B| = m−1, ˜ A = A\{a} ⊆ B\{a} = ˜ B and ˜ A = ˜ B (otherwise A = ˜ A∪{a} = ˜ B∪{a} = B). But then ˜ A and ˜ B are a counterexample of size m−1 < m, contradicting the choice of

• m. Therefore m > 1.

Hence m = 1, A = {a} and B = {b}. Then a ∈ B = {b}, so a = b and then A = B, contradiction.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Proof. If |A| = 0, then |B| = |A| = 0, so A = B = /
• 0. Thus, we

can assume that |A| ∈ N. Suppose for a contradiction that there are finite sets A and B so that A ⊆ B, |A| = |B| ∈ N and A = B. Let M ⊆ N be the set of all sizes |A| = |B| ∈ N of such sets. Then M has a smallest element m. Suppose for a contradiction that m was greater than 1. Let A and B be so that A ⊆ B, |A| = |B| = m and A = B. Choose a ∈ A, let ˜ A := A\{a} and let ˜ B := B\{a}. Then |˜ A| = |˜ B| = m−1, ˜ A = A\{a} ⊆ B\{a} = ˜ B and ˜ A = ˜ B (otherwise A = ˜ A∪{a} = ˜ B∪{a} = B). But then ˜ A and ˜ B are a counterexample of size m−1 < m, contradicting the choice of

• m. Therefore m > 1.

Hence m = 1, A = {a} and B = {b}. Then a ∈ B = {b}, so a = b and then A = B, contradiction.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

Theorem.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction (“strong induction”).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction (“strong induction”). Let

P(n) be a statement about the natural number n.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction (“strong induction”). Let

P(n) be a statement about the natural number n. If P(1) is true

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction (“strong induction”). Let

P(n) be a statement about the natural number n. If P(1) is true and if for all n ∈ N truth of P(1)∧···∧P(n) implies truth of P(n+1)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction (“strong induction”). Let

P(n) be a statement about the natural number n. If P(1) is true and if for all n ∈ N truth of P(1)∧···∧P(n) implies truth of P(n+1), then P(n) holds for all natural numbers.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction (“strong induction”). Let

P(n) be a statement about the natural number n. If P(1) is true and if for all n ∈ N truth of P(1)∧···∧P(n) implies truth of P(n+1), then P(n) holds for all natural numbers. Proof.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction (“strong induction”). Let

P(n) be a statement about the natural number n. If P(1) is true and if for all n ∈ N truth of P(1)∧···∧P(n) implies truth of P(n+1), then P(n) holds for all natural numbers.

• Proof. Let Q(n) := P(1)∧···∧P(n).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction (“strong induction”). Let

P(n) be a statement about the natural number n. If P(1) is true and if for all n ∈ N truth of P(1)∧···∧P(n) implies truth of P(n+1), then P(n) holds for all natural numbers.

• Proof. Let Q(n) := P(1)∧···∧P(n). We will prove that Q(n)

holds for all n ∈ N.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction (“strong induction”). Let

P(n) be a statement about the natural number n. If P(1) is true and if for all n ∈ N truth of P(1)∧···∧P(n) implies truth of P(n+1), then P(n) holds for all natural numbers.

• Proof. Let Q(n) := P(1)∧···∧P(n). We will prove that Q(n)

holds for all n ∈ N. Base step.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction (“strong induction”). Let

P(n) be a statement about the natural number n. If P(1) is true and if for all n ∈ N truth of P(1)∧···∧P(n) implies truth of P(n+1), then P(n) holds for all natural numbers.

• Proof. Let Q(n) := P(1)∧···∧P(n). We will prove that Q(n)

holds for all n ∈ N. Base step. By assumption, Q(1) = P(1) is true.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction (“strong induction”). Let

P(n) be a statement about the natural number n. If P(1) is true and if for all n ∈ N truth of P(1)∧···∧P(n) implies truth of P(n+1), then P(n) holds for all natural numbers.

• Proof. Let Q(n) := P(1)∧···∧P(n). We will prove that Q(n)

holds for all n ∈ N. Base step. By assumption, Q(1) = P(1) is true. Induction step.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction (“strong induction”). Let

P(n) be a statement about the natural number n. If P(1) is true and if for all n ∈ N truth of P(1)∧···∧P(n) implies truth of P(n+1), then P(n) holds for all natural numbers.

