Mathematical Induction Chapter 16 Language, Proof and Logic

If Person #1 knows the secret For all n, if person # n knows the secret, then so does person # n Then For all n, person # n knows the secret 16.0 P(1)  x ( P(x)  P(x+1) )  x P(x) Induction – a powerful method of proof

Inductive definitions and inductive proofs 16.1.a Example of an inductive definition: wff 1. Every atomic wff is a wff 2. If P is a wff, so is  P 3. If P 1,…,P n are wffs, so are (P 1  …  P n ) and (P 1  …  P n ) 4. If P and Q are wffs, so are (P  Q) and (P  Q) 5. If P is a wff and x is a variable,  xP and  xP are wffs; 6. Nothing is a wff unless it is generated by repeated applications of 1-5. REMEMBER An inductive definition of a set consists of: a base clause, which specifies the basic elements of the defined set, one or more inductive clauses, which tell us how to generate additional elements, and a final clause, which tells us that all the elements are either basic or generated by the inductive clauses.

Inductive definitions and inductive proofs 16.1.b Example of an inductive definition: pal 1. Each letter in the alphabet (a,b,...,z) is a pal. 2. If a string  is a pal, so is the result of putting any letter of the alphabet both in front and in back of  (e.g. a  a, b  b, etc.) 3. Nothing is a pal unless it is generated by repeated applications of 1-2. REMEMBER An inductive definition of a set consists of: a base clause, which specifies the basic elements of the defined set, one or more inductive clauses, which tell us how to generate additional elements, and a final clause, which tells us that all the elements are either basic or generated by the inductive clauses.

Inductive definitions and inductive proofs 16.1.c Example of an inductive definition: pal 1. Each letter in the alphabet (a,b,...,z) is a pal. 2. If a string  is a pal, so is the result of putting any letter of the alphabet both in front and in back of  (e.g. a  a, b  b, etc.) 3. Nothing is a pal unless it is generated by repeated applications of 1-2. REMEMBER Given an inductive definition of a set S, an inductive proof of the fact that a certain property holds of all elements of S requires: a basis step, which shows that the property holds of the basic elements, and an inductive step, which shows that if the property holds of some elements, then it holds of any elements generated from them by the inductive clauses. The assumption that begins the inductive step is called the inductive hypothesis.

Inductive definitions and inductive proofs 16.1.d Example of an inductive definition: pal 1. Each letter in the alphabet (a,b,...,z) is a pal. 2. If a string  is a pal, so is the result of putting any letter of the alphabet both in front and in back of  (e.g. a  a, b  b, etc.) 3. Nothing is a pal unless it is generated by repeated applications of 1-2. By induction, prove that: a) Every pal has an odd length b) Every pal is a palindrome

Inductive definitions in set theory 16.2 Making the final clause of an inductive definition more precise: The set P of pals is the smallest set --- i.e. the intersection of all sets --- such that: 1. Each letter in the alphabet (a,b,...,z) is in P. 2. If a string  is in P, so is the result of putting any letter of the alphabet both in front and in back of  (e.g. a  a, b  b, etc.)

Induction on the natural numbers 16.3.a The inductive definition of natural numbers: 1. 0 is a natural number. 2. If n is a natural number, then n+1 is a natural number. 3. Nothing is a natural number except in virtue of repeated applications of (1) and (2). OR: The set N of natural numbers is the smallest set satisfying: 1. 0  N 2. If n  N, then n+1  N. To prove by induction that P(x) is true of all natural numbers: 1.Prove P(0) (basis step) 2.Prove  x[P(x)  P(x+1)] (inductive step)

Induction on the natural numbers 16.3.b Proposition 4. For every natural number n, the sum of the first n natural numbers is n(n-1)/2. Proof. Basis: The sum of the first 0 natural numbers is indeed 0. Inductive step: Assume the sum of the first k natural numbers is k(k-1)/2 (inductive hypothesis). We want to show that then the same is true for k+1 instead of k, that is, the sum of the first k+1 natural numbers is (k+1)((k+1)-1)/2, i.e. it is k(k+1)/2. But indeed, the sum of the first k+1 natural numbers is X+k, where X is the sum of the fist k natural numbers. By the inductive hypothesis, X= k(k-1)/2. Thus, the sought X+k is k(k-1)/2+k = (k(k-1)+2k)/2 = k(k-1+2)/2 = k(k+1)/2, as desired.

Axiomatizing the natural numbers 16.4.a Peano Arithmetic PA Language: =, 0, s, +,  (s(a) means a+1) Axioms: 1.  x(s(x)  0) 2.  x  y (s(x)=s(y)  x=y) 3.  x (x+0 = x) 4.  x  y [x+s(y) = s(x+y)] 5.  x (x  0 = 0) 6.  x  y [x  s(y) = (x  y)+x] 7. [ Q(0)   x ( Q(x)  Q(s(x)) )]   xQ(x) Axiom 7 is a scheme, for every Q. If Q contains additional variables z 1,...,z n, then the whole thing should be prefixed with  z 1...  z n This axiom is called the induction scheme Gödel’s Incompleteness Theorem: These axioms are not sufficient to prove every true arithmetical sentence. Nor would any bigger set of axioms be sufficient.

Informal proof of  x(s(x)=s(0)+x) 16.4.b What we need: Axiom 3:  x (x+0 = x) Axiom 4:  x  y [x+s(y) = s(x+y)] Axiom 7: [ Q(0)   x ( Q(x)  Q(s(x)) )]   xQ(x) with s(x)=s(0)+x in the role of Q(x) Basis: Q(0), i.e. s(0)=s(0)+0. Follows from Axiom 3. Inductive step: Assume (induction hypothesis) Q(n), i.e. s(n)=s(0)+n We want to show that then Q(s(n)), i.e. s(s(n))=s(0)+s(n) But indeed: s(s(n)) = s(s(0)+n) by induction hypothesis. = s(0)+s(n) by Axiom 4.

Induction in Fitch 16.5 P(0) n P(n) … P(s(n))  xP(x) Peano Induction: Where n does not occur outside the subproof where it is introduced

Ordering the Natural Numbers 16.6 A binary relation R is said to be a total strict ordering iff it is: 1. Irreflexive: 2. Transitive: 3. Trichotomous:  x  y (xRy  x=y  yRx) The ordinary relation < on natural numbers is a total strict ordering. In PA, “x<y” can be treated as an abbreviation of “  z(x+s(z)=y)”. An alternative approach, taken in Fitch, is to treat < as a legitimate symbol of the language, defined by the (additional Peano) axiom  x  y[x<y   z(x+s(z)=y)]

Strong Induction 16.7 Strong induction (other names: complete induction; course of values induction) in Fitch takes the following form: n  x[x<n  P(x)] … P(n)  xP(x) Where n does not occur outside the subproof where it is introduced This principle does not allow us to prove anything new because it is equivalent to ordinary induction (either one follows from the other --- see the textbook). However, it can often offer greater convenience. Example: Using strong induction, prove the so called Fundamental Theorem of Arithmetic, according to which every natural number greater than 1 is either prime or the product of some primes.