Induction - Complete Inductioncommunity.wvu.edu/~krsubramani/courses/sp13/dm2/lecnotes/...Complete...

Outline

Induction - Complete Induction

K. Subramani1

1Lane Department of Computer Science and Electrical EngineeringWest Virginia University

April 1 2013

Subramani Mathematical Induction

Outline

Outline

1 Complete Induction


Complete Induction

Complete Induction or Strong Induction

Axiom Schema

[(∀n) (∀n′) ((n′ < n)→ P(n′))→ P(n)]→ (∀x) P(x).

Note

Do we need a base case? Has it been addressed?


Complete Induction


Axiom Schema

[(∀n) (∀n′) ((n′ < n)→ P(n′))→ P(n)]→ (∀x) P(x).

Note

Do we need a base case?

Has it been addressed?


Complete Induction


Axiom Schema

[(∀n) (∀n′) ((n′ < n)→ P(n′))→ P(n)]→ (∀x) P(x).

Note

Do we need a base case? Has it been addressed?


Complete Induction

Example

Show that every number greater than or equal to 8 can be expressed in the form5 · a + 3 · b, for suitably chosen a and b.

Proof.

The conjecture is clearly true for 8, 9 and 10. Assume that the conjecture holds for allr , 8 ≤ r ≤ k . Consider the integer k + 1. Without loss of generality, we assume that(k + 1) ≥ 11. Observe that (k + 1)− 3 = k − 2 is at least 8 and less than k . As perthe inductive hypothesis, k − 2 can be expressed in the form 3a + 5b, for suitablychosen a and b. It follows that (k + 1) = 3 · (a + 1) + 5 · b, can also be so expressed.Applying the second principle of mathematical induction, we conclude that theconjecture is true.


Complete Induction

Example


Proof.

The conjecture is clearly true for 8, 9 and 10.

Assume that the conjecture holds for allr , 8 ≤ r ≤ k . Consider the integer k + 1. Without loss of generality, we assume that(k + 1) ≥ 11. Observe that (k + 1)− 3 = k − 2 is at least 8 and less than k . As perthe inductive hypothesis, k − 2 can be expressed in the form 3a + 5b, for suitablychosen a and b. It follows that (k + 1) = 3 · (a + 1) + 5 · b, can also be so expressed.Applying the second principle of mathematical induction, we conclude that theconjecture is true.


Complete Induction

Example


Proof.

The conjecture is clearly true for 8, 9 and 10. Assume that the conjecture holds for allr , 8 ≤ r ≤ k .

Consider the integer k + 1. Without loss of generality, we assume that(k + 1) ≥ 11. Observe that (k + 1)− 3 = k − 2 is at least 8 and less than k . As perthe inductive hypothesis, k − 2 can be expressed in the form 3a + 5b, for suitablychosen a and b. It follows that (k + 1) = 3 · (a + 1) + 5 · b, can also be so expressed.Applying the second principle of mathematical induction, we conclude that theconjecture is true.


Complete Induction

Example


Proof.

The conjecture is clearly true for 8, 9 and 10. Assume that the conjecture holds for allr , 8 ≤ r ≤ k . Consider the integer k + 1.

Without loss of generality, we assume that(k + 1) ≥ 11. Observe that (k + 1)− 3 = k − 2 is at least 8 and less than k . As perthe inductive hypothesis, k − 2 can be expressed in the form 3a + 5b, for suitablychosen a and b. It follows that (k + 1) = 3 · (a + 1) + 5 · b, can also be so expressed.Applying the second principle of mathematical induction, we conclude that theconjecture is true.


Complete Induction

Example


Proof.

The conjecture is clearly true for 8, 9 and 10. Assume that the conjecture holds for allr , 8 ≤ r ≤ k . Consider the integer k + 1. Without loss of generality, we assume that(k + 1) ≥ 11.

Observe that (k + 1)− 3 = k − 2 is at least 8 and less than k . As perthe inductive hypothesis, k − 2 can be expressed in the form 3a + 5b, for suitablychosen a and b. It follows that (k + 1) = 3 · (a + 1) + 5 · b, can also be so expressed.Applying the second principle of mathematical induction, we conclude that theconjecture is true.


Complete Induction

Example


Proof.

The conjecture is clearly true for 8, 9 and 10. Assume that the conjecture holds for allr , 8 ≤ r ≤ k . Consider the integer k + 1. Without loss of generality, we assume that(k + 1) ≥ 11. Observe that (k + 1)− 3 = k − 2 is at least 8 and less than k .

As perthe inductive hypothesis, k − 2 can be expressed in the form 3a + 5b, for suitablychosen a and b. It follows that (k + 1) = 3 · (a + 1) + 5 · b, can also be so expressed.Applying the second principle of mathematical induction, we conclude that theconjecture is true.