• Proof. Let Q(n) := P(1)∧···∧P(n). We will prove that Q(n)

holds for all n ∈ N. Base step. By assumption, Q(1) = P(1) is true. Induction step. Let Q(n) be true.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction (“strong induction”). Let

P(n) be a statement about the natural number n. If P(1) is true and if for all n ∈ N truth of P(1)∧···∧P(n) implies truth of P(n+1), then P(n) holds for all natural numbers.

• Proof. Let Q(n) := P(1)∧···∧P(n). We will prove that Q(n)

holds for all n ∈ N. Base step. By assumption, Q(1) = P(1) is true. Induction step. Let Q(n) be true. Then P(1)∧···∧P(n) is true.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction (“strong induction”). Let

P(n) be a statement about the natural number n. If P(1) is true and if for all n ∈ N truth of P(1)∧···∧P(n) implies truth of P(n+1), then P(n) holds for all natural numbers.

• Proof. Let Q(n) := P(1)∧···∧P(n). We will prove that Q(n)

holds for all n ∈ N. Base step. By assumption, Q(1) = P(1) is true. Induction step. Let Q(n) be true. Then P(1)∧···∧P(n) is true. By assumption, this implies that P(n+1) is true.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction (“strong induction”). Let

P(n) be a statement about the natural number n. If P(1) is true and if for all n ∈ N truth of P(1)∧···∧P(n) implies truth of P(n+1), then P(n) holds for all natural numbers.

• Proof. Let Q(n) := P(1)∧···∧P(n). We will prove that Q(n)

holds for all n ∈ N. Base step. By assumption, Q(1) = P(1) is true. Induction step. Let Q(n) be true. Then P(1)∧···∧P(n) is true. By assumption, this implies that P(n+1) is true. Thus Q(n+1) =

• P(1)∧···∧P(n)
• ∧P(n+1) is true.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction (“strong induction”). Let

P(n) be a statement about the natural number n. If P(1) is true and if for all n ∈ N truth of P(1)∧···∧P(n) implies truth of P(n+1), then P(n) holds for all natural numbers.

• Proof. Let Q(n) := P(1)∧···∧P(n). We will prove that Q(n)

holds for all n ∈ N. Base step. By assumption, Q(1) = P(1) is true. Induction step. Let Q(n) be true. Then P(1)∧···∧P(n) is true. By assumption, this implies that P(n+1) is true. Thus Q(n+1) =

• P(1)∧···∧P(n)
• ∧P(n+1) is true.

Hence Q(n) holds for all n ∈ N.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction (“strong induction”). Let

P(n) be a statement about the natural number n. If P(1) is true and if for all n ∈ N truth of P(1)∧···∧P(n) implies truth of P(n+1), then P(n) holds for all natural numbers.

• Proof. Let Q(n) := P(1)∧···∧P(n). We will prove that Q(n)

holds for all n ∈ N. Base step. By assumption, Q(1) = P(1) is true. Induction step. Let Q(n) be true. Then P(1)∧···∧P(n) is true. By assumption, this implies that P(n+1) is true. Thus Q(n+1) =

• P(1)∧···∧P(n)
• ∧P(n+1) is true.

Hence Q(n) holds for all n ∈ N. So P(n) holds for all n ∈ N.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. Principle of Induction (“strong induction”). Let

P(n) be a statement about the natural number n. If P(1) is true and if for all n ∈ N truth of P(1)∧···∧P(n) implies truth of P(n+1), then P(n) holds for all natural numbers.

• Proof. Let Q(n) := P(1)∧···∧P(n). We will prove that Q(n)

holds for all n ∈ N. Base step. By assumption, Q(1) = P(1) is true. Induction step. Let Q(n) be true. Then P(1)∧···∧P(n) is true. By assumption, this implies that P(n+1) is true. Thus Q(n+1) =

• P(1)∧···∧P(n)
• ∧P(n+1) is true.

Hence Q(n) holds for all n ∈ N. So P(n) holds for all n ∈ N.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

Definition.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Definition. For each j ∈ N let aj ∈ N.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Definition. For each j ∈ N let aj ∈ N. Recursively define

C(a1,...,an) by C(a1) := a1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Definition. For each j ∈ N let aj ∈ N. Recursively define

C(a1,...,an) by C(a1) := a1 and C(a1,...,an,an+1) :=

• C(a1,...,an)
• +an+1.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Definition. For each j ∈ N let aj ∈ N. Recursively define

C(a1,...,an) by C(a1) := a1 and C(a1,...,an,an+1) :=

• C(a1,...,an)
• +an+1. We will call

C(a1,...,an) the canonical bracketing of the sum of a1,...,an.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Definition. For each j ∈ N let aj ∈ N. Recursively define

C(a1,...,an) by C(a1) := a1 and C(a1,...,an,an+1) :=

• C(a1,...,an)
• +an+1. We will call

C(a1,...,an) the canonical bracketing of the sum of a1,...,an. (Recursive definitions work because of induction.)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Definition. For each j ∈ N let aj ∈ N. Recursively define

C(a1,...,an) by C(a1) := a1 and C(a1,...,an,an+1) :=

• C(a1,...,an)
• +an+1. We will call

C(a1,...,an) the canonical bracketing of the sum of a1,...,an. (Recursive definitions work because of induction.) Definition.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Definition. For each j ∈ N let aj ∈ N. Recursively define

C(a1,...,an) by C(a1) := a1 and C(a1,...,an,an+1) :=

• C(a1,...,an)
• +an+1. We will call

C(a1,...,an) the canonical bracketing of the sum of a1,...,an. (Recursive definitions work because of induction.)

• Definition. For each j ∈ N let aj ∈ N.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Definition. For each j ∈ N let aj ∈ N. Recursively define

C(a1,...,an) by C(a1) := a1 and C(a1,...,an,an+1) :=

• C(a1,...,an)
• +an+1. We will call

C(a1,...,an) the canonical bracketing of the sum of a1,...,an. (Recursive definitions work because of induction.)

• Definition. For each j ∈ N let aj ∈ N. Define S(aj) := aj

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Definition. For each j ∈ N let aj ∈ N. Recursively define

C(a1,...,an) by C(a1) := a1 and C(a1,...,an,an+1) :=

• C(a1,...,an)
• +an+1. We will call

C(a1,...,an) the canonical bracketing of the sum of a1,...,an. (Recursive definitions work because of induction.)

• Definition. For each j ∈ N let aj ∈ N. Define S(aj) := aj and for

any ak,...,am, we call S(ak,...,am) a sum of these numbers iff there is an l ∈ {k,...,m−1} so that S(ak,...,am) =

• S(ak,...,al)
• +
• S(al+1,...,am)
• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Definition. For each j ∈ N let aj ∈ N. Recursively define

C(a1,...,an) by C(a1) := a1 and C(a1,...,an,an+1) :=

• C(a1,...,an)
• +an+1. We will call

C(a1,...,an) the canonical bracketing of the sum of a1,...,an. (Recursive definitions work because of induction.)

• Definition. For each j ∈ N let aj ∈ N. Define S(aj) := aj and for

any ak,...,am, we call S(ak,...,am) a sum of these numbers iff there is an l ∈ {k,...,m−1} so that S(ak,...,am) =

• S(ak,...,al)
• +
• S(al+1,...,am)
• , where

S(ak,...,al) and S(al+1,...,am) are shorter sums.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Definition. For each j ∈ N let aj ∈ N. Recursively define

C(a1,...,an) by C(a1) := a1 and C(a1,...,an,an+1) :=

• C(a1,...,an)
• +an+1. We will call

C(a1,...,an) the canonical bracketing of the sum of a1,...,an. (Recursive definitions work because of induction.)

• Definition. For each j ∈ N let aj ∈ N. Define S(aj) := aj and for

any ak,...,am, we call S(ak,...,am) a sum of these numbers iff there is an l ∈ {k,...,m−1} so that S(ak,...,am) =

• S(ak,...,al)
• +
• S(al+1,...,am)
• , where

S(ak,...,al) and S(al+1,...,am) are shorter sums. (This recursive definitions works because of strong induction.)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

Theorem.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. For each j ∈ N let aj ∈ N

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. For each j ∈ N let aj ∈ N and for each n ∈ N let

S(a1,...,an) be a sum of a1,...,an.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. For each j ∈ N let aj ∈ N and for each n ∈ N let

S(a1,...,an) be a sum of a1,...,an. Then for all n ∈ N we have that S(a1,...,an) = C(a1,...,an).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. For each j ∈ N let aj ∈ N and for each n ∈ N let

S(a1,...,an) be a sum of a1,...,an. Then for all n ∈ N we have that S(a1,...,an) = C(a1,...,an). Proof.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. For each j ∈ N let aj ∈ N and for each n ∈ N let

S(a1,...,an) be a sum of a1,...,an. Then for all n ∈ N we have that S(a1,...,an) = C(a1,...,an).