Complete Induction

Example


Proof.

The conjecture is clearly true for 8, 9 and 10. Assume that the conjecture holds for allr , 8 ≤ r ≤ k . Consider the integer k + 1. Without loss of generality, we assume that(k + 1) ≥ 11. Observe that (k + 1)− 3 = k − 2 is at least 8 and less than k . As perthe inductive hypothesis, k − 2 can be expressed in the form 3a + 5b, for suitablychosen a and b.

It follows that (k + 1) = 3 · (a + 1) + 5 · b, can also be so expressed.Applying the second principle of mathematical induction, we conclude that theconjecture is true.


Complete Induction

Example


Proof.

The conjecture is clearly true for 8, 9 and 10. Assume that the conjecture holds for allr , 8 ≤ r ≤ k . Consider the integer k + 1. Without loss of generality, we assume that(k + 1) ≥ 11. Observe that (k + 1)− 3 = k − 2 is at least 8 and less than k . As perthe inductive hypothesis, k − 2 can be expressed in the form 3a + 5b, for suitablychosen a and b. It follows that (k + 1) = 3 · (a + 1) + 5 · b, can also be so expressed.

Applying the second principle of mathematical induction, we conclude that theconjecture is true.


Complete Induction

Example


Proof.

The conjecture is clearly true for 8, 9 and 10. Assume that the conjecture holds for allr , 8 ≤ r ≤ k . Consider the integer k + 1. Without loss of generality, we assume that(k + 1) ≥ 11. Observe that (k + 1)− 3 = k − 2 is at least 8 and less than k . As perthe inductive hypothesis, k − 2 can be expressed in the form 3a + 5b, for suitablychosen a and b. It follows that (k + 1) = 3 · (a + 1) + 5 · b, can also be so expressed.Applying the second principle of mathematical induction, we conclude that theconjecture is true.


Complete Induction

The theory T ∗PA

Peano Arithmetic with division

Consider the theory of T∗PA, which is the theory of Peano arithmetic TPA, augmented bythe following axioms:

A1. (∀x)(∀y) (x < y)→ [quot(x , y) = 0].

A2. (∀x)(∀y) (y > 0)→ [quot(x + y , y) = quot(x , y) + 1].

A3. (∀x)(∀y) (x < y)→ [rem(x , y) = x ].

A4. (∀x)(∀y) (y > 0)→ [rem(x + y , y) = rem(x , y)].

Example

Argue that(∀x)(∀y) (y > 0)→ [rem(x , y) < y ].


Complete Induction

The theory T ∗PA



A1. (∀x)(∀y) (x < y)→ [quot(x , y) =

0].


A3. (∀x)(∀y) (x < y)→ [rem(x , y) = x ].


Example



Complete Induction

The theory T ∗PA



A1. (∀x)(∀y) (x < y)→ [quot(x , y) = 0].


A3. (∀x)(∀y) (x < y)→ [rem(x , y) = x ].


Example



Complete Induction

The theory T ∗PA



A1. (∀x)(∀y) (x < y)→ [quot(x , y) = 0].

A2. (∀x)(∀y) (y > 0)→ [quot(x + y , y) =

quot(x , y) + 1].

A3. (∀x)(∀y) (x < y)→ [rem(x , y) = x ].


Example



Complete Induction

The theory T ∗PA



A1. (∀x)(∀y) (x < y)→ [quot(x , y) = 0].


A3. (∀x)(∀y) (x < y)→ [rem(x , y) = x ].


Example



Complete Induction

The theory T ∗PA



A1. (∀x)(∀y) (x < y)→ [quot(x , y) = 0].


A3. (∀x)(∀y) (x < y)→ [rem(x , y) =

x ].


Example



Complete Induction

The theory T ∗PA



A1. (∀x)(∀y) (x < y)→ [quot(x , y) = 0].


A3. (∀x)(∀y) (x < y)→ [rem(x , y) = x ].


Example



Complete Induction

The theory T ∗PA



A1. (∀x)(∀y) (x < y)→ [quot(x , y) = 0].


A3. (∀x)(∀y) (x < y)→ [rem(x , y) = x ].

A4. (∀x)(∀y) (y > 0)→ [rem(x + y , y) =

rem(x , y)].

Example



Complete Induction

The theory T ∗PA



A1. (∀x)(∀y) (x < y)→ [quot(x , y) = 0].


A3. (∀x)(∀y) (x < y)→ [rem(x , y) = x ].


Example



Induction - Complete Inductioncommunity.wvu.edu/~krsubramani/courses/sp13/dm2/lecnotes/...Complete...

Documents

Transcript of Induction - Complete Inductioncommunity.wvu.edu/~krsubramani/courses/sp13/dm2/lecnotes/...Complete...