• Proof. Base step:

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. For each j ∈ N let aj ∈ N and for each n ∈ N let

S(a1,...,an) be a sum of a1,...,an. Then for all n ∈ N we have that S(a1,...,an) = C(a1,...,an).

• Proof. Base step: trivial.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. For each j ∈ N let aj ∈ N and for each n ∈ N let

S(a1,...,an) be a sum of a1,...,an. Then for all n ∈ N we have that S(a1,...,an) = C(a1,...,an).

• Proof. Base step: trivial.

Induction step

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. For each j ∈ N let aj ∈ N and for each n ∈ N let

S(a1,...,an) be a sum of a1,...,an. Then for all n ∈ N we have that S(a1,...,an) = C(a1,...,an).

• Proof. Base step: trivial.

Induction step {1,...,n} → n+1:

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. For each j ∈ N let aj ∈ N and for each n ∈ N let

S(a1,...,an) be a sum of a1,...,an. Then for all n ∈ N we have that S(a1,...,an) = C(a1,...,an).

• Proof. Base step: trivial.

Induction step {1,...,n} → n+1: By definition there is an m ∈ {1,...,n} so that S(a1,...,an+1) =

• S(a1,...,am)
• +
• S(am+1,...,an+1)
• .

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. For each j ∈ N let aj ∈ N and for each n ∈ N let

S(a1,...,an) be a sum of a1,...,an. Then for all n ∈ N we have that S(a1,...,an) = C(a1,...,an).

• Proof. Base step: trivial.

Induction step {1,...,n} → n+1: By definition there is an m ∈ {1,...,n} so that S(a1,...,an+1) =

• S(a1,...,am)
• +
• S(am+1,...,an+1)
• . By

induction hypothesis for m summands, S(a1,...,am) = C(a1,...,am).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. For each j ∈ N let aj ∈ N and for each n ∈ N let

S(a1,...,an) be a sum of a1,...,an. Then for all n ∈ N we have that S(a1,...,an) = C(a1,...,an).

• Proof. Base step: trivial.

Induction step {1,...,n} → n+1: By definition there is an m ∈ {1,...,n} so that S(a1,...,an+1) =

• S(a1,...,am)
• +
• S(am+1,...,an+1)
• . By

induction hypothesis for m summands, S(a1,...,am) = C(a1,...,am). If n > m, there is a sum T(am+2,...,an+1) so that S(am+1,...,an+1) = (am+1)+

• T(am+2,...,an+1)
• .

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. For each j ∈ N let aj ∈ N and for each n ∈ N let

S(a1,...,an) be a sum of a1,...,an. Then for all n ∈ N we have that S(a1,...,an) = C(a1,...,an).

• Proof. Base step: trivial.

Induction step {1,...,n} → n+1: By definition there is an m ∈ {1,...,n} so that S(a1,...,an+1) =

• S(a1,...,am)
• +
• S(am+1,...,an+1)
• . By

induction hypothesis for m summands, S(a1,...,am) = C(a1,...,am). If n > m, there is a sum T(am+2,...,an+1) so that S(am+1,...,an+1) = (am+1)+

• T(am+2,...,an+1)
• . We argue

as follows

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

• Theorem. For each j ∈ N let aj ∈ N and for each n ∈ N let

S(a1,...,an) be a sum of a1,...,an. Then for all n ∈ N we have that S(a1,...,an) = C(a1,...,an).

• Proof. Base step: trivial.

Induction step {1,...,n} → n+1: By definition there is an m ∈ {1,...,n} so that S(a1,...,an+1) =

• S(a1,...,am)
• +
• S(am+1,...,an+1)
• . By

induction hypothesis for m summands, S(a1,...,am) = C(a1,...,am). If n > m, there is a sum T(am+2,...,an+1) so that S(am+1,...,an+1) = (am+1)+

• T(am+2,...,an+1)
• . We argue

as follows (in case n = m we would leave out T).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

Proof (cont.).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

Proof (cont.). S(a1,...,an+1)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

Proof (cont.). S(a1,...,an+1) =

• S(a1,...,am)
• +
• S(am+1,...,an+1)
• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

Proof (cont.). S(a1,...,an+1) =

• S(a1,...,am)
• +
• S(am+1,...,an+1)
• =
• C(a1,...,am)
• +
• (am+1)+
• T(am+2,...,an+1)
• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

Proof (cont.). S(a1,...,an+1) =

• S(a1,...,am)
• +
• S(am+1,...,an+1)
• =
• C(a1,...,am)
• +
• (am+1)+
• T(am+2,...,an+1)
• =
• C(a1,...,am)
• +
• am+1 +
• T(am+2,...,an+1)
• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

Proof (cont.). S(a1,...,an+1) =

• S(a1,...,am)
• +
• S(am+1,...,an+1)
• =
• C(a1,...,am)
• +
• (am+1)+
• T(am+2,...,an+1)
• =
• C(a1,...,am)
• +
• am+1 +
• T(am+2,...,an+1)
• =
• C(a1,...,am)
• +am+1
• +
• T(am+2,...,an+1)
• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

Proof (cont.). S(a1,...,an+1) =

• S(a1,...,am)
• +
• S(am+1,...,an+1)
• =
• C(a1,...,am)
• +
• (am+1)+
• T(am+2,...,an+1)
• =
• C(a1,...,am)
• +
• am+1 +
• T(am+2,...,an+1)
• =
• C(a1,...,am)
• +am+1
• +
• T(am+2,...,an+1)
• =
• C(a1,...,am+1)
• +
• T(am+2,...,an+1)
• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

Proof (cont.). S(a1,...,an+1) =

• S(a1,...,am)
• +
• S(am+1,...,an+1)
• =
• C(a1,...,am)
• +
• (am+1)+
• T(am+2,...,an+1)
• =
• C(a1,...,am)
• +
• am+1 +
• T(am+2,...,an+1)
• =
• C(a1,...,am)
• +am+1
• +
• T(am+2,...,an+1)
• =
• C(a1,...,am+1)
• +
• T(am+2,...,an+1)
• =

C

• C(a1,...,am+1)
• ,am+2,...,an+1
• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

Proof (cont.). S(a1,...,an+1) =

• S(a1,...,am)
• +
• S(am+1,...,an+1)
• =
• C(a1,...,am)
• +
• (am+1)+
• T(am+2,...,an+1)
• =
• C(a1,...,am)
• +
• am+1 +
• T(am+2,...,an+1)
• =
• C(a1,...,am)
• +am+1
• +
• T(am+2,...,an+1)
• =
• C(a1,...,am+1)
• +
• T(am+2,...,an+1)
• =

C

• C(a1,...,am+1)
• ,am+2,...,an+1
• =

C(a1,...,an+1)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

Proof (cont.). S(a1,...,an+1) =

• S(a1,...,am)
• +
• S(am+1,...,an+1)
• =
• C(a1,...,am)
• +
• (am+1)+
• T(am+2,...,an+1)
• =
• C(a1,...,am)
• +
• am+1 +
• T(am+2,...,an+1)
• =
• C(a1,...,am)
• +am+1
• +
• T(am+2,...,an+1)
• =
• C(a1,...,am+1)
• +
• T(am+2,...,an+1)
• =

C

• C(a1,...,am+1)
• ,am+2,...,an+1
• =

C(a1,...,an+1)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

Proof (cont.). S(a1,...,an+1) =

• S(a1,...,am)
• +
• S(am+1,...,an+1)
• =
• C(a1,...,am)
• +
• (am+1)+
• T(am+2,...,an+1)
• =
• C(a1,...,am)
• +
• am+1 +
• T(am+2,...,an+1)
• =
• C(a1,...,am)
• +am+1
• +
• T(am+2,...,an+1)
• =
• C(a1,...,am+1)
• +
• T(am+2,...,an+1)
• =

C

• C(a1,...,am+1)
• ,am+2,...,an+1
• =

C(a1,...,an+1) We don’t want to do proofs like this too often.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction

logo1 The Principle of Induction Examples Strong Induction Associativity

Proof (cont.). S(a1,...,an+1) =

• S(a1,...,am)
• +
• S(am+1,...,an+1)
• =
• C(a1,...,am)
• +
• (am+1)+
• T(am+2,...,an+1)
• =
• C(a1,...,am)
• +
• am+1 +
• T(am+2,...,an+1)
• =
• C(a1,...,am)
• +am+1
• +
• T(am+2,...,an+1)
• =
• C(a1,...,am+1)
• +
• T(am+2,...,an+1)
• =

C

• C(a1,...,am+1)
• ,am+2,...,an+1
• =

C(a1,...,an+1) We don’t want to do proofs like this too often. A more abstract approach will help.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Induction